HC Verma

Solution : When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If 1 and 2 be the respective angular speeds, we have 1 r1 2 r2. (i) The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists. If this force f acts for a time t, the torque on the first cylinder is f r1 and that on the second is f r2. Assuming 1 r1 > 2 r2 , the corresponding angular impluses are f r1 t and f r2 t. We, therefore, have f r1 t I11 1 and f r2 t I22 2 or, I1 r1 1 1 I2 r2 2 2. (ii) Solving (i) and (ii), 1 I1 1 r2 I2 2 r1 I2 r1 2 I1 r2 2 r2 and 2 I1 1 r2 I2 2 r1 I2 r1 2 I1 r2 2 r1 . 26. A cylinder of mass m is suspended through two strings wrapped around it as shown in figure (10-W19). Find (a) the tension T in the string and (b) the speed of the cylinder as it falls through a distance h. Solution : The portion of the strings between the ceiling and the cylinder is at rest. Hence the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. Suppose the centre of the cylinder falls with an acceleration a. The angular acceleration of the cylinder about its axis is a/R, as the cylinder does not slip over the strings. The equation of motion for the centre of mass of the cylinder is mg 2T ma (i) and for the motion about the centre of mass, it is 2 Tr 1 2 mr 2 1 2 mra or, 2 T 1 2 ma. (ii) From (i) and (ii), a 2 3 g and T mg 6 As the centre of the cylinder starts moving from rest, the velocity after it has fallen through a distance h is given by v 2 2 2 3 g h or, v 4 gh 3 27. A force F acts tangentially at the highest point of a sphere of mass m kept on a rough horizontal plane. If the sphere rolls without slipping, find the acceleration of the centre of the sphere. Solution : The situation is shown in figure (10-W20). As the force F rotates the sphere, the point of contact has a tendency to slip towards left so that the static friction on the sphere will act towards right. Let r be the radius of the sphere and a be the linear acceleration of the centre of the sphere. The angular acceleration about the centre of the sphere is a/r, as there is no slipping. For the linear motion of the centre F f ma (i) and for the rotational motion about the centre, Fr f r I 2 5 mr 2 a r or, F f 2 5 ma. (ii) From (i) and (ii), 2F 7 5 ma or, a 10 F 7 m 28. A sphere of mass M and radius r shown in figure (10-W21) slips on a rough horizontal plane. At some instant it has translational velocity v0 and rotational velocity about the centre v0 2 r Find the translational velocity after the sphere starts pure rolling. Solution : Velocity of the centre v0 and the angular velocity about the centre 0 v0 2 r Thus, v0 > 0 r. The sphere slips forward and thus the friction by the plane  Figure 10-W19     Figure 10-W20 v 2r 0 v0 f r A Figure 10-W21 190 Concepts of Physics

on the sphere will act backward. As the friction is kinetic, its value is N Mg and the sphere will be decelerated by acm f/M. Hence, vt v0 f M t (i) This friction will also have a torque f r about the centre. This torque is clockwise and in the direction of 0. Hence the angular acceleration about the centre will be f r 2/5 Mr 2 5 f 2 Mr and the clockwise angular velocity at time t will be t 0 5 f 2 Mr t v0 2 r 5 f 2 Mr t Pure rolling starts when vt rt i.e., vt v0 2 5 f 2 M t (ii) Eliminating t from (i) and (ii), 5 2 vt vt 5 2 v0 v0 2 or, vt 2 7 3 v0 6 7 v0 . Thus, the sphere rolls with translational velocity 6v0 /7 in the forward direction. Alternative : Let us consider the torque about the initial point of contact A. The force of friction passes through this point and hence its torque is zero. The normal force and the weight balance each other. The net torque about A is zero. Hence the angular momentum about A is conserved. Initial angular momentum is, L Lcm Mrv0 Icm 0 Mrv0 2 5 Mr 2 v0 2 r Mrv0 6 5 Mrv0. Suppose the translational velocity of the sphare, after it starts rolling, is v. The angular velocity is v/r. The angular momentum about A is, L Lcm Mrv 2 5 Mr 2 v r Mrv 7 5 Mrv. Thus, 6 5 Mrv0 7 5 Mrv or, v 6 7 v0 . 29. The sphere shown in figure (10-W22) lies on a rough plane when a particle of mass m travelling at a speed v0 collides and sticks with it. If the line of motion of the particle is at a distance h above the plane, find (a) the linear speed of the combined system just after the collision, (b) the angular speed of the system about the centre of the sphere just after the collision and (c) the value of h for which the sphere starts pure rolling on the plane. Assume that the mass M of the sphere is large compared to the mass of the particle so that the centre of mass of the combined system is not appreciably shifted from the centre of the sphere. Solution : Take the particle plus the sphere as the system. (a) Using conservation of linear momentum, the linear speed of the combined system v is given by mv0 M mv or, v mv0 M m (i) (b) Next, we shall use conservation of angular momentum about the centre of mass, which is to be taken at the centre of the sphere M m. Angular momentum of the particle before collision is mv0h R. If the system rotates with angular speed after collision, the angular momentum of the system becomes 2 5 MR 2 mR 2 . Hence, mv0h R 2 5 M m R 2 or, mv0h R 2 5 M m R 2 (c) The sphere will start rolling just after the collision if v R, i.e., mv0 M m mv0h R 2 5 M m R giving, h 7 5 M 2 m M m R 7 5 R.    Figure 10-W22 Rotational Mechanics 191

QUESTIONS FOR SHORT ANSWER 1. Can an object be in pure translation as well as in pure rotation ? 2. A simple pendulum is a point mass suspended by a light thread from a fixed point. The particle is displaced towards one side and then released. It makes small oscillations. Is the motion of such a simple pendulum a pure rotation ? If yes, where is the axis of rotation ? 3. In a rotating body, a = αr and v = ω r. Thus a α = v ω ⋅ Can you use the theorems of ratio and proportion studied in algebra so as to write a + α a − α = v + ω v − ω 4. A ball is whirled in a circle by attaching it to a fixed point with a string. Is there an angular rotation of the ball about its centre ? If yes, is this angular velocity equal to the angular velocity of the ball about the fixed point ? 5. The moon rotates about the earth in such a way that only one hemisphere of the moon faces the earth (figure 10-Q1). Can we ever see the “other face” of the moon from the earth ? Can a person on the moon ever see all the faces of the earth ? 6. The torque of the weight of any body about any vertical axis is zero. Is it always correct ? 7. The torque of a force F → about a point is defined as Γ → = r → × F → . Suppose r → , F → and Γ → are all nonzero. Is r → × Γ → ||F → always true ? Is it ever true ? 8. A heavy particle of mass m falls freely near the earth’s surface. What is the torque acting on this particle about a point 50 cm east to the line of motion ? Does this torque produce any angular acceleration in the particle ? 9. If several forces act on a particle, the total torque on the particle may be obtained by first finding the resultant force and then taking torque of this resultant. Prove this. Is this result valid for the forces acting on different particles of a body in such a way that their lines of action intersect at a common point ? 10. If the sum of all the forces acting on a body is zero, is it necessarily in equilibrium ? If the sum of all the forces on a particle is zero, is it necessarily in equilibrium ? 11. If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different point ? 12. If the resultant torque of all the forces acting on a body is zero about a point, is it necessary that it will be zero about any other point ? 13. A body is in translational equilibrium under the action of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point ? 14. A rectangular brick is kept on a table with a part of its length projecting out. It remains at rest if the length projected is slightly less than half the total length but it falls down if the length projected is slightly more than half the total length. Give reason. 15. When a fat person tries to touch his toes, keeping the legs straight, he generally falls. Explain with reference to figure (10-Q2). 16. A ladder is resting with one end on a vertical wall and the other end on a horizontal floor. Is it more likely to slip when a man stands near the bottom or near the top ? 17. When a body is weighed on an ordinary balance we demand that the arm should be horizontal if the weights on the two pans are equal. Suppose equal weights are put on the two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (point of support) zero ? Is the total torque zero ? If so, why does the arm rotate and finally become horizontal ? 18. The density of a rod AB continuously increases from A to B. Is it easier to set it in rotation by clamping it at A and applying a perpendicular force at B or by clamping it at B and applying the force at A ? 19. When tall buildings are constructed on earth, the duration of day–night slightly increases. Is it true ? 20. If the ice at the poles melts and flows towards the equator, how will it affect the duration of day–night ? 21. A hollow sphere, a solid sphere, a disc and a ring all having same mass and radius are rolled down on an inclined plane. If no slipping takes place, which one will take the smallest time to cover a given length ? 22. A sphere rolls on a horizontal surface. Is there any point of the sphere which has a vertical velocity ? E Figure 10-Q1 Figure 10-Q2 192 Concepts of Physics

OBJECTIVE I 1. Let A → be a unit vector along the axis of rotation of a purely rotating body and B → be a unit vector along the velocity of a particle P of the body away from the axis. The value of A → . B → is (a) 1 (b) –1 (c) 0 (d) None of these. 2. A body is uniformly rotating about an axis fixed in an inertial frame of reference. Let A → be a unit vector along the axis of rotation and B → be the unit vector along the resultant force on a particle P of the body away from the axis. The value of A → . B → is (a) 1 (b) –1 (c) 0 (d) none of these. 3. A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin (a) is zero (b) remains constant (c) goes on increasing (d) goes on decreasing. 4. A body is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity ω of the body are related as ω = v r ⋅ Thus (a) ω ∝ 1 r (b) ω ∝ r (c) ω = 0 (d) ω is independent of r. 5. Figure (10-Q3) shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If x and y be the distances travelled by A and B in the same time interval, then (a) x = 2 y (b) x = y (c) y = 2 x (d) none of these. 6. A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is (a) vertical (b) horizontal and skew with the axis (c) horizontal and intersecting the axis (d) none of these. 7. A body is rotating nonuniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is (a) vertical (b) horizontal and skew with the axis (c) horizontal and intersecting the axis (d) none of these. 8. Let F → be a force acting on a particle having position vector r → . Let Γ → be the torque of this force about the origin, then (a) r → . Γ → = 0 and F → . Γ → = 0 (b) r → . Γ → = 0 but F → . Γ → ≠ 0 (c) r → . Γ → ≠ 0 but F → . Γ → = 0 (d) r → . Γ → ≠ 0 and F → . Γ → ≠ 0. 9. One end of a uniform rod of mass m and length l is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity ω. The force exerted by the clamp on the rod has a horizontal component (a) mω 2 l (b) zero (c) mg (d) 1 2 mω 2 l. 10. A uniform rod is kept vertically on a horizontal smooth surface at a point O. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain (a) at O (b) at a distance less than l/2 from O (c) at a distance l/2 from O (d) at a distance larger than l/2 from O. 11. A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is (a) IA > IB (b) IA = IB (c) IA < IB (d) depends on the actual values of t and r. 12. Equal torques act on the discs A and B of the previous problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of A and another point on the rim of B are vA and vB respectively. We have (a) vA > vB (b) vA = vB (c) vA < vB (d) the relation depends on the actual magnitude of the torques. 13. A closed cylindrical tube containing some water (not filling the entire tube) lies in a horizontal plane. If the tube is rotated about a perpendicular bisector, the moment of inertia of water about the axis (a) increases (b) decreases (c) remains constant (d) increases if the rotation is clockwise and decreases if it is anticlockwise. 14. The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is (a) Mr 2 (b) 1 2 Mr 2 (c) 1 4 Mr 2 (d) 2 5 Mr 2 . 15. Let I1 and I2 be the moments of inertia of two bodies of identical geometrical shape, the first made of aluminium and the second of iron. (a) I1 < I2 (b) I1 = I2 (c) I1 > I2 (d) relation between I1 and I2 depends on the actual shapes of the bodies. 16. A body having its centre of mass at the origin has three of its particles at (a,0,0), (0,a,0), (0,0,a). The moments of inertia of the body about the X and Y axes are 0. 20 kg–m 2 each. The moment of inertia about the Z-axis (a) is 0. 20 kg–m 2 (b) is 0. 40 kg–m 2 (c) is 0. 20√2 kg–m 2 (d) cannot be deduced with this information. 17. A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform A B Figure 10-Q3 Rotational Mechanics 193

velocity. The torque of the normal force on the block about its centre has a magnitude (a) zero (b) Mga (c) Mga sinθ (d) 1 2 Mga sinθ. 18. A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become (a) ωM M + m (b) ωM M + 2 m (c) ω(M − 2 m) M + 2 m (d) ω(M + 2 m) M ⋅ 19. A person sitting firmly over a rotating stool has his arms stretched. If he folds his arms, his angular momentum about the axis of rotation (a) increases (b) decreases (c) remains unchanged (d) doubles. 20. The centre of a wheel rolling on a plane surface moves with a speed v0. A particle on the rim of the wheel at the same level as the centre will be moving at speed (a) zero (b) v0 (c) √2v0 (d) 2v0. 21. A wheel of radius 20 cm is pushed to move it on a rough horizontal surface. It is found to move through a distance of 60 cm on the road during the time it completes one revolution about the centre. Assume that the linear and the angular accelerations are uniform. The frictional force acting on the wheel by the surface is (a) along the velocity of the wheel (b) opposite to the velocity of the wheel (c) perpendicular to the velocity of the wheel (d) zero. 22. The angular velocity of the engine (and hence of the wheel) of a scooter is proportional to the petrol input per second. The scooter is moving on a frictionless road with uniform velocity. If the petrol input is increased by 10%, the linear velocity of the scooter is increased by (a) 50% (b) 10% (c) 20% (d) 0%. 23. A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of a smooth incline and released. Least time will be taken in reaching the bottom by (a) the solid sphere (b) the hollow sphere (c) the disc (d) all will take same time. 24. A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by (a) the solid sphere (b) the hollow sphere (c) the disc (d) all will take same time. 25. In the previous question, the smallest kinetic energy at the bottom of the incline will be achieved by (a) the solid sphere (b) the hollow sphere (c) the disc (d) all will achieve same kinetic energy. 26. A string of negligible thickness is wrapped several times around a cylinder kept on a rough horizontal surface. A man standing at a distance l from the cylinder holds one end of the string and pulls the cylinder towards him (figure 10-Q4). There is no slipping anywhere. The length of the string passed through the hand of the man while the cylinder reaches his hands is (a) l (b) 2l (c) 3l (d) 4l. OBJECTIVE II 1. The axis of rotation of a purely rotating body (a) must pass through the centre of mass (b) may pass through the centre of mass (c) must pass through a particle of the body (d) may pass through a particle of the body. 2. Consider the following two equations (A) L = I ω (B) dL dt = Γ In noninertial frames (a) both A and B are true (b) A is true but B is false (c) B is true but A is false (d) both A and B are false. 3. A particle moves on a straight line with a uniform velocity. Its angular momentum (a) is always zero (b) is zero about a point on the straight line (c) is not zero about a point away from the straight line (d) about any given point remains constant. 4. If there is no external force acting on a nonrigid body, which of the following quantities must remain constant ? (a) angular momentum (b) linear momentum (c) kinetic energy (d) moment of inertia. 5. Let IA and IB be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not. (a) IA < IB (b) If IA < IB, the axes are parallel (c) If the axes are parallel, IA < IB (d) If the axes are not parallel, IA ≥ IB. 6. A sphere is rotating about a diameter. (a) The particles on the surface of the sphere do not have any linear acceleration. (b) The particles on the diameter mentioned above do not have any linear acceleration. (c) Different particles on the surface have different Figure 10-Q4 194 Concepts of Physics

angular speeds. (d) All the particles on the surface have same linear speed. 7. The density of a rod gradually decreases from one end to the other. It is pivoted at an end so that it can move about a vertical axis through the pivot. A horizontal force F is applied on the free end in a direction perpendicular to the rod. The quantities, that do not depend on which end of the rod is pivoted, are (a) angular acceleration (b) angular velocity when the rod completes one rotation (c) angular momentum when the rod completes one rotation (d) torque of the applied force. 8. Consider a wheel of a bicycle rolling on a level road at a linear speed v0 (figure 10-Q5). (a) the speed of the particle A is zero (b) the speed of B, C and D are all equal to v0 (c) the speed of C is 2 v0 (d) the speed of B is greater than the speed of O. 9. Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping, (a) the heavier sphere reaches the bottom first (b) the bigger sphere reaches the bottom first (c) the two spheres reach the bottom together (d) the information given is not sufficient to tell which sphere will reach the bottom first. 10. A hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane. (a) The hollow sphere reaches the bottom first. (b) The solid sphere reaches the bottom with greater speed. (c) The solid sphere reaches the bottom with greater kinetic energy. (d) The two spheres will reach the bottom with same linear momentum. 11. A sphere cannot roll on (a) a smooth horizontal surface (b) a smooth inclined surface (c) a rough horizontal surface (d) a rough inclined surface. 12. In rear-wheel drive cars, the engine rotates the rear wheels and the front wheels rotate only because the car moves. If such a car accelerates on a horizontal road, the friction (a) on the rear wheels is in the forward direction (b) on the front wheels is in the backward direction (c) on the rear wheels has larger magnitude than the friction on the front wheels (d) on the car is in the backward direction. 13. A sphere can roll on a surface inclined at an angle if the friction coefficient is more than 2 7 tan. Suppose the friction coefficient is 1 7 g tan. If a sphere is released from rest on the incline, (a) it will stay at rest (b) it will make pure translational motion (c) it will translate and rotate about the centre (d) the angular momentum of the sphere about its centre will remain constant. 14. A sphere is rolled on a rough horizontal surface. It gradually slows down and stops. The force of friction tries to (a) decrease the linear velocity (b) increase the angular velocity (c) increase the linear momentum (d) decrease the angular velocity. 15. Figure (10-Q6) shows a smooth inclined plane fixed in a car accelerating on a horizontal road. The angle of incline is related to the acceleration a of the car as a g tan. If the sphere is set in pure rolling on the incline, (a) it will continue pure rolling (b) it will slip down the plane (c) its linear velocity will increase (d) its linear velocity will slowly decrease. EXERCISES 1. A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds. 2. A wheel rotating with uniform angular acceleration covers 50 revolutions in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds. 3. A wheel starting from rest is uniformly accelerated at 4 rad/s 2 for 10 seconds. It is allowed to rotate uniformly       Figure 10-Q5 Figure 10-Q6 Rotational Mechanics 195

for the next 10 seconds and is finally brought to rest in the next 10 seconds. Find the total angle rotated by the wheel. 4. A body rotates about a fixed axis with an angular acceleration of one radian/second/second. Through what angle does it rotate during the time in which its angular velocity increases from 5 rad/s to 15 rad/s. 5. Find the angular velocity of a body rotating with an acceleration of 2 rev/s 2 as it completes the 5th revolution after the start. 6. A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rad/s. Find the linear speed of (a) a point on the rim, (b) the middle point of a radius. 7. A disc rotates about its axis with a constant angular acceleration of 4 rad/s2 . Find the radial and tangential accelerations of a particle at a distance of 1 cm from the axis at the end of the first second after the disc starts rotating. 8. A block hangs from a string wrapped on a disc of radius 20 cm free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is 10 rad/s at some instant, with what speed is the block going down at that instant ? 9. Three particles, each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertia of the system about an axis (a) joining two of the particles and (b) passing through one of the particles and perpendicular to the plane of the particles. 10. Particles of masses 1 g, 2 g, 3 g, ..., 100 g are kept at the marks 1 cm, 2 cm, 3 cm, ..., 100 cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale. 11. Find the moment of inertia of a pair of spheres, each having a mass m and radius r, kept in contact about the tangent passing through the point of contact. 12. The moment of inertia of a uniform rod of mass 0. 50 kg and length 1 m is 0. 10 kg-m2 about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod. 13. Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of its particles. 14. The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line from the centre. 15. Find the moment of inertia of a uniform square plate of mass m and edge a about one of its diagonals. 16. The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as ρ(r) = A + Br. Find its moment of inertia about the line perpendicular to the plane of the disc through its centre. 17. A particle of mass m is projected with a speed u at an angle θ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point. 18. A simple pendulum of length l is pulled aside to make an angle θ with the vertical. Find the magnitude of the torque of the weight w of the bob about the point of suspension. When is the torque zero ? 19. When a force of 6. 0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut, it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut ? 20. Calculate the total torque acting on the body shown in figure (10-E2) about the point O. 21. A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its centre. 22. A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts. 23. A square plate of mass 120 g and edge 5. 0 cm rotates about one of the edges. If it has a uniform angular acceleration of 0. 2 rad/s 2 , what torque acts on the plate ? 24. Calculate the torque on the square plate of the previous problem if it rotates about a diagonal with the same angular acceleration. 25. A flywheel of moment of inertia 5. 0 kg-m 2 is rotated at a speed of 60 rad/s. Because of the friction at the axle, it comes to rest in 5. 0 minutes. Find (a) the average torque of the friction, (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating. 26. Because of the friction between the water in oceans with the earth’s surface, the rotational kinetic energy of the earth is continuously decreasing. If the earth’s angular speed decreases by 0. 0016 rad/day in 100 years, find the      Figure 10-E1              Figure 10-E2 196 Concepts of Physics

average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is 6. 0 × 10 24 kg. 27. A wheel rotating at a speed of 600 rpm (revolutions per minute) about its axis is brought to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque. 28. A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed of 100 rev/min when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions. 29. A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ? 30. A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s 2 . At what time will the body have kinetic energy same as the initial value if the torque continues to act ? 31. A light rod of length 1 m is pivoted at its centre and two masses of 5 kg and 2 kg are hung from the ends as shown in figure (10-E3). Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning. 32. Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length. (a) Find the initial angular acceleration of the rod. (b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg. 33. Figure (10-E4) shows two blocks of masses m and M connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley has a radius r and moment of inertia I about its axis and it can freely rotate about this axis. Find the acceleration of the mass M assuming that the string does not slip on the pulley. 34. A string is wrapped on a wheel of moment of inertia 0. 20 kg-m 2 and radius 10 cm and goes through a light pulley to support a block of mass 2. 0 kg as shown in figure (10-E5). Find the acceleration of the block. 35. Suppose the smaller pulley of the previous problem has its radius 5. 0 cm and moment of inertia 0. 10 kg-m2 . Find the tension in the part of the string joining the pulleys. 36. The pulleys in figure (10-E6) are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M. 37. The descending pulley shown in figure (10-E7) has a radius 20 cm and moment of inertia 0. 20 kg-m2 . The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1. 0 kg. 38. The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0. 5 kg-m2 about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4. 0 kg block. 39. Solve the previous problem if the friction coefficient between the 2. 0 kg block and the plane below it is 0. 5 and the plane below the 4. 0 kg block is frictionless. 40. A uniform metre stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings. 41. A uniform ladder of length 10. 0 m and mass 16. 0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60. 0 kg climbs up the ladder. If he stays on the ladder at a point 8. 00 m from 2 kg 5 kg Figure 10-E3 m M Figure 10-E4 2 kg 10 cm Figure 10-E5 M m Figure 10-E6 Figure 10-E7 2.0 kg 4 kg 45° 45° Figure 10-E8 Rotational Mechanics 197

the lower end, what will be the normal force and the force of friction on the ladder by the ground ? What should be the minimum coefficient of friction for the elctrician to work safely ? 42. Suppose the friction coefficient between the ground and the ladder of the previous problem is 0. 540. Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder. 43. A 6. 5 m long ladder rests against a vertical wall reaching a height of 6. 0 m. A 60 kg man stands half way up the ladder. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder. 44. The door of an almirah is 6 ft high, 1. 5 ft wide and weighs 8 kg. The door is supported by two hinges situated at a distance of 1 ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude. 45. A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is θ. 46. A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2 rad/s about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation, (b) the speed of the centre of the rod and (c) its kinetic energy. 47. A uniform square plate of mass 2. 0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0. 10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start. 48. Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth = 6400 km and radius of the orbit of the earth about the sun = 1. 5 × 10 8 km. 49. Two particles of masses m1 and m2 are joined by a light rigid rod of length r. The system rotates at an angular speed ω about an axis through the centre of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is L = µ r 2 ω where µ is the reduced mass of the system defined as µ = m1m2 m1 + m2 ⋅ 50. A dumb-bell consists of two identical small balls of mass 1/2 kg each connected to the two ends of a 50 cm long light rod. The dumb-bell is rotating about a fixed axis through the centre of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5. 0 N acts on one of the masses in the direction of its velocity for 0. 10 s. Find the new angular velocity of the system. 51. A wheel of moment of inertia 0. 500 kg-m2 and radius 20. 0 cm is rotating about its axis at an angular speed of 20. 0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel. 52. A diver having a moment of inertia of 6. 0 kg-m2 about an axis through its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5. 0 kg-m2 , what will be the new angular speed ? 53. A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from 6 kg-m2 to 2 kg-m2 , what will be the new angular speed ? 54. A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella in his hand. The axis of the umbrella coincides with that of the platform. The moment of inertia of “the platform plus the boy system” is 3. 0 × 10 − 3 kg–m 2 and that of the umbrella is 2. 0 × 10 − 3 kg–m 2 . The boy starts spinning the umbrella about the axis at an angular speed of 2. 0 rev/s with respect to himself. Find the angular velocity imparted to the platform. 55. A wheel of moment of inertia 0. 10 kg-m2 is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel. 56. A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is v horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event. 57. Suppose the platform of the previous problem is brought to rest with the ball in the hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed v as seen by his friend. Find the angular velocity with which the platform will start rotating. 58. Suppose the platform with the kid in the previous problem is rotating in anticlockwise direction at an angular speed ω. The kid starts walking along the rim with a speed v relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform. h L Figure 10-E9 198 Concepts of Physics

59. A uniform rod of mass m and length l is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate (a) the speed of the centre of mass, (b) the angular speed of the rod about the centre of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts. 60. A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance travelled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic. 61. Suppose the particle of the previous problem has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of the centre of mass C of the system constituting “the rod plus the particle”. (b) Find the velocity of the particle with respect to C before the collision. (c) Find the velocity of the rod with respect to C before the collision. (d) Find the angular momentum of the particle and of the rod about the centre of mass C before the collision. (e) Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision. (f) Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision. 62. Two small balls A and B, each of mass m, are joined rigidly by a light horizontal rod of length L. The rod is clamped at the centre in such a way that it can rotate freely about a vertical axis through its centre. The system is rotated with an angular speed ω about the axis. A particle P of mass m kept at rest sticks to the ball A as the ball collides with it. Find the new angular speed of the rod. 63. Two small balls A and B, each of mass m, are joined rigidly to the ends of a light rod of lengh L (figure 10-E10). The system translates on a frictionless horizontal surface with a velocity v0 in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. Find (a) the linear speeds of the balls A and B after the collision, (b) the velocity of the centre of mass C of the system A + B + P and (c) the angular speed of the system about C after the collision. [Hint : The light rod will exert a force on the ball B only along its length.] 64. Suppose the rod with the balls A and B of the previous problem is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of the same mass m is dropped from a height h on the ball B. The particle collides with B and sticks to it. (a) Find the angular momentum and the angular speed of the system just after the collision. (b) What should be the minimum value of h so that the system makes a full rotation after the collision. 65. Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia 1. 6 × 10 − 4 kg–m 2 and a radius 2. 0 cm. Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant. 66. The pulley shown in figure (10-E11) has a radius of 20 cm and moment of inertia 0. 2 kg-m2 . The string going over it is attached at one end to a vertical spring of spring constant 50 N/m fixed from below, and supports a 1 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take g = 10 m/s 2 . 67. A metre stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor. 68. A metre stick weighing 240 g is pivoted at its upper end in such a way that it can freely rotate in a vertical plane through this end (figure 10-E12). A particle of mass 100 g is attached to the upper end of the stick through a light string of length l m. Initially, the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise. 69. A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37° with the vertical. 70. A cylinder rolls on a horizontal plane surface. If the speed of the centre is 25 m/s, what is the speed of the highest point ? B A v0 L v0 Figure 10-E10 1kg Figure 10-E11 1m Figure 10-E12 Rotational Mechanics 199

71. A sphere of mass m rolls on a plane surface. Find its kinetic energy at an instant when its centre moves with speed v. 72. A string is wrapped over the edge of a uniform disc and the free end is fixed with the ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc. 73. A small spherical ball is released from a point at a height h on a rough track shown in figure (10-E13). Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track. 74. A small disc is set rolling with a speed v on the horizontal part of the track of the previous problem from right to left. To what height will it climb up the curved part ? 75. A sphere starts rolling down an incline of inclination . Find the speed of its centre when it has covered a distance l. 76. A hollow sphere is released from the top of an inclined plane of inclination . (a) What should be the minimum coefficient of friction between the sphere and the plane to prevent sliding ? (b) Find the kinetic energy of the ball as it moves down a length l on the incline if the friction coefficient is half the value calculated in part (a). 77. A solid sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup. 78. Figure (10-E14) shows a rough track, a portion of which is in the form of a cylinder of radius R. With what minimum linear speed should a sphere of radius r be set rolling on the horizontal part so that it completely goes round the circle on the cylindrical part. 79. Figure (10-E15) shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R. (a) Find the kinetic energy of the ball when it is at a point A where the radius makes an angle with the horizontal. (b) Find the radial and the tangential accelerations of the centre when the ball is at A. (c) Find the normal force and the frictional force acting on the ball if H 60 cm, R 10 cm, 0 and m = 70 g. 80. A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. Where should it be hit so that the shell does not slip on the surface ? 81. A uniform wheel of radius R is set into rotation about its axis at an angular speed . This rotating wheel is now placed on a rough horizontal surface with its axis horizontal. Because of friction at the contact, the wheel accelerates forward and its rotation decelerates till the wheel starts pure rolling on the surface. Find the linear speed of the wheel after it starts pure rolling. 82. A thin spherical shell lying on a rough horizontal surface is hit by a cue in such a way that the line of action passes through the centre of the shell. As a result, the shell starts moving with a linear speed v without any initial angular velocity. Find the linear speed of the shell after it starts pure rolling on the surface. 83. A hollow sphere of radius R lies on a smooth horizontal surface. It is pulled by a horizontal force acting tangentially from the highest point. Find the distance travelled by the sphere during the time it makes one full rotation. 84. A solid sphere of mass 0. 50 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface ? 85. A solid sphere is set into motion on a rough horizontal surface with a linear speed v in the forward direction and an angular speed v/R in the anticlockwise direction as shown in figure (10-E16). Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts. 86. A solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction. Figure 10-E13  Figure 10-E14 H A R m Figure 10-E15 v = v/R Figure 10-E16 200 Concepts of Physics

ANSWERS OBJECTIVE I 1. (c) 2. (c) 3. (b) 4. (d) 5. (c) 6. (c) 7. (b) 8. (a) 9. (d) 10. (c) 11. (c) 12. (a) 13. (a) 14. (a) 15. (a) 16. (d) 17. (d) 18. (b) 19. (b) 20. (c) 21. (a) 22. (d) 23. (d) 24. (d) 25. (b) 26. (b) OBJECTIVE II 1. (b), (d) 2. (b) 3. (b), (c), (d) 4. (a), (b) 5. (c) 6. (b) 7. (d) 8. (a), (c), (d) 9. (c) 10. (b) 11. (b) 12. (a), (b), (c) 13. (c) 14. (a), (b) 15. (a) EXERCISES 1. 25 rev/s 2 , 400 rad 2. 4 rev/s 2 , 20 rev/s 3. 800 rad 4. 100 rad 5. 2 5 rev/s 6. 2 m/s, 1 m/s 7. 16 cm/s 2 , 4 cm/s 2 8. 2 m/s 9. 1. 5 10 3 kgm 2 , 4. 0 10 3 kgm 2 10. 0. 43 kgm 2 11. 14 mr 2 5 12. 0. 34 m 13. 2 r 14. r /2 15. ma 2 /12 16. 2 Aa 4 4 Ba 5 5 17. mu 2 sin cos perpendicular to the plane of motion 18. wl sin, when the bob is at the lowest point 19. 1. 5 N 20. 0. 54 N–m 21. 1 2 mg a sin 22. 3 Ft 2 2 ml 23. 2. 0 10 5 N–m 24. 0. 5 10 5 N–m 25. (a) 1. 0 N–m (b) 9. 0 kJ (c) 60 kg–m 2 /s 26. 5. 8 10 20 N–m 27. 1 rev/s 2 , 5 rev/s 28. 0. 87 N 29. 25 s 30. 20 s 31. 8. 4 rad/s 2 32. 8. 0 rad/s 2 , 27. 6 N, 29 N 33. Mg M m I/r 2 34. 0. 89 m/s 2 35. 6. 3 N 36. M mg M m 2 I/R 2 37. 10 m/s 2 38. 0. 25 m/s 2 39. 0. 125 m/s 2 40. 1. 04 N in the left string and 1. 12 N in the right 41. 745 N, 412 N, 0. 553 42. 44. 0 kg 43. (a) 740 N-m (b) 590 N vertical and 120 N horizontal 44. 43 N 45. L cos sin 2 2 h L cos 2 sin 46. (a) 0. 05 kg–m 2 /s (b) 50 cm/s (c) 0. 05 J 47. 0. 5 kg–m 2 /s, 75 J 48. 2. 66 10 7 50. 12 rad/s 51. 19. 7 rad/s 52. 2. 4 rad/s 53. 360 rev/minute 54. 0. 8 rev/s 55. 0. 04 kg–m 2 56. mvR I M mR 2 57. mvR I MR 2 58. MvR I MR 2 59. (a) Ft m (b) 6 Ft ml (c) 2 F 2 t 2 m (d) Flt 2 60. L/12 61. (a) mv M m (b) Mv M m (c) mv M m (d) M 2 mvl 2M m 2 , Mm 2 vl 2M m 2 (e) MM 4 mL2 12M m (f) mv M m , 6 mv M 4 mL 62. 2 /3 Rotational Mechanics 201

63. (a) v0 2 , v0 (b) 2 3 v0 along the initial motion of the rod (c) v0 2 L 64. (a) mL √gh √2 , √ 8 gh 3 L (b) 3 2 L 65. (a) 0. 98 J (b) 1. 4 m/s 66. 0. 5 m/s 67. 5. 4 rad/s 68. 41° 69. 0. 9 √2 dm g 70. 50 m/s 71. 7 10 mv 2 72. 2 3 g 73. √ 10 gh/7 74. 3 v 2 4 g 75. √ 10 7 g l sinθ 76. (a) 2 5 tanθ (b) 7 8 mgl sinθ 77. 17 mg/7 78. √ 27 7 g(R − r) 79. (a) mg(H − R − R sinθ), (b) 10 7 g    H R − 1 − sinθ    , − 5 7 g cosθ (c) 4. 9 N, 0. 196 N upward 80. 2 R/3 above the centre 81. ωR/3 82. 3 v/5 83. 4 π R/3 84. 3. 3 N 85. (a) 3 v/5 (b) 3 v/7 86. 3 v/7 202 Concepts of Physics

11.1 HISTORICAL INTRODUCTION The motion of celestial bodies such as the moon, the earth, the planets, etc., has been a subject of great interest for a long time. Famous Indian astronomer and mathematician, Aryabhat, studied these motions in great detail, most likely in the 5th century A.D., and wrote his conclusions in his book Aryabhatiya. He established that the earth revolves about its own axis. He also gave description of motion of other celestial bodies as seen from the earth. About a thousand years after Aryabhat, the brilliant combination of Tycho Brahe (1546–1601) and Johannes Kepler (1571–1630) studied the planetary motion in great detail. Kepler formulated his important findings in his three laws of planetary motion : The year 1665 was very fruitful for Isaac Newton aged 23. He was forced to take rest at his home in Lincolnshire after his college at Cambridge was closed for an indefinite period due to plague. In this year, he performed brilliant theoretical and experimental tasks mainly in the field of mechanics and optics. In this same year he focussed his attention on the motion of the moon about the earth. The moon makes a revolution about the earth in T = 27. 3 days. The distance of the moon from the earth is R = 3. 85 × 10 5 km. The acceleration of the moon is, therefore, a = ω 2 R = 4π 2 × (3. 85 × 10 5 km) (27. 3 days) 2 = 0. 0027 m s−2 ⋅ The first question before Newton was that what is the force that produces this acceleration. The acceleration is towards the centre of the orbit, that is towards the centre of the earth. Hence the force must act towards the centre of the earth. A natural guess was that the earth is attracting the moon. The saying goes that Newton was sitting under an apple tree when an apple fell down from the tree on the earth. This sparked the idea that the earth attracts all bodies towards its centre. The next question was what is the law governing this force. Newton had to make several daring assumptions which proved to be turning points in science and philosophy. He declared that the laws of nature are the same for earthly and celestial bodies. The force operating between the earth and an apple and that operating between the earth and the moon, must be governed by the same laws. This statement may look very obvious today but in the era before Newton, there was a general belief in the western countries that the earthly bodies are governed by certain rules and the heavenly bodies are governed by different rules. In particular, this heavenly structure was supposed to be so perfect that there could not be any change in the sky. This distinction was so sharp that when Tycho Brahe saw a new star in the sky, he did not believe 1. All planets move in elliptical orbits with the sun at a focus. 2. The radius vector from the sun to the planet sweeps equal area in equal time. 3. The square of the time period of a planet is proportional to the cube of the semimajor axis of the ellipse. Figure 11.1 CHAPTER 11 GRAVITATION

his eyes as there could be no change in the sky. So the Newton’s declaration was indeed revolutionary. The acceleration of a body falling near the earth’s surface is about 9. 8 ms−2 . Thus, aapple amoon = 9. 8 m s−2 0. 0027 m s−2 = 3600. Also, distance of the moon from the earth distance of the apple from the earth = dmoon dapple = 3. 85 × 10 5 km 6400 km = 60 Thus, aapple amoon =    dmoon dapple    2 ⋅ Newton guessed that the acceleration of a body towards the earth is inversely proportional to the square of the distance of the body from the centre of the earth. Thus, a ∝ 1 r 2 ⋅ Also, the force is mass times acceleration and so it is proportional to the mass of the body. Hence, F∝ m r 2 ⋅ By the third law of motion, the force on a body due to the earth must be equal to the force on the earth due to the body. Therefore, this force should also be proportional to the mass of the earth. Thus, the force between the earth and a body is F ∝ Mm r 2 or, F = GMm r 2 ⋅ … (11.1) Newton further generalised the law by saying that not only the earth but all material bodies in the universe attract each other according to equation (11.1) with same value of G. The constant G is called universal constant of gravitation and its value is found to be 6. 67 × 10 – 11 N–m 2 /kg 2 . Equation (11.1) is known as the universal law of gravitation. In this argument, the distance of the apple from the earth is taken to be equal to the radius of the earth. This means we have assumed that earth can be treated as a single particle placed at its centre. This is of course not obvious. Newton had spent several years to prove that indeed this can be done. A spherically symmetric body can be replaced by a point particle of equal mass placed at its centre for the purpose of calculating gravitational force. In the process he discovered the methods of calculus that we have already learnt in Chapter 2. There is evidence that quite a bit of differential calculus was known to the ancient Indian mathematicians but this literature was almost certainly not known to Newton or other scientists of those days. Example 11.1 Two particles of masses 1. 0 kg and 2. 0 kg are placed at a separation of 50 cm. Assuming that the only forces acting on the particles are their mutual gravitation, find the initial accelerations of the two particles. Solution : The force of gravitation exerted by one particle on another is F = Gm1m2 r 2 = 6. 67 × 10 − 11 N−m 2 kg 2 × (1. 0 kg) × (2. 0 kg) (0. 5 m) 2 = 5. 3 × 10 − 10 N. The acceleration of 1. 0 kg particle is a1 = F m1 = 5. 3 × 10 − 10 N 1. 0 kg = 5. 3 × 10 − 10 m s−2 . This acceleration is towards the 2. 0 kg particle. The acceleration of the 2. 0 kg particle is a2 = F m2 = 5. 3 × 10 − 10 N 2. 0 kg = 2. 65 × 10 − 10 m s−2 . This acceleration is towards the 1. 0 kg particle. 11.2 MEASUREMENT OF GRAVITATIONAL CONSTANT G The gravitational constant G is a small quantity and its measurement needs very sensitive arrangement. The first important successful measurement of this quantity was made by Cavendish in 1736 about 71 years after the law was formulated. In this method, two small balls of equal mass are attached at the two ends of a light rod to form a dumb-bell. The rod is suspended vertically by a fine quartz wire. Two large spheres of equal mass are placed near the smaller spheres in such a way that all the four spheres are on a horizontal circle. The centre of the circle is at the middle point of the rod (figure 11.2). 204 Concepts of Physics

Two larger spheres lie on the opposite sides of the smaller balls at equal distance. A small plane mirror is attached to the vertical wire. A light beam, incident on the mirror, falls on a scale after reflection. If the wire rotates by an angle θ, the reflected beam rotates by 2θ and the spot on the scale moves. By measuring this movement of the spot on the scale and the distance between the mirror and the scale, the angle of deviation can be calculated. When the heavy balls are placed close to the small balls, a torque acts on the dumb-bell to rotate it. As the dumb-bell rotates, the suspension wire gets twisted and produces a torque on the dumb-bell in opposite direction. This torque is proportional to the angle rotated. The dumb-bell stays in equilibrium where the two torques have equal magnitude. Let the mass of a heavy ball = M, the mass of a small ball = m, the distance between the centres of a heavy ball and the small ball placed close to it = r, the deflection of the dumb-bell as it comes to equilibrium = θ, the torsional constant of the suspension wire = k, the length of the rod = l and the distance between the scale and the mirror = D. The force acting on each of the small balls is F = G Mm r 2 ⋅ Here we have used the fact that the gravitational force due to a uniform sphere is same as that due to a single particle of equal mass placed at the centre of the sphere. As the four balls are on the same horizontal circle and the heavy balls are placed close to the smaller balls, this force acts in a horizontal direction perpendicular to the length of the dumb-bell. The torque due to each of these gravitational forces about the suspension wire is F(l/2). The total gravitational torque on the dumb-bell is, therefore, Γ = 2F(l/2) = Fl . The opposing torque produced by the suspension wire is kθ. For rotational equilibrium, Fl = kθ or, GMml r 2 = kθ G = kθr 2 Mml ⋅ … (i) In an experiment, the heavy balls are placed close to the smaller balls as shown in the figure and the dumb-bell is allowed to settle down. The light beam is adjusted so that the beam reflected by the plane mirror falls on the scale. Now the heavy balls are shifted in such a way that they are placed on the same horizontal circle at same distance from the smaller balls but on the opposite side. In figure (11.3), the original positions of the heavy balls are shown by A, B and the shifted positions by A′, B′. As the heavy balls are shifted to the new position, the dumb-bell rotates. If it was settled previously at an angle θ deviated from the mean position, it will now settle at the same angle θ on the other side. Thus, the total deflection of the dumb-bell due to the change in the positions of the heavy balls is 2θ. The reflected light beam deviates by an angle of 4θ. If the linear displacement of the light spot is d, we have (figure 11.4) 4θ = d D or, θ = d 4D ⋅ Figure 11.2

Figure 11.3     Figure 11.4 Gravitation 205

Substituting in (i), G = kdr 2 4MmlD ⋅ All the quantities on the right hand side are experimentally known and hence the value of G may be calculated. 11.3 GRAVITATIONAL POTENTIAL ENERGY The concept of potential energy of a system was introduced in Chapter-8. The potential energy of a system corresponding to a conservative force was defined as Uf − Ui = − ∫ i f F → . dr → . The change in potential energy is equal to the negative of the work done by the internal forces. We also calculated the change in gravitational potential energy of the earth–particle system when the particle was raised through a small height over earth’s surface. In this case the force mg may be treated as constant and the change in potential energy is Uf − Ui = mgh where the symbols have their usual meanings. We now derive the general expression for the change in gravitational potential energy of a two-particle system. Let a particle of mass m1 be kept fixed at a point A (figure 11.5) and another particle of mass m2 is taken from a point B to a point C. Initially, the distance between the particles is AB = r1 and finally it becomes AC = r2. We have to calculate the change in potential energy of the system of the two particles as the distance changes from r1 to r2. Consider a small displacement when the distance between the particles changes from r to r + dr. In the figure, this corresponds to the second particle going from D to E. The force on the second particle is F = Gm1 m2 r 2 along DA → . The work done by the gravitational force in the displacement is dW = − Gm1 m2 r 2 dr . The increase in potential energy of the two-particle system during this displacement is dU = − dW = Gm1 m2 r 2 dr. The increase in potential energy as the distance between the particles changes from r1 to r2 is U(r2) − U(r1) = ∫ dU = ∫ r1 r2 Gm1 m2 r 2 dr = Gm1 m2 ∫ r1 r2 1 r 2 dr = Gm1 m2    − 1 r    r1 r2 = Gm1 m2    1 r1 − 1 r2    ⋅ … (11.2) We choose the potential energy of the two-particle system to be zero when the distance between them is infinity. This means that we choose U(∞) = 0. By (11.2) the potential energy U(r), when the separation between the particles is r, is U(r) = U(r) − U(∞) = Gm1m2    1 ∞ − 1 r    = − Gm1m2 r ⋅ The gravitational potential energy of a two-particle system is U(r) = − Gm1 m2 r … (11.3) where m1 and m2 are the masses of the particles, r is the separation between the particles and the potential energy is chosen to be zero when the separation is infinite. We have proved this result by assuming that one of the particles is kept at rest and the other is displaced. However, as the potential energy depends only on the separation and not on the location of the particles, equation (11.3) is general. Equation (11.3) gives the potential energy of a pair of particles. If there are three particles A, B and C, there are three pairs AB, AC and BC. The potential energy of the three-particle system is equal to the sum of the potential energies of the three pairs. For an Nparticle system there are N(N − 1)/2 pairs and the potential energy is calculated for each pair and added to get the total potential energy of the system.  
       Figure 11.5 206 Concepts of Physics

Example 11.2 Find the work done in bringing three particles, each having a mass of 100 g, from large distances to the vertices of an equilateral triangle of side 20 cm. Solution : When the separations are large, the gravitational potential energy is zero. When the particles are brought at the vertices of the triangle ABC, three pairs AB, BC and CA are formed. The potential energy of each pair is − Gm1 m2/r and hence the total potential energy becomes U = 3 ×    − Gm1 m2 r    = 3 ×    − 6. 67 × 10 − 11 N−m 2 /kg 2 × (0. 1 kg) × (0. 1 kg) 0. 20 m    = − 1. 0 × 10 − 11 J . The work done by the gravitational forces is W = − U = 1. 0 × 10 − 11 J. If the particles are brought by some external agency without changing the kinetic energy, the work done by the external agency is equal to the change in potential energy = − 1. 0 × 10 − 11 J. 11.4 GRAVITATIONAL POTENTIAL Suppose a particle of mass m is taken from a point A to a point B while keeping all other masses fixed. Let UA and UB denote the gravitational potential energy when the mass m is at point A and point B respectively. We define the “change in potential” VB − VA between the two points as VB − VA = UB − UA m ⋅ … (11.4) The equation defines only the change in potential. We can choose any point to have zero potential. Such a point is called a reference point. If A be the reference point, VA = 0 and VB = UB − UA m ⋅ … (11.5) Thus, gravitational potential at a point is equal to the change in potential energy per unit mass, as the mass is brought from the reference point to the given point. If the particle is slowly brought without increasing the kinetic energy, the work done by the external agent equals the change in potential energy. Thus, the potential at a point may also be defined as the work done per unit mass by an external agent in bringing a particle slowly from the reference point to the given point. Generally the reference point is chosen at infinity so that the potential at infinity is zero. The SI unit of gravitational potential is J kg–1. 11.5 CALCULATION OF GRAVITATIONAL POTENTIAL (A) Potential due to a Point Mass Suppose a particle of mass M is kept at a point A (figure 11.6) and we have to calculate the potential at a point P at a distance r away from A. The reference point is at infinity. From equation (11.5), the potential at the point P is V(r) = U(r) − U(∞) m ⋅ But U(r) − U(∞) = − GMm r ⋅ so that, V = − GM r ⋅ … (11.6) The gravitational potential due to a point mass M at a distance r is − GM r ⋅ (B) Potential due to a Uniform Ring at A Point on its Axis Let the mass of the ring be M and its radius be a. We have to calculate the gravitational potential at a point P on the axis of the ring (figure 11.7). The centre is at O and OP = r. Consider any small part of the ring of mass dm. The point P is at a distance z = √ a 2 + r 2 from dm. The potential at P due to dm is dV = − G dm z = − G dm √ a 2 + r 2 ⋅ The potential V due to the whole ring is obtained by summing the contributions from all the parts. As the potential is a scalar quantity, we have   Figure 11.6     Figure 11.7 Gravitation 207

V dV G dm a 2 r 2 G a 2 r 2 dm GM a 2 r 2 (11.7) In terms of the distance z between the point P and any point of the ring, the expression for the potential is given by V GM z (11.8) (C) Potential due to a Uniform Thin Spherical Shell Let the mass of the given spherical shell be M and the radius a. We have to calculate the potential due to this shell at a point P. The centre of the shell is at O and OP r (figure 11.8). Let us draw a radius OA making an angle with OP. Let us rotate this radius about OP keeping the angle AOP fixed at value . The point A traces a circle on the surface of the shell. Let us now consider another radius at an angle d and likewise rotate it about OP. Another circle is traced on the surface of the shell. The part of the shell included between these two circles (shown shaded in the figure) may be treated as a ring. The radius of this ring is a sin and hence the perimeter is 2 a sin. The width of the ring is ad. The area of the ring is 2 a sin ad 2 a 2 sin d. The total area of the shell is 4 a 2 . As the shell is uniform, the mass of the ring enclosed is dm M 4 a 2 2a 2 sin d M 2 sin d . Let the distance of any point of the ring from P be AP z. From the triangle OAP z 2 a 2 r 2 2ar cos or, 2z dz 2ar sin d or, sin d z dz ar Thus, the mass of the ring is dm M 2 sin d M 2ar z dz. As the distance of any point of the ring from P is z, the potential at P due to the ring is dV G dm z GM 2ar dz . As we vary from 0 to , the rings formed on the shell cover up the whole shell. The potential due to the whole shell is obtained by integrating dV within the limits 0 to . Case I : P is outside the shell r > a As figure (11.8) shows, when 0, the distance z AP r a. When , it is z r a. Thus, as varies from 0 to , the distance z varies from r a to r a. Thus, V dV GM 2ar r a r a dz GM 2ar [z] r a r a GM 2ar [r a r a] GM r (11.9) To calculate the potential at an external point, a uniform spherical shell may be treated as a point particle of equal mass placed at its centre. Case II : P is inside the shell r < a In this case when 0, the distance z AP a r and when it is z a r (figure 11.9). Thus, as varies from 0 to , the distance z varies from a r to a r. Thus, the potential due to the shell is V dV GM 2ar [z] a r a r GM 2ar [a r a r] GM a (11.10) Figure 11.8 208 Concepts of Physics

This does not depend on r. Thus, the potential due to a uniform spherical shell is constant throughout the cavity of the shell. Figure (11.10) shows graphically the variation of potential with the distance from the centre of the shell. Example 11.3 A particle of mass M is placed at the centre of a uniform spherical shell of equal mass and radius a. Find the gravitational potential at a point P at a distance a/2 from the centre. Solution : The gravitational potential at the point P due to the particle at the centre is V1 = − GM a/2 = − 2GM a ⋅ The potential at P due to the shell is V2 = − GM a ⋅ The net potential at P is V1 + V2 = − 3GM a ⋅ (D) Potential due to a Uniform Solid Sphere The situation is shown in figure (11.11). Let the mass of the sphere be M and its radius a. We have to calculate the gravitational potential at a point P. Let OP = r. Let us draw two spheres of radii x and x + dx concentric with the given sphere. These two spheres enclose a thin spherical shell of volume 4 π x 2 dx. The volume of the given sphere is 4 3 π a 3 . As the sphere is uniform, the mass of the shell is dm = M 4 3 πa 3 4π x 2 dx = 3M a 3 x 2 dx. The potential due to this shell at the point P is dV = − G dm r if x < r and dV = − G dm x if x > r. Case I : Potential at an external point Suppose the point P is outside the sphere (figure 11.11). The potential at P due to the shell considered is dV = − G dm r ⋅ Thus, the potential due to the whole sphere is V = ∫ dV = − G r ∫ dm = − GM r ⋅ … (11.11) The gravitational potential due to a uniform sphere at an external point is same as that due to a single particle of equal mass placed at its centre. Case II : Potential at an internal point Let us divide the sphere in two parts by imagining a concentric spherical surface passing through P. The inner part has a mass M′ = M 4 3 π a 3 × 4 3 π r 3 = Mr 3 a 3 ⋅ The potential at P due to this inner part is by equation (11.11) V1 = − GM′ r = − GMr 2 a 3 ⋅ … (i)     Figure 11.9     Figure 11.10      Figure 11.11      Figure 11.12 Gravitation 209

To get the potential at P due to the outer part of the sphere, we divide this part in concentric shells. The mass of the shell between radii x and x + dx is dm = M 4 3 π a 3 4 π x 2 dx = 3Mx 2 dx a 3 ⋅ The potential at P due to this shell is, − G dm x = − 3 GM a 3 xdx ⋅ The potential due to the outer part is V2 = ∫ r a − 3GM a 3 xdx = − 3GM a 3    x 2 2    r a = − 3GM 2a 3 (a 2 − r 2 ). … (ii) By (i) and (ii) the total potential at P is V = V1 + V2 = − GMr 2 a 3 − 3GM 2a 3 (a 2 − r 2 ) = − GM 2a 3 (3a 2 − r 2 ). … (11.12) At the centre of the sphere the potential is V = − 3GM 2a ⋅ 11.6 GRAVITATIONAL FIELD We have been saying all through that a body A exerts a force of gravitation on another body B kept at a distance. This is called action at a distance viewpoint. However, this viewpoint creates certain problems when one deals with objects separated by large distances. It is now assumed that a body can not directly interact with another body kept at a distance. The force between two objects is seen to be a two-step process. In the first step, it is assumed that the body A creates a gravitational field in the space around it. The field has its own existence and has energy and momentum. This field has a definite direction at each point of the space and its intensity varies from point to point. In the second step, it is assumed that when a body B is placed in a gravitational field, this field exerts a force on it. The direction and the intensity of the field is defined in terms of the force it exerts on a body placed in it. We define the intensity of gravitational field E → at a point by the equation E → = F → m … (11.13) where F → is the force exerted by the field on a body of mass m placed in the field. Quite often the intensity of gravitational field is abbreviated as gravitational field. Its SI unit is N kg–1. Gravitational field adds according to the rules of vector addition. If E → 1 is the field due to a source S1 and E → 2 is the field at the same point due to another source S2, the resultant field when both the sources are present is E → 1 + E → 2. If a mass m is placed close to the surface of the earth, the force on it is mg. We say that the earth has set up a gravitational field and this field exerts a force on the mass. The intensity of the field is E → = F → m = mg→ m = g → . Thus, the intensity of the gravitational field near the surface of the earth is equal to the acceleration due to gravity. It should be clearly understood that the intensity of the gravitational field and the acceleration due to gravity are two separate physical quantities having equal magnitudes and directions. Example 11.4 A particle of mass 50 g experiences a gravitational force of 2. 0 N when placed at a particular point. Find the gravitational field at that point. Solution : The gravitational field has a magnitude E = F m = 2. 0 N (50 × 10 − 3 kg) = 40 N kg−1 . This field is along the direction of the force. 11.7 RELATION BETWEEN GRAVITATIONAL FIELD AND POTENTIAL Suppose the gravitational field at a point r → due to a given mass distribution is E → . By definition (equation 11.13), the force on a particle of mass m when it is at r → is F → = mE→ . As the particle is displaced from r → to r → + dr → the work done by the gravitational force on it is dW = F → . dr → = m E → . dr → . The change in potential energy during this displacement is dU = − dW = − mE→ . dr → . The change in potential is, by equation (11.4), 210 Concepts of Physics

dV = dU m = − E → . dr → . … (11.14) Integrating between r → 1 and r → 2 V(r → 2) − V(r → 1) = − ∫ r → 1 r → 2 E → . dr → . … (11.15) If r → 1 is taken at the reference point, V(r → 1) = 0. The potential V(r → ) at any point r → is, therefore, V(r → ) = − ∫ r → 0 r → E → . dr → … (11.16) where r → 0 denotes the reference point. If we work in Cartesian coordinates, we can write E → = i → Ex + j → Ey + k →Ez and dr → = i → dx + j → dy + k →dz so that E → . dr → = Ex dx + Ey dy + Ez dz . Equation (11.14) may be written as dV = − Ex dx − Ey dy − Ez dz. If y and z remain constant, dy = dz = 0 . Thus, Ex = − ∂V ∂x … (11.17) Similarly, Ey = − ∂V ∂y and Ez = − ∂V ∂z ⋅ The symbol ∂ ∂x means partial differentiation with respect to x treating y and z to be constants. Similarly for ∂ ∂y and ∂ ∂z ⋅ If the field is known, the potential may be obtained by integrating the field according to equation (11.16) and if the potential is known, the field may be obtained by differentiating the potential according to equation (11.17). Example 11.5 The gravitational field due to a mass distribution is given by E = K/x 3 in X-direction. Taking the gravitational potential to be zero at infinity, find its value at a distance x. Solution : The potential at a distance x is V(x) = − ∫ ∞ x E dx = − ∫ ∞ x K x 3 dx =    K 2x 2    ∞ x = K 2x 2 ⋅ Example 11.6 The gravitational potential due to a mass distribution is V = A √ x  2 + a 2 ⋅ Find the gravitational field. Solution : V = A √ x  2 + a 2 = A(x 2 + a 2 ) − 1/2 . If the gravitational field is E, Ex = − ∂V ∂x = − A   − 1 2    (x 2 + a 2 ) − 3/2 (2x) = Ax (x 2 + a 2 ) 3/2 Ey = − ∂V ∂y = 0 and Ez = − ∂V ∂z = 0. The gravitational field is Ax (x 2 + a 2 ) 3/2 in the x-direction. 11.8 CALCULATION OF GRAVITATIONAL FIELD (A) Field due to a Point Mass Suppose a particle of mass M is placed at a point O (figure 11.13) and a second particle of mass m is placed at a point P. Let OP = r. The mass M creates a field E → at the site of mass m and this field exerts a force F → = mE→ on the mass m. But the force F → on the mass m due to the mass M is F = GMm r 2 acting along PO → . Thus, the gravitational field at P is E = GM r 2 … (11.18) along PO → . If O is taken as the origin, the position vector of mass m is r → = OP → . Equation (11.18) may be rewritten in vector form as = − GM r 2 e → r … (11.19) where e → r is the unit vector along r → . (B) Field due to a Uniform Circular Ring at a Point on its Axis Figure (11.14) shows a uniform circular ring of radius a and mass M. Let P be a point on its axis at    Figure 11.13 Gravitation 211

a distance r from the centre. We have to obtain the gravitational field at P due to the ring. By symmetry the field must be towards the centre that is along PO → . Consider any particle of mass dm on the ring, say at point A. The distance of this particle from P is AP = z = √ a 2 + r 2 . The gravitational field at P due to dm is along PA → and its magnitude is dE = G dm z 2 ⋅ The component along PO is dE cosα = G dm z 2 cosα . The net gravitational field at P due to the ring is E = ∫ G dm z 2 cosα = G cosα z 2 ∫ dm = GM cosα z 2 = GMr (a 2 + r 2 ) 3/2 ⋅ … (11.20) The field is directed towards the centre of the ring. (C) Field due to a Uniform Disc at a Point on its Axis The situation is shown in figure (11.15). Let the mass of the disc be M and its radius be a. Let O be the centre of the disc and P be a point on its axis at a distance r from the centre. We have to find the gravitational field at P due to the disc. Let us draw a circle of radius x with the centre at O. We draw another concentric circle of radius x + dx. The part of the disc enclosed between these two circles can be treated as a uniform ring of radius x. The point P is on its axis at a distance r from the centre. The area of this ring is 2π x dx. The area of the whole disc is π a 2 . As the disc is uniform, the mass of this ring is dm = M π a 2 2π x dx = 2M x dx a 2 ⋅ The gravitational field at P due to the ring is, by equation (11.20), dE = G    2M x dx a 2    r (r 2 + x 2 ) 3/2 = 2GMr a 2 x dx (r 2 + x 2 ) 3/2 ⋅ As x varies from 0 to a, the rings cover up the whole disc. The field due to each of these rings is in the same direction PO. Thus, the net field due to the whole disc is along PO and its magnitude is E = ∫ 0 a 2GMr a 2 x dx (r 2 + x 2 ) 3/2 = 2GMr a 2 ∫ 0 a x dx (r 2 + x 2 ) 3/2 ⋅ … (i) Let r 2 + x 2 = z 2 . Then 2x dx = 2z dz and ∫ x dx (r 2 + x 2 ) 3/2 = ∫ z dz z 3 = ∫ 1 z 2 dz = − 1 z = − 1 √ r 2 + x 2 ⋅ From (i), E = 2GMr a 2    − 1 √ r 2 + x 2    0 a = 2GMr a 2    1 r − 1 √ r 2 + a 2    … (11.21) Equation (11.21) may be expressed in terms of the angle θ subtended by a radius of the disc at P as, E = 2GM a 2 (1 − cosθ) . (D) Field due to a Uniform Thin Spherical Shell We can use the construction of figure (11.8) to find the gravitational field at a point due to a uniform thin spherical shell. The figure is reproduced here (figure 11.16) with symbols having same meanings. The shaded ring has mass dm = M 2 sinθ dθ. The field at P due to this ring is dE = Gdm z 2 cosα = GM 2 sinθ dθ cosα z 2 ⋅ … (i)       Figure 11.14      Figure 11.15 212 Concepts of Physics

From the triangle OAP, z 2 a 2 r 2 2ar cos or, 2z dz 2ar sin d or, sin d z dz ar (ii) Also from the triangle OAP, a 2 z 2 r 2 2zr cos or, cos z 2 r 2 a 2 2zr (iii) Putting from (ii) and (iii) in (i), dE GM 4ar 2 1 a 2 r 2 z 2 dz or, dE GM 4ar 2 z a 2 r 2 z Case I : P is outside the shell r > a In this case z varies from r a to r a. The field due to the whole shell is E GM 4ar 2 z a 2 r 2 z r a r a GM r 2 We see that the shell may be treated as a point particle of the same mass placed at its centre to calculate the gravitational field at an external point. Case II : P is inside the shell r < a In this case z varies from a r to a r (figure 11.9). The field at P due to the whole shell is E GM 4ar 2 z a 2 r 2 z a r a r 0. Hence the field inside a uniform spherical shell is zero. (E) Gravitational Field due to a Uniform Solid Sphere Case I : Field at an external point Let the mass of the sphere be M and its radius be a. We have to calculate the gravitational field due to the sphere at a point outside the sphere at a distance r from the centre. Figure (11.17) shows the situation. The centre of the sphere is at O and the field is to be calculated at P. Let us divide the sphere into thin spherical shells each centred at O. Let the mass of one such shell be dm. To calculate the gravitational field at P, we can replace the shell by a single particle of mass dm placed at the centre of the shell that is at O. The field at P due to this shell is then dE G dm r 2 towards PO. The field due to the whole sphere may be obtained by summing the fields of all the shells making the solid sphere. Thus, E dE G dm r 2 G r 2 dm GM r 2 (11.23) Thus, a uniform sphere may be treated as a single particle of equal mass placed at its centre for calculating the gravitational field at an external point. This allows us to treat the earth as a point particle placed at its centre while calculating the force between the earth and an apple. Case II : Field at an internal point Suppose the point P is inside the solid sphere (figure 11.18). In this case r a. The sphere may be divided into thin spherical shells all centered at O. Suppose the mass of such a shell is dm. If the radius      Figure 11.16     Figure 11.17    Figure 11.18 Gravitation 213

of the shell is less than r, the point P is outside the shell and the field due to the shell is dE = G dm r 2 along PO . If the radius of the shell considered is greater than r, the point P is internal and the field due to such a shell is zero. The total field due to the whole sphere is obtained by summing the fields due to all the shells. As all these fields are along the same direction, the net field is E = ∫ dE = ∫ G dm r 2 = G r 2 ∫ dm . … (i) Only the masses of the shells with radii less than r should be added to get ∫ dm. These shells form a solid sphere of radius r. The volume of this sphere is 4 3 π r 3 . The volume of the whole sphere is 4 3 π a 3 . As the given sphere is uniform, the mass of the sphere of radius r is M 4 3 π a 3 ⋅    4 3 π r 3   = Mr 3 a 3 ⋅ Thus, ∫ dm = Mr 3 a 3 and by (i) E = G r 2 Mr 3 a 3 = GM a 3 r. … (11.24) The gravitational field due to a uniform sphere at an internal point is proportional to the distance of the point from the centre of the sphere. At the centre itself, r = 0 and the field is zero. This is also expected from symmetry because any particle at the centre is equally pulled from all sides and the resultant must be zero. At the surface of the sphere, r = a and E = GM a 2 ⋅ The formula (11.23) for the field at an external point also gives E = GM a 2 at the surface of the sphere. The two formulae agree at r = a. Figure (11.19) shows graphically the variation of gravitational field due to a solid sphere with the distance from its centre. Example 11.7 Find the gravitational field due to the moon at its surface. The mass of the moon is 7. 36 × 10 22 kg and the radius of the moon is 1. 74 × 10 6 m. Assume the moon to be a spherically symmetric body. Solution : To calculate the gravitational field at an external point, the moon may be replaced by a single particle of equal mass placed at its centre. Then the field at the surface is E = GM a 2 = 6. 67 × 10 − 11N−m 2 /kg 2 × 7. 36 × 10 22 kg (1. 74 × 10 6 m) 2 = 1. 62 N kg−1 . This is about one sixth of the gravitational field due to the earth at its surface. 11.9 VARIATION IN THE VALUE OF g The acceleration due to gravity is given by g = F m where F is the force exerted by the earth on an object of mass m. This force is affected by a number of factors and hence g also depends on these factors. (a) Height from the Surface of the Earth If the object is placed at a distance h above the surface of the earth, the force of gravitation on it due to the earth is F = GMm (R + h) 2 where M is the mass of the earth and R is its radius. Thus, g = F m = GM (R + h) 2 ⋅ We see that the value of g decreases as one goes up. We can write, g = GM R 2   1 + h R    2 = g0    1 + h R    2 where g0 = GM R 2 is the value of g at the surface of the earth. If h << R, g = g0    1 + h R    − 2 ≈ g0    1 − 2 h R    ⋅ If one goes a distance h inside the earth such as in mines, the value of g again decreases. The force by the     Figure 11.19 214 Concepts of Physics

earth is, by equation (11.24), F = GMm R 3 (R − h) or, g = F m = GM R 2    R − h R    = g0    1 − h R    ⋅ The value of g is maximum at the surface of the earth and decreases with the increase in height as well as with depth similar to that shown in figure (11.19). Example 11.8 Calculate the value of acceleration due to gravity at a point (a) 5. 0 km above the earth’s surface and (b) 5. 0 km below the earth’s surface. Radius of earth = 6400 km and the value of g at the surface of the earth is 9. 80 m s–2 . Solution : (a) The value of g at a height h is (for h << R) g = g0    1 − 2 h R    = (9. 80 m s−2 )    1 − 2 × 5. 0 km 6400 km    = 9. 78 m s−2 . (b) The value at a depth h is g = g0    1 − h R    = (9. 8 m s−2 )    1 − 5. 0 km 6400 km    = 9. 79 m s−2 . (b) Rotation of the Earth As the earth rotates about its own axis the frame attached to the earth is noninertial. If we wish to use the familiar Newton’s laws, we have to include pseudo forces. For an object at rest with respect to the earth, a centrifugal force mω 2 r is to be added where m is the mass of the object, ω is the angular velocity of the earth and r is the radius of the circle in which the particle rotates. If the colatitude of the location of the particle is θ (figure 11.20), r = R sin θ where R is the radius of the earth. Acceleration of an object falling near the earth’s surface, as measured from the earth frame, is F/m where F is the vector sum of the gravitational force GMm R 2 = mg and the centrifugal force mω 2 r = mω 2 R sin θ. The acceleration F/m = g′ is the apparent value of the acceleration due to gravity. At the equator, θ = π/2 and the centrifugal force is just opposite to the force of gravity. The resultant of these two is F = mg − mω 2 R or, g′ = g − ω 2 R. At the poles, θ = 0 and the centrifugal force mω 2 R sin θ = 0. Thus, F = mg and g′ = g. Thus, the observed value of the acceleration due to gravity is minimum at the equator and is maximum at the poles. This effect had been discussed in the chapter on circular motion. (c) Nonsphericity of the Earth All formulae and equations have been derived by assuming that the earth is a uniform solid sphere. The shape of the earth slightly deviates from the perfect sphere. The radius in the equatorial plane is about 21 km larger than the radius along the poles. Due to this the force of gravity is more at the poles and less at the equator. The value of g is accordingly larger at the poles and less at the equator. Note that due to rotation of earth also, the value of g is smaller at the equator than that at the poles. (d) Nonuniformity of the Earth The earth is not a uniformly dense object. There are a variety of minerals, metals, water, oil, etc., inside the earth. Then at the surface there are mountains, seas, etc. Due to these nonuniformities in the mass distribution, the value of g is locally affected. “Weighing” the Earth The force exerted by the earth on a body is called the weight of the body. In this sense ‘‘weight of the earth’’ is a meaningless concept. However, the mass of the earth can be determined by noting the acceleration due to gravity near the surface of the earth. We have, g = GM R 2 or, M = gR 2 /G Putting g = 9. 8 m s−2 , R = 6400 km
   Figure 11.20 Gravitation 215

and G = 6. 67 × 10 − 11 N−m 2 kg 2 the mass of the earth comes out to be 5. 98 × 10 24 kg. 11.10 PLANETS AND SATELLITES Planets Planets move round the sun due to the gravitational attraction of the sun. The path of these planets are elliptical with the sun at a focus. However, the difference in major and minor axes is not large. The orbits can be treated as nearly circular for not too sophisticated calculations. Let us derive certain characteristics of the planetary motion in terms of the radius of the orbit assuming it to be perfectly circular. Let the mass of the sun be M and that of the planet under study be m. The mass of the sun is many times larger than the mass of the planet. The sun may, therefore, be treated as an inertial frame of reference. Speed Let the radius of the orbit be a and the speed of the planet in the orbit be v. By Newton’s second law, the force on the planet equals its mass times the acceleration. Thus, GMm a 2 = m    v 2 a    or, v = √GM a ⋅ … (11.25) The speed of a planet is inversely proportional to the square root of the radius of its orbit. Time period The time taken by a planet in completing one revolution is its time period T. In one revolution it covers a linear distance of 2πa at speed v. Thus, T = 2πa v = 2πa √GM a = 2π √ GM a 3/2 or, T 2 = 4π 2 GM a 3 . … (11.26) Energy The kinetic energy of the planet is K = 1 2 mv 2 . Using (11.25), K = 1 2 m GM a = GMm 2a ⋅ The gravitational potential energy of the sun–planet system is U = − GMm a ⋅ The total mechanical energy of the sun–planet system is E = K + U = GMm 2a − GMm a = − GMm 2a ⋅ The total energy is negative. This is true for any bound system if the potential energy is taken to be zero at infinite separation. Satellite Satellites are launched from the earth so as to move round it. A number of rockets are fired from the satellite at proper time to establish the satellite in the desired orbit. Once the satellite is placed in the desired orbit with the correct speed for that orbit, it will continue to move in that orbit under gravitational attraction of the earth. All the equations derived above for planets are also true for satellites with M representing the mass of the earth and m representing the mass of the satellite. Example 11.9 A satellite is revolving round the earth at a height of 600 km. Find (a) the speed of the satellite and (b) the time period of the satellite. Radius of the earth = 6400 km and mass of the earth = 6 × 10 24 kg. Solution : The distance of the satellite from the centre of the earth is 6400 km + 600 km = 7000 km. The speed of the satellite is v = √GM a = √ 6. 67 × 10 − 11 N−m 2 /kg 2 × 6 × 10 24 kg 7000 × 10 3 m = 7. 6 × 10 3 m s−1 = 7. 6 km s−1 .  ffi ffl Figure 11.21 216 Concepts of Physics

The time period is T = 2π a v = 2π × 7000 × 10 3 m 7. 6 × 10 3 m s−1 = 5. 8 × 10 3 s. Geostationary Satellite The earth rotates about its own axis (the line joining the north pole and the south pole) once in 24 hours. Suppose a satellite is established in an orbit in the plane of the equator. Suppose the height is such that the time period of the satellite is 24 hours and it moves in the same sense as the earth. The satellite will always be overhead a particular place on the equator. As seen from the earth, this satellite will appear to be stationary. Such a satellite is called a geostationary satellite. Such satellites are used for telecommunication, weather forecasting and other applications. According to equation (11.26), T 2 = 4π 2 GM a 3 or, a =    GMT 2 4π 2    1/3 . Putting the values of G, M = (6 × 10 24 kg) and T = (24 hours); the radius of the geostationary orbit comes out to be a = 4. 2 × 10 4 km. The height above the surface of the earth is about 3. 6 × 10 4 km. 11.11 KEPLER’S LAWS From the observations of Tycho Brahe, Kepler formulated the laws of planetary motion which we have listed in the first section of this chapter. The first law states that the path of a planet is elliptical with the sun at a focus. Circular path is a special case of an ellipse when the major and minor axes are equal. For a circular path, the planet should have velocity perpendicular to the line joining it with the sun and the magnitude should satisfy equation (11.25), that is v = √GM a ⋅ If these conditions are not satisfied, the planet moves in an ellipse. The second law states that the radius vector from the sun to the planet sweeps out equal area in equal time. For a circular orbit, this is obvious because the speed of the particle remains constant. The third law of Kepler states that the square of the time period of a planet is proportional to the cube of the semimajor axis. For a circular orbit semimajor axis is same as the radius. We have already proved this law for circular orbits in equation (11.26). As M denotes mass of the sun, 4π 2 GM is fixed for all planets and T 2 ∝ a 3 . 11.12 WEIGHTLESSNESS IN A SATELLITE A satellite moves round the earth in a circular orbit under the action of gravity. The acceleration of the satellite is GM R 2 towards the centre of the earth, where M is the mass of the earth and R is the radius of the orbit of the satellite. Consider a body of mass m placed on a surface inside a satellite moving round the earth. The forces on the body are (a) the gravitational pull of the earth = GMm R 2 , (b) the contact force N by the surface. By Newton’s law, G Mm R 2 − N = m   GM R 2    or, N = 0. Thus, the surface does not exert any force on the body and hence its apparent weight is zero. No support is needed to hold a body in the satellite. All positions shown in figure (11.22) are equally comfortable. One can analyse the situation from the frame of the satellite. Working in the satellite frame we have to add a centrifugal force on all bodies. If the mass of a body is m, the centrifugal force is m    GM R 2    away from the centre of the earth. This pseudo force exactly balances the weight of the body which is GMm R 2 towards the centre of the earth. A body needs no support to stay at rest in the satellite and hence all positions are equally comfortable. Water will not fall down from the glass even if it is inverted. It will act like a “gravityfree hall”. Such a state is called weightlessness. It should be clear that the earth still attracts a body with the same force GMm R 2 ⋅ The feeling of weightlessness arises because one stays in a rotating frame. 11.13 ESCAPE VELOCITY When a stone is thrown up it goes up to a maximum height and then returns. As the particle Figure 11.22 Gravitation 217

goes up, the gravitational potential energy increases and the kinetic energy of the particle decreases. The particle will continue to go up till its kinetic energy becomes zero and will return from there. Let the initial velocity of the particle be u. The kinetic energy of the particle is K = 1 2 mu 2 and the gravitational potential energy of the earth–particle system is U = − GMm R , where M is the mass of the earth, m is the mass of the particle and R is the radius of the earth. When it reaches a height h above the earth’s surface, its speed becomes v. The kinetic energy there is 1 2 mv 2 and the gravitational potential energy is − GMm R + h ⋅ By conservation of energy 1 2 mu 2 − GMm R = 1 2 mv 2 − GMm R + h or, 1 2 mv 2 =    1 2 mu 2 − GMm R    + GMm R + h ⋅ … (i) The particle will reach the maximum height when v becomes zero. If 1 2 mu 2 − GMm R ≥ 0, the right-hand side of (i) is greater than zero for all values of h. Thus, 1 2 mv 2 never becomes zero. The particle’s velocity never reaches zero and so the particle will continue to go farther and farther away from the earth. Thus, the particle will never return to the earth if 1 2 mu 2 − GMm R ≥ 0 or, u ≥ √ 2GM R ⋅ … (11.27) This critical initial velocity is called the escape velocity. Putting the values of G, M and R, the escape velocity from the earth comes out to be 11. 6 km s–1. In this we have neglected the effect of other planets, stars and other objects in space. In fact, even if the initial velocity is somewhat less than the escape velocity, the particle may get attracted by some other celestial object and land up there. Equation (11.27) is valid for any celestial object. For example, if something is thrown up from the surface of the moon, it will never return to the moon if the initial velocity is greater than √ 2GM R , where M is the mass of the moon and R is the radius of the moon. Example 11.10 Calculate the escape velocity from the moon. The mass of the moon = 7. 4 × 10 22 kg and radius of the moon = 1740 km. Solution : The escape velocity is v = √ 2GM R = √ 2 × 6. 67 × 10 − 11 N−m 2 /kg 2 × 7. 4 × 10 22 kg 1740 × 10 3 m = 2. 4 km s−1 . 11.14 GRAVITATIONAL BINDING ENERGY We have seen that if a particle of mass m placed on the earth is given an energy 1 2 mu 2 = GMm R or more, it finally escapes from earth. The minimum energy needed to take the particle infinitely away from the earth is called the binding energy of the earth–particle system. Thus, the binding energy of the earth–particle system is GMm R ⋅ 11.15 BLACK HOLES Consider a spherical body of mass M and radius R. Suppose, due to some reason the volume goes on decreasing while the mass remains the same. The escape velocity √ 2GM R from such a dense material will be very high. Suppose the radius is so small that √ 2GM R ≥ c where c = 3 × 10 8 m s−1 is the speed of light. The escape velocity for such an object is equal to or greater than the speed of light. This means, anything starting from the object with a speed less than the speed of light will return to the object (neglecting the effect of other objects in space). According to the theory of relativity it is not possible to achieve a velocity greater than c for any material object. Thus, nothing can escape from such a dense material. Such objects are known as black holes. A number of such black holes exist in space. Even light cannot escape from a black hole. 11.16 INERTIAL AND GRAVITATIONAL MASS Given two objects A and B, how can we determine the ratio of the mass of A to the mass of B. One way is to use Newton’s second law of motion. If we apply equal forces F on each of the two objects, F = mA aA and also F = mB aB. 218 Concepts of Physics

Thus, mA mB = aB aA or, mA = aB aA mB . … (i) This equation may be used to ‘‘define the mass’’ of an object. Taking the object B to be the standard kilogram (mB = 1 kg), mass of any object may be obtained by measuring their accelerations under equal force and using (i). The mass so defined is called inertial mass. Another way to compare masses of two objects is based on the law of gravitation. The gravitational force exerted by a massive body on an object is proportional to the mass of the object. If FA and FB be the forces of attraction on the two objects due to the earth, FA = GmA M R 2 and FB = GmB M R 2 ⋅ Thus, mA mB = FA FB or, mA = FA FB mB . … (ii) We can use this equation to ‘‘define the mass’’ of an object. If B is a standard unit mass, by measuring the gravitational forces FA and FB we can obtain the mass of the object A. The mass so defined is called gravitational mass. When we measure the mass using a spring balance, we actually measure the gravitational mass. Equivalence of Inertial and Gravitational Mass The two definitions of mass, described above, are quite independent of each other. There is no obvious reason why the two should be identical. However, they happen to be identical. Several sophisticated experiments have been performed to test this equivalence and none of them has supplied any evidence against it. The general theory of relativity is based on the principle of equivalence of inertial and gravitational mass. 11.17 POSSIBLE CHANGES IN THE LAW OF GRAVITATION There is some indication that the force between two masses is not as described in this chapter. The deviation from the simple law F = GMm R 2 is being taken as an indication of the existence of a fifth interaction besides gravitational, electromagnetic, nuclear and weak. It has been reported (Phys. Rev. Lett. Jan 6, 1986) that the force between two masses may be better represented by F = G∞ m1 m2 r 2    1 +    1 + r λ    α e − r λ    with α ≈ − 0. 007 and λ ≈ 200 m. As α is negative, the second term in the square bracket represents a repulsive force. For r >> 200 m F = G∞ m1 m2 r 2 which is the force operative between the earth and other objects. For r << 200 m F = G∞ m1 m2(1 + α) r 2 = G′m1 m2 r 2 where G′ = G∞(1 + α). This is the force we measure in a Cavendishexperiment. The value of G for small distances is about 1% less than the value of G for large distances. Worked Out Examples 1. Three particles A, B and C, each of mass m, are placed in a line with AB = BC = d. Find the gravitational force on a fourth particle P of same mass, placed at a distance d from the particle B on the perpendicular bisector of the line AC. Solution : The force at P due to A is FA = G m 2 (AP) 2 = G m 2 2 d 2 along PA. The force at P due to C is FC = G m 2 (CP) 2 = G m 2 2 d 2 along PC. The force at P due to B is FB = G m 2 d 2 along PB. The resultant of FA, FB and FC will be along PB. Clearly ∠ APB = ∠ BPC = 45°. Component of FA along PB = FA cos45° = G m 2 2 √2 d 2 ⋅  
  Figure 11-W1 Gravitation 219

Component of FC along PB = FC cos45° = G m 2 2 √2 d 2 ⋅ Component of FB along PB = G m 2 d 2 ⋅ Hence, the resultant of the three forces is G m 2 d 2    1 2 √2 + 1 2 √2 + 1   = G m 2 d 2    1 + 1 √ 2    along PB. 2. Find the distance of a point from the earth’s centre where the resultant gravitational field due to the earth and the moon is zero. The mass of the earth is 6. 0 × 10 24 kg and that of the moon is 7. 4 × 10 22 kg. The distance between the earth and the moon is 4. 0 × 10 5 km. Solution : The point must be on the line joining the centres of the earth and the moon and in between them. If the distance of the point from the earth is x, the distance from the moon is (4. 0 × 10 5 km – x). The magnitude of the gravitational field due to the earth is E1 = GMe x 2 = G × 6 × 10 24 kg x 2 and magnitude of the gravitational field due to the moon is E2 = GMm (4. 0 × 10 5 km − x) 2 = G × 7. 4 × 10 22 kg (4. 0 × 10 5 km − x) 2 ⋅ These fields are in opposite directions. For the resultant field to be zero E1 = E2 , or, 6 × 10 24 kg x 2 = 7. 4 × 10 22 kg (4. 0 × 10 5 km − x) 2 or, x 4. 0 × 10 5 km − x = √6 × 10 24 7. 4 × 10 22 = 9 or, x = 3. 6 × 10 5 km. 3. Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle. Solution : The particles will always remain diametrically opposite so that the force on each particle will be directed along the radius. Consider the motion of one of the particles. The force on the particle is F = G m 2 4 R 2 ⋅ If the speed is v, its acceleration is v 2 /R. Thus, by Newton’s law, G m 2 4 R 2 = m v 2 R or, v = √G m 4 R ⋅ 4. Two particles A and B of masses 1 kg and 2 kg respectively are kept 1 m apart and are released to move under mutual attraction. Find the speed of A when that of B is 3. 6 cm/hour. What is the separation between the particles at this instant ? Solution : The linear momentum of the pair A + B is zero initially. As only mutual attraction is taken into account, which is internal when A + B is taken as the system, the linear momentum will remain zero. The particles move in opposite directions. If the speed of A is v when the speed of B is 3. 6 cm/hour = 10 – 5 m s–1 , (1 kg) v = (2 kg) (10 − 5 m s−1 ) or, v = 2 × 10 − 5 m s−1 . The potential energy of the pair is − G mA mB R with usual symbols. Initial potential energy = − 6. 67 × 10 − 11 N−m 2 /kg 2 × 2 kg × 1 kg 1 m = − 13. 34 × 10 − 11 J. If the separation at the given instant is d, using conservation of energy, − 13. 34 × 10 − 11 J + 0 = − 13. 34 × 10 − 11 J−m d + 1 2 (2 kg) (10 − 5 m s−1 ) 2 + 1 2 (1 kg) (2 × 10 − 5 m s−1 ) 2 Solving this, d = 0. 31 m. 5. The gravitational field in a region is given by E → = (10 N kg−1 ) ( i → + j → ). Find the work done by an external agent to slowly shift a particle of mass 2 kg from the point (0,0) to a point (5 m, 4 m). Solution : As the particle is slowly shifted, its kinetic energy remains zero. The total work done on the particle is thus zero. The work done by the external agent should be negative of the work done by the gravitational field. The work done by the field is ∫ i f F → ⋅ dr → Consider figure (11-W2). Suppose the particle is taken from O to A and then from A to B. The force on the particle is F → = mE→ = (2 kg) (10 N kg−1 ) ( i → + j → ) = (20 N) ( i → + j → ). The work done by the field during the displacement OA is
      Figure 11-W2 220 Concepts of Physics

W1 = ∫ 0 5m Fx dx = ∫ 0 5m (20 N) dx = 20 N × 5 m = 100 J. Similarly, the work done in displacement AB is W2 = ∫ 0 4m Fy dy = ∫ 0 4m (20 N) dy = (20 N) (4 m) = 80 J. Thus, the total work done by the field, as the particle is shifted from O to B, is 180 J. The work done by the external agent is –180 J. Note that the work is independent of the path so that we can choose any path convenient to us from O to B. 6. A uniform solid sphere of mass M and radius a is surrounded symmetrically by a uniform thin spherical shell of equal mass and radius 2 a. Find the gravitational field at a distance (a) 3 2 a from the centre, (b) 5 2 a from the centre. Solution : Figure (11-W3) shows the situation. The point P1 is at a distance 3 2 a from the centre and P2 is at a distance 5 2 a from the centre. As P1 is inside the cavity of the thin spherical shell, the field here due to the shell is zero. The field due to the solid sphere is E = GM    3 2 a    2 = 4 GM 9 a 2 ⋅ This is also the resultant field. The direction is towards the centre. The point P2 is outside the sphere as well as the shell. Both may be replaced by single particles of the same mass at the centre. The field due to each of them is E′ = GM    5 2 a    2 = 4 GM 25 a 2 ⋅ The resultant field is E = 2 E′ = 8 GM 25 a 2 towards the centre. 7. The density inside a solid sphere of radius a is given by ρ = ρ0 a/r, where ρ0 is the density at the surface and r denotes the distance from the centre. Find the gravitational field due to this sphere at a distance 2 a from its centre. Solution : The field is required at a point outside the sphere. Dividing the sphere in concentric shells, each shell can be replaced by a point particle at its centre having mass equal to the mass of the shell. Thus, the whole sphere can be replaced by a point particle at its centre having mass equal to the mass of the given sphere. If the mass of the sphere is M, the gravitational field at the given point is E = GM (2a) 2 = GM 4a 2 ⋅ … (i) The mass M may be calculated as follows. Consider a concentric shell of radius r and thickness dr. Its volume is dV = (4πr 2 ) dr and its mass is dM = ρdV =    ρ0 a r    (4πr 2 dr) = 4πρ0 ar dr. The mass of the whole sphere is M = ∫ 0 a 4πρ0 ar dr = 2πρ0a 3 . Thus, by (i) the gravitational field is E = 2πGρ0a 3 4a 2 = 1 2 πGρ0a. 8. A uniform ring of mass m and radius a is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is at a distance √3 a from the centre of the sphere. Find the gravitational force exerted by the sphere on the ring. Solution : The gravitational field at any point on the ring due to the sphere is equal to the field due to a single particle of mass M placed at the centre of the sphere. Thus, the force on the ring due to the sphere is also equal to the force on it by a particle of mass M placed at this point. By Newton’s third law it is equal to the force on the particle by the ring. Now the gravitational field due to the ring at a distance d = √3 a on its axis is E = Gmd (a 2 + d 2 ) 3/2 = √3 Gm 8a 2 ⋅       Figure 11-W3     Figure 11-W4 Gravitation 221

The force on a particle of mass M placed here is F = ME = √3GMm 8a 2 ⋅ This is also the force due to the sphere on the ring. 9. A particle is fired vertically upward with a speed of 9. 8 km s–1. Find the maximum height attained by the particle. Radius of earth = 6400 km and g at the surface = 9. 8 m s–2. Consider only earth’s gravitation. Solution : At the surface of the earth, the potential energy of the earth–particle system is − G Mm R with usual symbols. The kinetic energy is 1 2 m v0 2 where v0 = 9. 8 km s−1 . At the maximum height the kinetic energy is zero. If the maximum height reached is H, the potential energy of the earth–particle system at this instant is − G M m R + H ⋅ Using conservation of energy, − G M m R + 1 2 m v0 2 = − G M m R + H ⋅ Writing GM = gR 2 and dividing by m, − gR + v0 2 2 = − gR 2 R + H or, R 2 R + H = R − v0 2 2 g or, R + H = R 2 R − v0 2 2 g . Putting the values of R, v0 and g on the right side, R + H = (6400 km) 2 6400 km − (9. 8 km s−1 ) 2 2 × 9. 8 m s−2 = (6400 km) 2 1500 km = 27300 km or, H = (27300 − 6400) km = 20900 km. 10. A particle hanging from a spring stretches it by 1 cm at earth’s surface. How much will the same particle stretch the spring at a place 800 km above the earth’s surface ? Radius of the earth = 6400 km. Solution : Suppose the mass of the particle is m and the spring constant of the spring is k. The acceleration due to gravity at earth’s surface is g = GM R 2 with usual symbols. The extension in the spring is mg/k. Hence, 1 cm = GMm k R 2 ⋅ … (i) At a height h = 800 km, the extension is given by x = GMm k (R + h) 2 ⋅ … (ii) By (i) and (ii), x 1 cm = R 2 (R + h) 2 (6400 km) 2 (7200 km) 2 = 0. 79. Hence, x = 0. 79 cm. 11. A simple pendulum has a time period exactly 2 s when used in a laboratory at north pole. What will be the time period if the same pendulum is used in a laboratory at equator ? Account for the earth’s rotation only. Take g = GM R 2 = 9. 8 m s−2 and radius of earth = 6400 km. Solution : Consider the pendulum in its mean position at the north pole. As the pole is on the axis of rotation, the bob is in equilibrium. Hence in the mean position, the tension T is balanced by earth’s attraction. Thus, T = G M m R 2 = mg⋅ The time period t is t = 2π √l T/m = 2 π √l g ⋅ … (1) At equator, the lab and the pendulum rotate with the earth at angular velocity ω = 2 π radian 24 hour in a circle of radius equal to 6400 km. Using Newton’s second law, G M m R 2 − T ′ = m ω 2 R or, T ′ = m (g − ω 2 R) where T ′ is the tension in the string. The time period will be t ′ = 2 π √l (T ′/m) = 2 π √ l g − ω 2 R ⋅ … (ii) By (i) and (ii) t ′ t = √g g − ω 2 R =    1 − ω 2 R g    − 1/2 or, t ′ ≈ t    1 + ω 2 R 2g    ⋅ Putting the values, t ′ = 2. 004 seconds. 12. A satellite is to revolve round the earth in a circle of radius 8000 km. With what speed should this satellite be projected into orbit ? What will be the time period ? Take g at the surface = 9. 8 m s–2 and radius of the earth = 6400 km. Solution : Suppose, the speed of the satellite is v. The acceleration of the satellite is v 2 /r, where r is the radius of the orbit. The force on the satellite is GMm r 2 with usual symbols. Using Newton’s second law, GM m r 2 = m v 2 r 222 Concepts of Physics

or, v 2 = GM r = g R 2 r = (9. 8 m s−2 ) (6400 km) 2 (8000 km) giving v = 7. 08 km s−1 . The time period is 2 π r v = 2 π (8000 km) (7. 08 km s−1 ) ≈ 118 minutes. 13. Two satellites S1 and S2 revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 h and 8 h respectively. The radius of the orbit of S1 is 10 4 km. When S2 is closest to S1 , find (a) the speed of S2 relative to S1 and (b) the angular speed of S2 as observed by an astronaut in S1 . Solution : Let the mass of the planet be M, that of S1 be m1 and of S2 be m2 . Let the radius of the orbit of S1 be R1 ( = 10 4 km) and of S2 be R2 . Let v1 and v2 be the linear speeds of S1 and S2 with respect to the planet. Figure (11-W5) shows the situation. As the square of the time period is proportional to the cube of the radius,    R2 R1    3 =    T2 T1    2 =    8 h 1 h    2 = 64 or, R2 R1 = 4 or, R2 = 4R1 = 4 × 10 4 km. Now the time period of S1 is 1 h. So, 2 π R1 v1 = 1 h or, v1 = 2 π R1 1 h = 2 π × 10 4 km h−1 similarly, v2 = 2 π R2 8 h = π × 10 4 km h−1 . (a) At the closest separation, they are moving in the same direction. Hence the speed of S2 with respect to S1 is |v2 − v1| = π × 10 4 km h−1 . (b) As seen from S1, the satellite S2 is at a distance R2 − R1 = 3 × 10 4 km at the closest separation. Also, it is moving at π × 10 4 km h−1 in a direction perpendicular to the line joining them. Thus, the angular speed of S2 as observed by S1 is ω = π × 10 4 km h−1 3 × 10 4 km = π 3 rad h−1 . QUESTIONS FOR SHORT ANSWERS 1. Can two particles be in equilibrium under the action of their mutual gravitational force ? Can three particles be ? Can one of the three particles be ? 2. Is there any meaning of “Weight of the earth” ? 3. If heavier bodies are attracted more strongly by the earth, why don’t they fall faster than the lighter bodies ? 4. Can you think of two particles which do not exert gravitational force on each other ? 5. The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun ? Or does it ? 6. At noon, the sun and the earth pull the objects on the earth’s surface in opposite directions. At midnight, the sun and the earth pull these objects in same direction. Is the weight of an object, as measured by a spring balance on the earth’s surface, more at midnight as compared to its weight at noon ? 7. An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g. Who exerts the force needed to accelerate the earth with this acceleration g ? 8. Suppose the gravitational potential due to a small system is k/r 2 at a distance r from it. What will be the gravitational field ? Can you think of any such system ? What happens if there were negative masses ? 9. The gravitational potential energy of a two-particle system is derived in this chapter as U = − Gm1m2 r ⋅ Does it follow from this equation that the potential energy for r = ∞ must be zero ? Can we choose the potential energy for r = ∞ to be 20 J and still use this formula ? If no, what formula should be used to calculate the gravitational potential energy at separation r ? 10. The weight of an object is more at the poles than at the equator. Is it beneficial to purchase goods at equator and sell them at the pole ? Does it matter whether a spring balance is used or an equal-beam balance is used ? 11. The weight of a body at the poles is greater than the weight at the equator. Is it the actual weight or the apparent weight we are talking about ? Does your

        Figure 11-W5 Gravitation 223

answer depend on whether only the earth’s rotation is taken into account or the flattening of the earth at the poles is also taken into account ? 12. If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease ? If so, by what per cent ? 13. A nut becomes loose and gets detached from a satellite revolving around the earth. Will it land on the earth ? If yes, where will it land ? If no, how can an astronaut make it land on the earth ? 14. Is it necessary for the plane of the orbit of a satellite to pass through the centre of the earth ? 15. Consider earth satellites in circular orbits. A geostationary satellite must be at a height of about 36000 km from the earth’s surface. Will any satellite moving at this height be a geostationary satellite ? Will any satellite moving at this height have a time period of 24 hours ? 16. No part of India is situated on the equator. Is it possible to have a geostationary satellite which always remains over New Delhi ? 17. As the earth rotates about its axis, a person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. Why does he/she not feel weightless as a satellite passenger does ? 18. Two satellites going in equatorial plane have almost same radii. As seen from the earth one moves from east to west and the other from west to east. Will they have the same time period as seen from the earth ? If not, which one will have less time period ? 19. A spacecraft consumes more fuel in going from the earth to the moon than it takes for a return trip. Comment on this statement. OBJECTIVE I 1. The acceleration of moon with respect to earth is 0. 0027 m s–2 and the acceleration of an apple falling on earth’s surface is about 10 m s−2 . Assume that the radius of the moon is one fourth of the earth’s radius. If the moon is stopped for an instant and then released, it will fall towards the earth. The initial acceleration of the moon towards the earth will be (a) 10 m s−2 (b) 0. 0027 m s−2 (c) 6. 4 m s−2 (d) 5. 0 m s−2 . 2. The acceleration of the moon just before it strikes the earth in the previous question is (a) 10 m s−2 (b) 0. 0027 m s−2 (c) 6. 4 m s−2 (d) 5. 0 m s−2 3. Suppose, the acceleration due to gravity at the earth’s surface is 10 m s–2 and at the surface of Mars it is 4. 0 m s–2. A 60 kg passenger goes from the earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of figure (11-Q1) best represents the weight (net gravitational force) of the passenger as a function of time ? (a) A (b) B (c) C (d) D. 4. Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weigh (a) W (b) 2 W (c) W/2 (d) 2 1/3 W at the planet. 5. If the acceleration due to gravity at the surface of the earth is g, the work done in slowly lifting a body of mass m from the earth’s surface to a height R equal to the radius of the earth is (a) 1 2 mgR (b) 2mgR (c) mgR (d) 1 4 mgR. 6. A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at rest but it moves at a speed of 2 m s–1 as it reaches A. The work done by the person on the mass is –3 J. The potential at A is (a) –3 J kg–1 (b) –2 J kg–1 (c) –5 J kg–1 (d) none of these. 7. Let V and E be the gravitational potential and gravitational field at a distance r from the centre of a uniform spherical shell. Consider the following two statements : (A) The plot of V against r is discontinuous. (B) The plot of E against r is discontinuous. (a) Both A and B are correct. (b) A is correct but B is wrong. (c) B is correct but A is wrong. (d) Both A and B are wrong. 8. Let V and E represent the gravitational potential and field at a distance r from the centre of a uniform solid sphere. Consider the two statements: (A) the plot of V against r is discontinuous. (B) The plot of E against r is discontinuous. (a) Both A and B are correct. (b) A is correct but B is wrong. (c) B is correct but A is wrong. (d) Both A and B are wrong. 9. Take the effect of bulging of earth and its rotation in account. Consider the following statements: (A) There are points outside the earth where the value of g is equal to its value at the equator. (B) There are points outside the earth where the value of g is equal to its value at the poles.           Figure 11-Q2 224 Concepts of Physics

(a) Both A and B are correct. (b) A is correct but B is wrong. (c) B is correct but A is wrong. (d) Both A and B are wrong. 10. The time period of an earth-satellite in circular orbit is independent of (a) the mass of the satellite (b) radius of the orbit (c) none of them (d) both of them. 11. The magnitude of gravitational potential energy of the moon–earth system is U with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K. (a) U < K (b) U > K (c) U = K. 12. Figure (11-Q2) shows the elliptical path of a planet about the sun. The two shaded parts have equal area. If t1 and t2 be the time taken by the planet to go from a to b and from c to d respectively, (a) t1 < t2 (b) t1 = t2 (c) t1 > t2 (d) insufficient information to deduce the relation between t1 and t2 . 13. A person sitting in a chair in a satellite feels weightless because (a) the earth does not attract the objects in a satellite (b) the normal force by the chair on the person balances the earth’s attraction (c) the normal force is zero (d) the person in satellite is not accelerated. 14. A body is suspended from a spring balance kept in a satellite. The reading of the balance is W1 when the satellite goes in an orbit of radius R and is W2 when it goes in an orbit of radius 2 R. (a) W1 = W2 (b) W1 < W2 (c) W1 > W2 (d) W1 ≠ W2 . 15. The kinetic energy needed to project a body of mass m from the earth’s surface to infinity is (a) 1 4 mgR (b) 1 2 mgR (c) mgR (d) 2 mgR. 16. A particle is kept at rest at a distance R (earth’s radius) above the earth’s surface. The minimum speed with which it should be projected so that it does not return is (a) √GM 4R (b) √GM 2R (c) √GM R (d) √2GM R ⋅ 17. A satellite is orbiting the earth close to its surface. A particle is to be projected from the satellite to just escape from the earth. The escape speed from the earth is ve . Its speed with respect to the satellite (a) will be less than ve (b) will be more than ve (c) will be equal to ve (d) will depend on direction of projection. OBJECTIVE II 1. Let V and E denote the gravitational potential and gravitational field at a point. It is possible to have (a) V = 0 and E = 0 (b) V = 0 and E ≠ 0 (c) V ≠ 0 and E = 0 (d) V ≠ 0 and E ≠ 0. 2. Inside a uniform spherical shell (a) the gravitational potential is zero (b) the gravitational field is zero (c) the gravitational potential is same everywhere (d) the gravitational field is same everywhere. 3. A uniform spherical shell gradually shrinks maintaining its shape. The gravitational potential at the centre (a) increases (b) decreases (c) remains constant (d) oscillates. 4. Consider a planet moving in an elliptical orbit round the sun. The work done on the planet by the gravitational force of the sun (a) is zero in any small part of the orbit (b) is zero in some parts of the orbit (c) is zero in one complete revolution (d) is zero in no part of the motion. 5. Two satellites A and B move round the earth in the same orbit. The mass of B is twice the mass of A. (a) Speeds of A and B are equal. (b) The potential energy of earth+A is same as that of earth+B. (c) The kinetic energy of A and B are equal. (d) The total energy of earth+A is same as that of earth+B. 6. Which of the following quantities remain constant in a planetary motion (consider elliptical orbits) as seen from the sun ? (a) Speed (b) Angular speed (c) Kinetic Energy (d) Angular momentum. EXERCISES 1. Two spherical balls of mass 10 kg each are placed 10 cm apart. Find the gravitational force of attraction between them. 2. Four particles having masses m, 2m, 3m and 4m are placed at the four corners of a square of edge a. Find ! "  #  Figure 11-Q2 Gravitation 225

the gravitational force acting on a particle of mass m placed at the centre. 3. Three equal masses m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at (a) the mid-point of a side, (b) at the centre of the triangle. 4. Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two. 5. Four particles of equal masses M move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle. 6. Find the acceleration due to gravity of the moon at a point 1000 km above the moon’s surface. The mass of the moon is 7. 4 10 22 kg and its radius is 1740 km. 7. Two small bodies of masses 10 kg and 20 kg are kept a distance 1. 0 m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0. 5 m. 8. A semicircular wire has a length L and mass M. A particle of mass m is placed at the centre of the circle. Find the gravitational attraction on the particle due to the wire. 9. Derive an expression for the gravitational field due to a uniform rod of length L and mass M at a point on its perpendicular bisector at a distance d from the centre. 10. Two concentric spherical shells have masses M1 , M2 and radii R1, R2 R1 R2 . What is the force exerted by this system on a particle of mass m if it is placed at a distance R1 R2/2 from the centre ? 11. A tunnel is dug along a diameter of the earth. Find the force on a particle of mass m placed in the tunnel at a distance x from the centre. 12. A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth’s centre. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on a particle of mass m when it is at a distance x from the centre of the tunnel. 13. A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure (11-E1). A particle of mass m is placed on the line joining the two centres at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if (a) r < x < 2r, (b) 2r < x < 2R and (c) x > 2R. 14. A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a (figure 11-E2). The centre of the shell falls on the surface of the inner sphere. Find the gravitational field at the points P1 and P2 shown in the figure. 15. A thin spherical shell having uniform density is cut in two parts by a plane and kept separated as shown in figure (11-E3). The point A is the centre of the plane section of the first part and B is the centre of the plane section of the second part. Show that the gravitational field at A due to the first part is equal in magnitude to the gravitational field at B due to the second part. 16. Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2. 0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies ? The particle is placed at this point. What is the gravitational potential energy of the system of three particles with usual reference level ? 17. Three particles of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a. 18. A particle of mass 100 g is kept on the surface of a uniform sphere of mass 10 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle away from the sphere. 19. The gravitational field in a region is given by E 5 N kg1 i 12 N kg1 j . (a) Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin. (b) Find the potential at the points (12 m, 0) and (0, 5 m) if the potential at the origin is taken to be zero. (c) Find the change in gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12 m, 5 m). (d) Find the change in potential energy if the particle is taken from (12 m, 0) Figure 11-E1 to (0, 5 m).      Figure 11-E2   Figure 11-E3 226 Concepts of Physics

20. The gravitational potential in a region is given by V = 20 N kg–1 (x + y). (a) Show that the equation is dimensionally correct. (b) Find the gravitational field at the point (x, y). Leave your answer in terms of the unit vectors i , j , k . (c) Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin. 21. The gravitational field in a region is given by E 2 i 3 j N kg1 . Show that no work is done by the gravitational field when a particle is moved on the line 3 y + 2 x 5. [Hint : If a line y mx c makes angle with the X-axis, m tan.] 22. Find the height over the earth’s surface at which the weight of a body becomes half of its value at the surface. 23. What is the acceleration due to gravity on the top of Mount Everest ? Mount Everest is the highest mountain peak of the world at the height of 8848 m. The value at sea level is 9. 80 m s–2 . 24. Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9. 800 m s–2. The radius of the earth is 6400 km. 25. A body is weighed by a spring balance to be 1. 000 kg at the north pole. How much will it weigh at the equator ? Account for the earth’s rotation only. 26. A body stretches a spring by a particular length at the earth’s surface at equator. At what height above the south pole will it stretch the same spring by the same length ? Assume the earth to be spherical. 27. At what rate should the earth rotate so that the apparent g at the equator becomes zero ? What will be the length of the day in this situation ? 28. A pendulum having a bob of mass m is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is T0. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between T0 and the earth’s attraction on the bob. (c) If the ship sails at speed v, what is the tension in the string ? Angular speed of earth’s rotation is and radius of the earth is R. 29. The time taken by Mars to revolve round the sun is 1. 88 years. Find the ratio of average distance between Mars and the sun to that between the earth and the sun. 30. The moon takes about 27. 3 days to revolve round the earth in a nearly circular orbit of radius 3. 84 10 5 km. Calculate the mass of the earth from these data. 31. A Mars satellite moving in an orbit of radius 9. 4 10 3 km takes 27540 s to complete one revolution. Calculate the mass of Mars. 32. A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth’s surface. Find (a) its speed in the orbit, (b) its kinetic energy, (c) the potential energy of the earth–satellite system and (d) its time period. Mass of the earth 6 10 24 kg. 33. (a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth’s rotation. (b) If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth 6 10 24 kg. 34. What is the true weight of an object in a geostationary satellite that weighed exactly 10. 0 N at the north pole ? 35. The radius of a planet is R1 and a satellite revolves round it in a circle of radius R2 . The time period of revolution is T. Find the acceleration due to the gravitation of the planet at its surface. 36. Find the minimum colatitude which can directly receive a signal from a geostationary satellite. 37. A particle is fired vertically upward from earth’s surface and it goes up to a maximum height of 6400 km. Find the initial speed of particle. 38. A particle is fired vertically upward with a speed of 15 km s–1. With what speed will it move in intersteller space. Assume only earth’s gravitational field. 39. A mass of 6 10 24 kg (equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is 3 10 8 m s–1. What should be the radius of the sphere ? ANSWERS OBJECTIVE I 1. (b) 2. (c) 3. (c) 4. (d) 5. (a) 6. (c) 7. (c) 8. (d) 9. (b) 10. (a) 11. (b) 12. (b) 13. (c) 14. (a) 15. (c) 16. (c) 17. (d) OBJECTIVE II 1. all 2. (b), (c), (d) 3. (b) 4. (b), (c) 5. (a) 6. (d) EXERCISES 1. 6. 67 10 7 N 2. 42 Gm 2 a 2 3. (a) 4Gm 2 3a 2 , (b) zero 4. 3 GM 2 4a 2 Gravitation 227

5. GM R 22 1 4 6. 0. 65 m s2 7. 4. 2 10 5 m s1 and 2. 1 10 5 m s1 8. 2 GMm L 2 9. 2 Gm d L 2 4 d 2 10. 4GM1m R1 R2 2 11. GMe m R 3 x 12. GMe m 2R 2 13. (a) Gmmx r r 3 (b) Gmm x r 2 (c) GMm x R 2 Gmm x r 2 14. GM 16a 2 , 61 GM 900a 2 16. 0. 83 m from the 2.00 kg body towards the other body, 3. 06 10 10 J 17. 3Gm 2 2a 18. 6. 67 10 10 J 19. (a) 26 N (b) 60 J kg1 , 60 J kg1 (c) 240 J (d) zero 20. (b) 20 i j Nkg1 (c) 102 N 22. 2 1 times the radius of the earth 23. 9. 77 m s2 24. 9. 799 m s2 25. 0. 997 kg 26. 10 km approx. 27. 1. 237 10 3 rads–1, 1. 41 h 28. (a) R (b) m 2 R (c) T0 2 mv approx. 29. 1. 52 30. 6. 02 10 24 kg 31. 6. 5 10 23 kg 32. (a) 6. 90 kms–1 (b) 2. 38 10 10 J (c) 4. 76 10 10 J with usual reference (d) 2. 12 hours 33. (a) 42300 km (b) 6 hours 34. 0. 23 N 35. 4 2 R2 3 T 2 R1 2 36. sin 1 0. 15 37. 7. 9 km s–1 38. 10. 0 km s–1 39. 9 mm 228 Concepts of Physics

12.1 SIMPLE HARMONIC MOTION When a body repeats its motion after regular time intervals we say that it is in harmonic motion or periodic motion. The time interval after which the motion is repeated is called the time period. If a body moves to and fro on the same path, it is said to perform oscillations. Simple harmonic motion (SHM) is a special type of oscillation in which the particle oscillates on a straight line, the acceleration of the particle is always directed towards a fixed point on the line and its magnitude is proportional to the displacement of the particle from this point. This fixed point is called the centre of oscillation. Taking this point as the origin and the line of motion as the X-axis, we can write the defining equation of a simple harmonic motion as a = − ω2 x … (12.1) where ω2 is a positive constant. If x is positive, a is negative and if x is negative, a is positive. This means that the acceleration is always directed towards the centre of oscillation. If we are looking at the motion from an inertial frame, a = F/m. The defining equation (12.1) may thus be written as F/m = − ω2 x or, F = − mω2 x or, F = − kx. … (12.2) We can use equation (12.2) as the definition of SHM. A particle moving on a straight line executes simple harmonic motion if the resultant force acting on it is directed towards a fixed point on the line and is proportional to the displacement of the particle from this fixed point. The constant k = mω2 is called the force constant or spring constant. The resultant force on the particle is zero when it is at the centre of oscillation. The centre of oscillation is, therefore, the equilibrium position. A force which takes the particle back towards the equilibrium position is called a restoring force. Equation (12.2) represents a restoring force which is linear. Figure (12.1) shows the linear restoring force graphically. Example 12.1 The resultant force acting on a particle executing simple harmonic motion is 4 N when it is 5 cm away from the centre of oscillation. Find the spring constant. Solution : The simple harmonic motion is defined as F = – k x. The spring constant is k =    F x    = 4 N 5 cm = 4 N 5 × 10 − 2 m = 80 N m−1 . 12.2 QUALITATIVE NATURE OF SIMPLE HARMONIC MOTION Let us consider a small block of mass m placed on a smooth horizontal surface and attached to a fixed wall through a spring as shown in figure (12.2). Let the spring constant of the spring be k.

 Figure 12.1      Figure 12.2 CHAPTER 12 SIMPLE HARMONIC MOTION

The block is at a position O when the spring is at its natural length. Suppose the block is taken to a point P stretching the spring by the distance OP = A and is released from there. At any point on its path the displacement x of the particle is equal to the extension of the spring from its natural length. The resultant force on the particle is given by F = – kx and hence by definition the motion of the block is simple harmonic. When the block is released from P, the force acts towards the centre O. The block is accelerated in that direction. The force continues to act towards O until the block reaches O. The speed thus increases all the time from P to O. When the block reaches O, its speed is maximum and it is going towards left. As it moves towards left from O, the spring becomes compressed. The spring pushes the block towards right and hence its speed decreases. The block moves to a point Q when its speed becomes zero. The potential energy of the system (block + spring), when the block is at P, is 1 2 k (OP) 2 and when the block is at Q it is 1 2 k (OQ) 2 . Since the block is at rest at P as well as at Q, the kinetic energy is zero at both these positions. As we have assumed frictionless surface, principle of conservation of energy gives 1 2 k (OP) 2 = 1 2 k (OQ) 2 or, OP = OQ. The spring is now compressed and hence it pushes the block towards right. The block starts moving towards right, its speed increases upto O and then decreases to zero when it reaches P. Thus the particle oscillates between P and Q. As OP = OQ, it moves through equal distances on both sides of the centre of oscillation. The maximum displacement on either side from the centre of oscillation is called the amplitude. Example 12.2 A particle of mass 0. 50 kg executes a simple harmonic motion under a force F = – (50 N m−1 )x. If it crosses the centre of oscillation with a speed of 10 m s–1, find the amplitude of the motion. Solution : The kinetic energy of the particle when it is at the centre of oscillation is E = 1 2 m v 2 = 1 2 (0. 50 kg) (10 m s−1 ) 2 = 25 J. The potential energy is zero here. At the maximum displacement x = A, the speed is zero and hence the kinetic energy is zero. The potential energy here is 1 2 k A 2 . As there is no loss of energy, 1 2 k A 2 = 25 J. … (i) The force on the particle is given by F = – (50 N m−1 )x. Thus, the spring constant is k = 50 N m–1 . Equation (i) gives 1 2 (50 N m−1 ) A 2 = 25 J or, A = 1 m. 12.3 EQUATION OF MOTION OF A SIMPLE HARMONIC MOTION Consider a particle of mass m moving along the X-axis. Suppose, a force F = – kx acts on the particle where k is a positive constant and x is the displacement of the particle from the assumed origin. The particle then executes a simple harmonic motion with the centre of oscillation at the origin. We shall calculate the displacement x and the velocity v as a function of time. Suppose the position of the particle at t = 0 is x0 and its velocity is v0. Thus, at t = 0, x = x0 and v = v0. The acceleration of the particle at any instant is a = F m = − k m x = − ω 2 x where ω = √ km−1 . Thus, dv dt = − ω 2 x … (12.3) or, dv dx dx dt = − ω 2 x or, v dv dx = − ω 2 x or, vdv = − ω 2 x dx. The velocity of the particle is v0 when the particle is at x = x0. It becomes v when the displacement becomes x. We can integrate the above equation and write ∫ v0 v v dv = ∫ x0 x − ω 2 x dx         Figure 12.3 230 Concepts of Physics

or,    v 2 2    v0 v = − ω 2    x 2 2    x0 x or, v 2 − v0 2 = − ω 2 (x 2 − x0 2) or, v 2 = (v0 2 + ω 2 x0 2 − ω 2 x 2 ) or, v = √ (v0 2 + ω 2 x0 2) − ω 2 x 2 or, v = ω√    v0 2 ω 2 + x0 2   − x 2 . Writing    v0 ω    2 + x0 2 = A 2 … (12.4) the above equation becomes v = ω √ A  2 − x 2 . … (12.5) We can write this equation as dx dt = ω √ A  2 − x 2 or, dx √ A  2 − x 2 = ω dt. At time t = 0 the displacement is x = x0 and at time t the displacement becomes x. The above equation can be integrated as ∫ x0 x dx √ A 2 − x 2 = ∫ 0 t ω dt or,   sin − 1 x A   x0 x = [ωt] 0 t or, sin − 1 x A − sin − 1 x0 A = ωt. Writing sin − 1 x0 A = δ, this becomes sin − 1 x A = ωt + δ or, x = A sin(ωt + δ). … (12.6) The velocity at time t is v = dx dt = A ω cos(ωt + δ). … (12.7) 12.4 TERMS ASSOCIATED WITH SIMPLE HARMONIC MOTION (a) Amplitude Equation (12.6) gives the displacement of a particle in simple harmonic motion. As sin(ωt + δ) can take values between – 1 and + 1, the displacement x can take values between – A and + A. This gives the physical significance of the constant A. It is the maximum displacement of the particle from the centre of oscillation, i.e, the amplitude of oscillation. (b) Time Period A particle in simple harmonic motion repeats its motion after a regular time interval. Suppose the particle is at a position x and its velocity is v at a certain time t. After some time the position of the particle will again be x and its velocity will again be v in the same direction. This part of the motion is called one complete oscillation and the time taken in one complete oscillation is called the time period T. Thus, in figure (12.4) Q to P and then back to Q is a complete oscillation, R to P to Q to R is a complete oscillation, O to P to Q to O is a complete oscillation, etc. Both the position and the velocity (magnitude as well as direction) repeat after each complete oscillation. We have, x = A sin(ωt + δ). If T be the time period, x should have same value at t and t + T. Thus, sin(ωt + δ) = sin[ω(t + T) + δ]. Now the velocity is (equation 12.7) v = A ω cos(ωt + δ). As the velocity also repeats its value after a time period, cos(ωt + δ) = cos[ω(t + T) + δ]. Both sin(ωt + δ) and cos(ωt + δ) will repeat their values if the angle (ωt + δ) increases by 2π or its multiple. As T is the smallest time for repetition, ω(t + T) + δ = (ωt + δ) + 2π or, ω T = 2π or, T = 2π ω ⋅ Remembering that ω = √ km−1 , we can write for the time period, T = 2π ω = 2π √m k … (12.8) where k is the force constant and m is the mass of the particle. Example 12.3 A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by a spring of spring constant 80 N m–1. Find the time period. Solution : The time period is T = 2π √ m k   Figure 12.4 Simple Harmonic Motion 231

2 200 10 3 kg 80 N m1 2 0. 05 s 0. 31 s. (c) Frequency and Angular Frequency The reciprocal of time period is called the frequency. Physically, the frequency represents the number of oscillations per unit time. It is measured in cycles per second also known as hertz and written in symbols as Hz. Equation (12.8) shows that the frequency is 1 T 2 (12.9) 1 2 k m (12.10) The constant is called the angular frequency. (d) Phase The quantity t is called the phase. It determines the status of the particle in simple harmonic motion. If the phase is zero at a certain instant, x A sint 0 and v A cost A . This means that the particle is crossing the mean position and is going towards the positive direction. If the phase is /2, we get x A, v 0 so that the particle is at the positive extreme position. Figure (12.5) shows the status of the particle at different phases. We see that as time increases the phase increases. An increase of 2 brings the particle to the same status in the motion. Thus, a phase t is equivalent to a phase t 2. Similarly, a phase change of 4, 6, 8, , etc., are equivalent to no phase change. Figure (12.6) shows graphically the variation of position and velocity as a function of the phase. (e) Phase constant The constant appearing in equation (12.6) is called the phase constant. This constant depends on the choice of the instant t 0. To describe the motion quantitatively, a particular instant should be called t 0 and measurement of time should be made from this instant. This instant may be chosen according to the convenience of the problem. Suppose we choose t 0 at an instant when the particle is passing through its mean position and is going towards the positive direction. The phase t should then be zero. As t 0 this means will be zero. The equation for displacement can then be written as x A sint. If we choose t 0 at an instant when the particle is at its positive extreme position, the phase is /2 at this instant. Thus t /2 and hence /2. The equation for the displacement is x A sint /2 or, x A cost. Any instant can be chosen as t 0 and hence the phase constant can be chosen arbitrarily. Quite often we shall choose 0 and write the equation for displacement as x A sint. Sometimes we may have to consider two or more simple harmonic motions together. The phase constant of any one can be chosen as 0. The phase constants of the rest will be determined by the actual situation. The general equation for displacement may be written as x A sint A sin t 2 A cost where is another arbitrary constant. The sine form and the cosine form are, therefore, equivalent. The value of phase constant, however, depends on the form chosen. Example 12.4 A particle executes simple harmonic motion of amplitude A along the X-axis. At t 0, the position of the particle is x A/2 and it moves along the positive x-direction. Find the phase constant if the equation is written as x A sint . Solution : We have x A sint . At t 0, x A/2. Thus, A/2 A sin or, sin 1/2 or, /6 or 5/6. The velocity is v dx dt A cost . At t 0, v A cos. Now, cos 6 3 2 and cos 5 6 3 2 As v is positive at t 0, must be equal to /6.  Figure 12.5 x A A 0 v 0 v 0 0 v Figure 12.6 232 Concepts of Physics

12.5 SIMPLE HARMONIC MOTION AS A PROJECTION OF CIRCULAR MOTION Consider a particle P moving on a circle of radius A with a constant angular speed ω (figure 12.7). Let us take the centre of the circle as the origin and two perpendicular diameters as the X and Y-axes. Suppose the particle P is on the X-axis at t = 0. The radius OP will make an angle θ = ωt with the X-axis at time t. Drop perpendicular PQ on X-axis and PR on Y-axis. The x and y-coordinates of the particle at time t are x = OQ = OP cos ωt or, x = A cos ωt … (12.11) and y = OR = OP sin ωt or, y = A sin ωt. … (12.12) Equation (12.11) shows that the foot of perpendicular Q executes a simple harmonic motion on the X-axis. The amplitude is A and the angular frequency is ω. Similarly, equation (12.12) shows that the foot of perpendicular R executes a simple harmonic motion on the Y-axis. The amplitude is A and the angular frequency is ω. The phases of the two simple harmonic motions differ by π/2 [remember cosωt = sin(ωt+ π/2)]. Thus, the projection of a uniform circular motion on a diameter of the circle is a simple harmonic motion. 12.6 ENERGY CONSERVATION IN SIMPLE HARMONIC MOTION Simple harmonic motion is defined by the equation F = − kx. The work done by the force F during a displacement from x to x + dx is dW = F dx = − kx dx. The work done in a displacement from x = 0 to x is W = ∫ 0 x (− kx)dx = − 1 2 kx 2 . Let U(x) be the potential energy of the system when the displacement is x. As the change in potential energy corresponding to a force is negative of the work done by this force, U(x) − U(0) = − W = 1 2 kx 2 . Let us choose the potential energy to be zero when the particle is at the centre of oscillation x = 0. Then U(0) = 0 and U(x) = 1 2 kx 2 . This expression for potential energy is same as that for a spring and has been used so far in this chapter. As ω = √⎯⎯ k m , k = m ω 2 we can write U(x) = 1 2 m ω 2 x 2 . … (12.13) The displacement and the velocity of a particle executing a simple harmonic motion are given by x = A sin(ωt + δ) and v = A ω cos(ωt + δ). The potential energy at time t is, therefore, U = 1 2 m ω 2 x 2 = 1 2 m ω 2 A 2 sin 2 (ωt + δ), and the kinetic energy at time t is K = 1 2 m v 2 = 1 2 m A 2 ω 2 cos 2 (ωt + δ). The total mechanical energy at time t is E = U + K = 1 2 m ω 2 A 2 [sin 2 (ωt + δ) + (cos 2 (ωt + δ)] = 1 2 m ω 2 A 2 . … (12.14) We see that the total mechanical energy at time t is independent of t. Thus, the mechanical energy remains constant as expected. As an example, consider a small block of mass m placed on a smooth horizontal surface and attached to a fixed wall through a spring of spring constant k (figure 12.8).      Figure 12.7              Figure 12.8 Simple Harmonic Motion 233

When displaced from the mean position (where the spring has its natural length), the block executes a simple harmonic motion. The spring is the agency exerting a force F = − kx on the block. The potential energy of the system is the elastic potential energy stored in the spring. At the mean position x = 0, the potential energy is zero. The kinetic energy is 1 2 m v0 2 = 1 2 m ω 2 A 2 . All the mechanical energy is in the form of kinetic energy here. As the particle is displaced away from the mean position, the kinetic energy decreases and the potential energy increases. At the extreme positions x = ± A, the speed v is zero and the kinetic energy decreases to zero. The potential energy is increased to its maximum value 1 2 kA 2 = 1 2 m ω 2 A 2 . All the mechanical energy is in the form of potential energy here. Example 12.5 A particle of mass 40 g executes a simple harmonic motion of amplitude 2. 0 cm. If the time period is 0. 20 s, find the total mechanical energy of the system. Solution : The total mechanical energy of the system is E = 1 2 m ω 2 A 2 = 1 2 m   2π T    2 A 2 = 2π 2 m A 2 T 2 = 2 π 2 (40 × 10 − 3 kg) (2. 0 × 10 − 2 m) 2 (0. 20 s) 2 = 7. 9 × 10 − 3 J. 12.7 ANGULAR SIMPLE HARMONIC MOTION A body free to rotate about a given axis can make angular oscillations. For example, a hanging umbrella makes angular oscillations when it is slightly pushed aside and released. The angular oscillations are called angular simple harmonic motion if (a) there is a position of the body where the resultant torque on the body is zero, this position is the mean position θ = 0, (b) when the body is displaced through an angle from the mean position, a resultant torque acts which is proportional to the angle displaced, and (c) this torque has a sense (clockwise or anticlockwise) so as to bring the body towards the mean position. If the angular displacement of the body at an instant is θ, the resultant torque acting on the body in angular simple harmonic motion should be Γ = − k θ. If the moment of inertia is I, the angular acceleration is α = Γ I = − k I θ or, d 2 θ dt 2 = − ω 2 θ … (12.15) where ω = √kI−1 . Equation (12.15) is identical to equation (12.3) except for the symbols. The linear displacement x in (12.3) is replaced here by the angular displacement θ. Thus, equation (12.15) may be integrated in the similar manner and we shall get an equation similar to (12.6), i.e., θ = θ0 sin(ωt + δ) … (12.16) where θ0 is the maximum angular displacement on either side. The angular velocity at time t is given by, Ω = dθ dt = θ0 ω cos(ωt + δ). … (12.17) The time period of oscillation is T = 2 π ω = 2 π √I k … (12.18) and the frequency of oscillation is ν = 1 T = 1 2 π √ k I ⋅ … (12.19) The quantity ω = √kI −1 is the angular frequency. Example 12.6 A body makes angular simple harmonic motion of amplitude π/10 rad and time period 0. 05 s. If the body is at a displacement θ = π/10 rad at t = 0, write the equation giving the angular displacement as a function of time. Solution : Let the required equation be θ = θ 0 sin(ωt + δ). Here θ0 = amplitude = π 10 rad ω = 2 π T = 2 π 0. 05 s = 40 π s − 1 so that θ =    π 10 rad   sin     40 π s − 1  t + δ  . … (i) At t = 0, θ = π/10 rad. Putting in (i), π 10 =    π 10    sinδ or, sinδ = 1 or, δ = π/2. Thus by (i), θ =    π 10 rad   sin   (40 π s − 1)t + π 2    =    π 10 rad   cos[(40 π s − 1) t ]. 234 Concepts of Physics

Energy The potential energy is U = 1 2 k θ 2 = 1 2 I ω 2 θ 2 and the kinetic energy is K = 1 2 I Ω 2 . The total energy is E = U + K = 1 2 I ω2 θ 2 + 1 2 I Ω 2 . Using θ = θ0 sin(ωt + δ) E = 1 2 I ω 2 θ0 2 sin 2 (ωt + δ) + 1 2 I θ 0 2 ω 2 cos 2 (ωt + δ) = 1 2 I ω 2 θ0 2 . … (12.20) 12.8 SIMPLE PENDULUM A simple pendulum consists of a heavy particle suspended from a fixed support through a light inextensible string. Simple pendulum is an idealised model. In practice, one takes a small metallic sphere and suspends it through a string. Figure (12.9) shows a simple pendulum in which a particle of mass m is suspended from the fixed support O through a light string of length l. The system can stay in equilibrium if the string is vertical. This is the mean or equilibrium position. If the particle is pulled aside and released, it oscillates in a circular arc with the centre at the point of suspension O. The position of the particle at any time can be described by the angle θ between the string and the vertical. The mean position or the equilibrium position corresponds to θ = 0. The particle makes pure rotation about the horizontal line OA (figure 12.9) which is perpendicular to the plane of motion. Let us see whether the motion of the particle is simple harmonic or not and find out its time period of oscillation. Let the particle be at P at a time t when the string OP makes an angle θ with the vertical (figure 12.10). Let OQ be the horizontal line in the plane of motion. Let PQ be the perpendicular to OQ. Forces acting on the particle are (a) the weight mg and (b) the tension T. The torque of T about OA is zero as it intersects OA. The magnitude of the torque of mg about OA is  Γ  = (mg) (OQ) = mg (OP) sinθ = mgl sinθ. Also, the torque tries to bring the particle back towards θ = 0. Thus, we can write Γ = − mgl sinθ. … (12.21) We see that the resultant torque is not proportional to the angular displacement and hence the motion is not angular simple harmonic. However, if the angular displacement is small, sinθ is approximately equal to θ (expressed in radians) and equation (12.21) may be written as Γ = − mgl θ. … (12.22) Thus, if the amplitude of oscillation is small, the motion of the particle is approximately angular simple harmonic. The moment of inertia of the particle about the axis of rotation OA is I = m(OP) 2 = ml 2 . The angular acceleration is α = Γ I = − mgl θ ml 2 = − g l θ or, α = − ω 2 θ where ω = √gl −1 . This is the equation of an angular simple harmonic motion. The constant ω = √gl −1 represents the angular frequency. The time period is T = 2π ω = 2π √l/g . … (12.23) Example 12.7 Calculate the time period of a simple pendulum of length one meter. The acceleration due to gravity at the place is π 2 m s −2 .    Figure 12.9     Figure 12.10 Simple Harmonic Motion 235

Solution : The time period is T = 2π √lg−1 = 2π √1. 00 m π 2 m s −2 = 2. 0 s. Simple Pendulum as a Linear Simple Harmonic Oscillator If the amplitude of oscillation is small, the path of the particle is approximately a straight line and the motion can be described as a linear simple harmonic motion. We rederive expression (12.23) for the time period using this approach. Consider the situation shown in figure (12.11). Suppose the string makes an angle θ with the vertical at time t. The distance of the particle from the equilibrium position along the arc is x = lθ. The speed of the particle at time t is v = dx dt and the tangential acceleration is at = dv dt = d 2 x dt 2 ⋅ … (i) Forces acting on the particle are (a) the weight mg and (b) the tension T. The component of mg along the tangent to the path is −mgsinθ and that of T is zero. Thus, the total tangential force on the particle is −mgsinθ. Using (i) we get −mgsinθ = m d 2 x dt 2 or, d 2 x dt 2 = −g sin θ. … (ii) If the amplitude of oscillation is small, sinθ ≈ θ = xl −1 . Equation (ii) above thus becomes (for small oscillations) d 2 x dt 2 = − g l x or, d 2 x dt 2 = − ω 2 x where ω = √gl−1 . This equation represents a simple harmonic motion of the particle along the arc of the circle in which it moves. The angular frequency is ω = √gl−1 and the time period is T = 2π ω = 2π √lg−1 which is same as in equation (12.23). Determination of g in Laboratory A simple pendulum provides an easy method to measure the value of ‘g’ in a laboratory. A small spherical ball with a hook is suspended from a clamp through a light thread as shown in figure (12.12). The lengths AC and BD are measured with slide callipers. The length OA of the thread is measured with a meter scale. The effective length is OP = OA + AP = OA + AC − BD 2 ⋅ The bob is slightly pulled aside and gently released from rest. The pendulum starts making oscillations. The time for a number of oscillations (say 20 or 50) is measured with a stop watch and the time period is obtained. The value of g is calculated by equation (12.23). The length of the thread is varied and the experiment is repeated a number of times to minimise the effect of random errors. Example 12.8 In a laboratory experiment with simple pendulum it was found that it took 36 s to complete 20 oscillations when the effective length was kept at 80 cm. Calculate the acceleration due to gravity from these data. Solution : The time period of a simple pendulum is given by T = 2π √lg−1 or, g = 4π 2 l T 2 ⋅ … (i) In the experiment described in the question, the time period is T = 36 s 20 = 1. 8 s.  fl flffiffl    Figure 12.11   ff fi Figure 12.12 236 Concepts of Physics

Thus, by (i), g 4 2 0. 80 m 1. 8 s 2 9. 75 m s2 . 12.9 PHYSICAL PENDULUM Any rigid body suspended from a fixed support constitutes a physical pendulum. A circular ring suspended on a nail in a wall, a heavy metallic rod suspended through a hole in it, etc., are examples of physical pendulum. Figure (12.13) shows a physical pendulum. A rigid body is suspended through a hole at O. When the centre of mass C is vertically below O, the body may remain at rest. We call this position 0. When the body is pulled aside and released, it executes oscillations. The body rotates about a horizontal axis through O and perpendicular to the plane of motion. Let this axis be OA. Suppose the angular displacement of the body is at time t. The line OC makes an angle with the vertical at this instant. Forces on the body are (a) the weight mg and (b) the contact force N by the support at O. The torque of N about OA is zero as the force N acts through the point O. The torque of mg has magnitude mg OD mg OC sin mgl sin where l OC is the separation between the point of suspension and the centre of mass. This torque tries to bring the body back towards 0. Thus, we can write mglsin. If the moment of inertia of the body about OA is I, the angular acceleration becomes I mgl I sin. (i) We see that the angular acceleration is not proportional to the angular displacement and the motion is not strictly simple harmonic. However, for small displacements sin so that equation (i) becomes 2 where 2 mglI 1 . Thus, for small oscillations, the motion is nearly simple harmonic. The time period is T 2 2 I mgl (12.24) Example 12.9 A uniform rod of length 1. 00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation. Solution : For small amplitude the angular motion is nearly simple harmonic and the time period is given by T 2I mgl 2 mL 2 /3 mgL/2 2 2L 3g 2 2 1. 00 m 3 9. 80 m s2 1. 64 s. 12.10 TORSIONAL PENDULUM In torsional pendulum, an extended body is suspended by a light thread or a wire. The body is rotated through an angle about the wire as the axis of rotation (figure 12.14). The wire remains vertical during this motion but a twist is produced in the wire. The lower end of the wire is rotated through an angle with the body but the upper end remains fixed with the support. Thus, a twist is produced. The twisted wire exerts a restoring torque on the body to bring it back to its original position in which the twist in the wire is zero. This torque has a magnitude proportional to the angle of twist which is equal to the angle rotated by the body. The proportionality constant is called the torsional constant of the wire. Thus, if the torsional constant of the wire is k and the body is rotated through an angle , the torque produced is k. If I be the moment of inertia of the body about the vertical axis, the angular acceleration is I k I Figure 12.13 Figure 12.14 Simple Harmonic Motion 237

2 where k I Thus, the motion of the body is simple harmonic and the time period is T 2 2 I k (12.25) Example 12.10 A uniform disc of radius 5. 0 cm and mass 200 g is fixed at its centre to a metal wire, the other end of which is fixed with a clamp. The hanging disc is rotated about the wire through an angle and is released. If the disc makes torsional oscillations with time period 0. 20 s, find the torsional constant of the wire. Solution : The situation is shown in figure (12.15). The moment of inertia of the disc about the wire is I mr 2 2 0. 200 kg 5. 0 10 2 m 2 2 2. 5 10 4 kg m 2 . The time period is given by T 2 I k or, k 4 2 I T 2 4 2 2. 5 10 4 kg m 2 0. 20 s 2 0. 25 kg m 2 s 2 12.11 COMPOSITION OF TWO SIMPLE HARMONIC MOTIONS A simple harmonic motion is produced when a restoring force proportional to the displacement acts on a particle. If the particle is acted upon by two separate forces each of which can produce a simple harmonic motion, the resultant motion of the particle is a combination of two simple harmonic motions. Let r1 denote the position of the particle at time t if the force F1 alone acts on it. Similarly, let r2 denote the position at time t if the force F2 alone acts on it. Newton’s second law gives, m d 2 r1 d t 2 F1 and m d 2 r2 d t 2 F2 . Adding them, m d 2 r1 dt 2 m d 2 r2 dt 2 F1 F2 or, m d 2 dt 2 r1 r2 F1 F2 . (i) But F1 F2 is the resultant force acting on the particle and so the position r of the particle when both the forces act, is given by m d 2 r dt 2 F1 F2 . (ii) Comparing (i) and (ii) we can show that r r1 r2 and u u 1 u 2 if these conditions are met at t 0. Thus, if two forces F1 and F2 act together on a particle, its position at any instant can be obtained as follows. Assume that only the force F1 acts and find the position r1 at that instant. Then assume that only the force F2 acts and find the position r2 at that same instant. The actual position will be the vector sum of r1 and r2 . (A) Composition of two Simple Harmonic Motions in Same Direction Suppose two forces act on a particle, the first alone would produce a simple harmonic motion given by x1 A1 sin t and the second alone would produce a simple harmonic motion given by x2 A2 sint . Both the motions are along the x-direction. The amplitudes may be different and their phases differ by . Their frequency is assumed to be same. The resultant position of the particle is then given by x x1 x2 A1 sin t A2 sint A1 sin t A2 sin t cos A2 cos t sin A1 A2 cos sin t A2 sin cos t Figure 12.15 238 Concepts of Physics

C sint D cost C2 D2 C C 2 D 2 sint D C 2 D 2 cost (i) where C A1 A2 cos and D A2 sin. Now C C 2 D 2 and D C 2 D 2 both have magnitudes less than 1 and the sum of their squares is 1. Thus, we can find an angle between 0 and 2 such that sin D C 2 D 2 and cos C C 2 D 2 Equation (i) then becomes x C 2 D 2 cos sint sin cost or, x A sint (12.26) where A C 2 D 2 A1 A2 cos 2 A2 sin 2 A1 2 2A1 A2 cos A2 2 cos 2 A2 2 sin 2 A1 2 2 A1A2 cos A2 2 (12.27) and tan D C A2 sin A1 A2 cos (12.28) Equation (12.26) shows that the resultant of two simple harmonic motions along the same direction is itself a simple harmonic motion. The amplitude and phase of the resultant simple harmonic motion depend on the amplitudes of the two component simple harmonic motions as well as the phase difference between them. Amplitude of The Resultant Simple Harmonic Motion The amplitude of the resultant simple harmonic motion is given by equation (12.27), A A1 2 2 A1A2 cos A2 2 . If 0, the two simple harmonic motions are in phase A A1 2 2 A1A2 A2 2 A1 A2 . The amplitude of the resultant motion is equal to the sum of amplitudes of the individual motions. This is the maximum possible amplitude. If , the two simple harmonic motions are out of phase and A A1 2 2 A1A2 A2 2 A1 A2 or A2 A1. As the amplitude is always positive we can write A A1 A2 . If A1 A2 the resultant amplitude is zero and the particle does not oscillate at all. For any value of other than 0 and the resultant amplitude is between A1 A2 and A1 A2. Example 12.11 Find the amplitude of the simple harmonic motion obtained by combining the motions x1 2. 0 cm sint and x2 2. 0 cm sint /3. Solution : The two equations given represent simple harmonic motions along X-axis with amplitudes A1 2. 0 cm and A2 2. 0 cm. The phase difference between the two simple harmonic motions is /3. The resultant simple harmonic motion will have an amplitude A given by A A1 2 A2 2 2 A1A2 cos 2. 0 cm 2 2. 0 cm 2 2 2. 0 cm 2 cos 3 3. 5 cm. Vector Method of Combining Two Simple Harmonic Motions There is a very useful method to remember the equations of resultant simple harmonic motion when two simple harmonic motions of same frequency and in same direction combine. Suppose the two individual motions are represented by x1 A1 sint and x2 A2 sint . Let us for a moment represent the first simple harmonic motion by a vector of magnitude A1 and the second simple harmonic motion by another vector of magnitude A2. We draw these vectors in figure (12.16). The vector A2 is drawn at an angle with A1 to represent that the second simple harmonic motion has a phase difference of with the first simple harmonic motion. The resultant A of these two vectors will represent the resultant simple harmonic motion. As we know from vector algebra, the magnitude of the resultant vector is A A1 2 2 A1A2 cos A2 2 A A A 2 1 Figure 12.16 Simple Harmonic Motion 239