HC Verma

Block B is placed on the incline and the angle of the incline is gradually increased. The angle of the incline is so adjusted that the block just starts to slide. The height h and the horizontal distance D between the two ends of the plank are measured. The angle of incline θ satisfies tanθ = h/D. Let m be the mass of the block. The forces on the block in case of limiting equilibrium are (figure 6.14) (i) weight of the block mg, (ii) the normal contact force N , and (iii) the force of static friction fs. Taking components along the incline and applying Newton’s first law, fs = mg sinθ. Taking components along the normal to the incline, N = mg cosθ. Thus, the coefficient of static friction between the block and the plank is µs = fs N = mg sinθ mg cosθ = tanθ = h D ⋅ To obtain the kinetic friction, the inclination is reduced slightly and the block is made to move on the plank by gently pushing it with a finger. The inclination is so adjusted that once started, the block continues with uniform velocity on the plank. The height h′ and the distance D′ are noted. An identical analysis shows that the force of kinetic friction is fk = mg sinθ and the normal contact force is N = mg cosθ so that the coefficient of kinetic friction between the block and the plank is µk = fk N = tanθ = h′/D′. Example 6.4 A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually increased. It is found that the block starts slipping when the plank makes an angle of 18° with the horizontal. However, once started the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficients of static and kinetic friction between the block and the plank. Solution : The coefficient of static friction is µs = tan 18° and the coefficient of kinetic friction is µk = tan 15° Rolling Friction It is quite difficult to pull a heavy iron box on a rough floor. However, if the box is provided with four wheels, also made of iron, it becomes easier to move the box on the same floor. The wheel does not slide on the floor rather it rolls on the floor. The surfaces at contact do not rub each other. The velocity of the point of contact of the wheel with respect to the floor remains zero all the time although the centre of the wheel moves forward. The friction in the case of rolling is quite small as compared to kinetic friction. Quite often the rolling friction is negligible in comparison to the static or kinetic friction which may be present simultaneously. To reduce the wear and tear and energy loss against friction, small steel balls are kept between the rotating parts of machines which are known as ball bearings (figure 6.16). B h D Figure 6.13 fs mg D h N Figure 6.14 Figure 6.15 Fixed Part Rotating Part Ball Bearing Figure 6.16 90 Concepts of Physics

As one part moves with respect to the other, the balls roll on the two parts. No kinetic friction is involed and rolling friction being very small causes much less energy loss. Worked Out Examples 1. The coefficient of static friction between a block of mass m and an incline is µs = 0. 3. (a) What can be the maximum angle θ of the incline with the horizontal so that the block does not slip on the plane ? (b) If the incline makes an angle θ/2 with the horizontal, find the frictional force on the block. Solution : The situation is shown in figure (6-W1). (a) the forces on the block are (i) the weight mg downward by the earth, (ii) the normal contact force N by the incline, and (iii) the friction f parallel to the incline up the plane, by the incline. As the block is at rest, these forces should add up to zero. Also, since θ is the maximum angle to prevent slipping, this is a case of limiting equilibrium and so f = µs N . Taking components prependicular to the incline, N − mg cosθ = 0 or, N = mg cosθ. … (i) Taking components parallel to the incline, f − mg sinθ = 0 or, f = mg sinθ or, µs N = mg sinθ. … (ii) Dividing (ii) by (i) µs = tanθ or, θ = tan− 1 µs = tan− 1 (0. 3). (b) If the angle of incline is reduced to θ/2, the equilibrium is not limiting, and hence the force of static friction f is less than µs N . To know the value of f, we proceed as in part (a) and get the equations N = mg cos(θ/2) and f = mg sin(θ/2). Thus, the force of friction is mg sin(θ/2). 2. A horizontal force of 20 N is applied to a block of mass 4 kg resting on a rough horizontal table. If the block does not move on the table, how much frictional force the table is applying on the block ? What can be said about the coefficient of static friction between the block and the table ? Take g = 10 m/s 2 . Solution : The situation is shown in figure (6-W2). The forces on the block are (a) 4 kg × 10 m/s 2 = 40 N downward by the earth, (b) N upward by the table, (c) 20 N towards right by the experimenter and (d) f towards left by the table (friction). As the block is at rest, these forces should add up to zero. Taking horizontal and vertical components, f = 20 N and N = 40 N. Thus, the table exerts a frictional (static) force of 20 N on the block in the direction opposite to the applied force. Since it is a case of static friction, f ≤ µs N , or, µs ≥ f/N or, µs ≥ 0. 5. 3. The coefficient of static friction between the block of 2 kg and the table shown in figure (6-W3) is µs = 0. 2. What should be the maximum value of m so that the blocks do not move ? Take g = 10 m/s 2 . The string and the pulley are light and smooth. Solution : Consider the equilibrium of the block of mass m. The forces on this block are mg N f Figure 6-W1 40 N 20 N f N Figure 6-W2 mg m T T 20 N f 2 kg N Figure 6-W3 Friction 91

(a) mg downward by the earth and (b) T upward by the string. Hence, T − mg = 0 or, T = mg. … (i) Now consider the equilibrium of the 2 kg block. The forces on this block are (a) T towards right by the string, (b) f towards left (friction) by the table, (c) 20 N downward (weight) by the earth and (d) N upward (normal force) by the table. For vertical equilibrium of this block, N = 20 N. … (ii) As m is the largest mass which can be used without moving the system, the friction is limiting. Thus, f = µs N . … (iii) For horizontal equilibrium of the 2 kg block, f = T. … (iv) Using equations (i), (iii) and (iv) µs N = mg or, 0. 2 × 20 N = mg or, m = 0. 2 × 20 10 kg = 0. 4 kg. 4. The coefficient of static friction between the two blocks shown in figure (6-W4) is µ and the table is smooth. What maximum horizontal force F can be applied to the block of mass M so that the blocks move together ? Solution : When the maximum force F is applied, both the blocks move together towards right. The only horizontal force on the upper block of mass m is that due to the friction by the lower block of mass M. Hence this force on m should be towards right. The force of friction on M by m should be towards left by Newton’s third law. As we are talking of the maximum possible force F that can be applied, the friction is limiting and hence f = µ N , where N is the normal force between the blocks. Consider the motion of m. The forces on m are (figure 6-W5), (a) mg downward by the earth (gravity), (b) N upward by the block M (normal force) and (c) f = µ N (friction) towards right by the block M. In the vertical direction, there is no acceleration. This gives N = mg. … (i) In the horizontal direction, let the acceleration be a, then µ N = m a or, µ mg = ma or, a = µ g. … (ii) Next, consider the motion of M (figure 6-W6). The forces on M are (a) Mg downward by the earth (gravity), (b) N 1 upward by the table (normal force), (c) N downward by m (normal force), (d) f = µ N (friction) towards left by m and (e) F (applied force) by the experimenter. The equation of motion is F − µ N = M a or, F − µ mg = M µ g [Using (i) and (ii)] or, F = µ g (M + m). 5. A block slides down an incline of angle 30° with an acceleration g/4. Find the kinetic friction coeffcient. Solution : Let the mass of the block be m. The forces on the block are (Figure 6-W7), (a) mg downward by the earth (gravity), (b) N normal force by the incline and (c) f up the plane, (friction) by the incline. Taking components parallel to the incline and writing Newton’s second law, mg sin 30° − f = mg/4 or, f = mg/4. m M F Figure 6-W4 mg N f = N Figure 6-W5 Mg f = N F N N 1 Figure 6-W6 mg f 30° N Figure 6-W7 92 Concepts of Physics

There is no acceleration perpendicular to the incline. Hence, N = mg cos 30° = mg ⋅ √3 2 ⋅ As the block is slipping on the incline, friction is f = µk N . So, µk = f N = mg 4 mg √3/2 = 1 2 √3 ⋅ 6. A block of mass 2. 5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is found that it takes 5 seconds to slide through the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking g = 10 m/s 2 , calculate the coefficients of static and kinetic friction between the block and the surface. Solution : The forces acting on the block are shown in figure (6-W8). Here M = 2. 5 kg and F = 15 N. When F = 15 N is applied to the block, the block remains in limiting equilibrium. The force of friction is thus f = µs N . Applying Newton’s first law, f = µs N and N = mg so that F = µs Mg or, µs = F mg = 15 N (2. 5 kg) (10 m/s 2 ) = 0. 60. When the block is gently pushed to start the motion, kinetic friction acts between the block and the surface. Since the block takes 5 second to slide through the first 10 m, the acceleration a is given by 10 m = 1 2 a (5 s) 2 or, a = 20 25 m/s 2 = 0. 8 m/s 2 . The frictional force is f = µk N = µk Mg. Applying Newton’s second law F − µk Mg = Ma or, µk = F − Ma Mg = 15 N − (2. 5 kg) (0. 8 m/s 2 ) (2. 5 kg) (10 m/s 2 ) = 0. 52 . 7. A block placed on a horizontal surface is being pushed by a force F making an angle θ with the vertical. If the friction coefficient is µ , how much force is needed to get the block just started. Discuss the situation when tanθ < µ. Solution : The situation is shown in figure (6-W9). In the limiting equilibrium the frictional force f will be equal to µ N . For horizontal equilibrium F sinθ = µ N For vertical equilibrium F cosθ + mg = N . Eliminating N from these equations, F sinθ = µ F cosθ + µ mg or, F = µ mg sinθ − µ cosθ ⋅ If tanθ < µ we have (sinθ − µ cosθ) < 0 and then F is negative. So for angles less than tan− 1 µ, one cannot push the block ahead, however large the force may be. 8. Find the maximum value of M/m in the situation shown in figure (6-W10) so that the system remains at rest. Friction coefficient at both the contacts is µ . Discuss the situation when tanθ < µ. Solution : Figure (6-W11) shows the forces acting on the two blocks. As we are looking for the maximum value of M/m, the equilibrium is limiting. Hence, the frictional forces are equal to µ times the corresponding normal forces. F Mg s N N Figure 6-W8 mg f =s F N N Figure 6-W9 m M Figure 6-W10 mg Mg T T m M N1 N2 N1 N2 Figure 6-W11 Friction 93

Equilibrium of the block m gives T = µ N 1 and N1 = mg which gives T = µ mg. … (i) Next, consider the equilibrium of the block M. Taking components parallel to the incline T + µ N 2 = Mg sinθ. Taking components normal to the incline N 2 = Mg cosθ. These give T = Mg (sinθ − µ cosθ). … (ii) From (i) and (ii), µ mg = Mg (sinθ − µ cosθ) or, M/m = µ sinθ − µ cosθ If tanθ < µ , (sinθ − µ cosθ) < 0 and the system will not slide for any value of M/m. 9. Consider the situation shown in figure (6-W12). The horizontal surface below the bigger block is smooth. The coefficient of friction between the blocks is µ. Find the minimum and the maximum force F that can be applied in order to keep the smaller block at rest with respect to the bigger block. Solution : If no force is applied, the block A will slip on C towards right and the block B will move downward. Suppose the minimum force needed to prevent slipping is F. Taking A + B + C as the system, the only external horizontal force on the system is F. Hence the acceleration of the system is a = F M + 2m ⋅ … (i) Now take the block A as the system. The forces on A are (figure 6-W13), (i) tension T by the string towards right, (ii) friction f by the block C towards left, (iii) weight mg downward and (iv) normal force N upward. For vertical equilibrium N = mg. As the minimum force needed to prevent slipping is applied, the friction is limiting. Thus, f = µ N = µ mg. As the block moves towards right with an acceleration a, T − f = ma or, T − µ mg = ma. … (ii) Now take the block B as the system. The forces are (figure 6-W14), (i) tension T upward, (ii) weight mg downward, (iii) normal force N ′ towards right, and (iv) friction f ′ upward. As the block moves towards right with an acceleration a, N ′ = ma. As the friction is limiting, f ′ = µ N ′ = µ ma. For vertical equilibrium T + f ′ = mg or, T + µ ma = mg. … (iii) Eliminating T from (ii) and (iii) amin = 1 − µ 1 + µ g . When a large force is applied the block A slips on C towards left and the block B slips on C in the upward direction. The friction on A is towards right and that on B is downwards. Solving as above, the acceleration in this case is amax = 1 + µ 1 − µ g . Thus, a lies between 1 − µ 1 + µ g and 1 + µ 1 − µ g. From (i) the force F should be between 1 − µ 1 + µ (M + 2m) g and 1 + µ 1 − µ (M + 2m) g. 10. Figure (6-W15) shows two blocks connected by a light string placed on the two inclined parts of a triangular structure. The coefficients of static and kinetic friction are 0. 28 and 0. 25 respectively at each of the surfaces. (a) Find the minimum and maximum values of m for which the system remains at rest. (b) Find the acceleration of either block if m is given the minimum value calculated in the first part and is gently pushed up the incline for a short while. m m F A C B Figure 6-W12 f T mg N Figure 6-W13 T mg f N Figure 6-W14 m 2 kg 45° 45° Figure 6-W15 94 Concepts of Physics

Solution : (a) Take the 2 kg block as the system. The forces on this block are shown in figure (6-W16) with M = 2 kg. It is assumed that m has its minimum value so that the 2 kg block has a tendency to slip down. As the block is in equilibrium, the resultant force should be zero. Taking components ⊥ to the incline N = Mg cos 45° = Mg/√2. Taking components | | to the incline T + f = Mg sin 45° = Mg/√2 or, T = Mg/√2 − f. As it is a case of limiting equilibrium, f = μs N or, T = Mg √2 − μs Mg √2 = Mg √2 (1 − μs). … (i) Now consider the other block as the system. The forces acting on this block are shown in figure (6-W17). Taking components ⊥ to the incline, N ′ = mg cos 45° = mg/√2. Taking components | | to the incline T = mg sin 45° + f ′ = mg √2 + f ′. As it is the case of limiting equilibrium f ′ = μs N ′ = μs mg √2 ⋅ Thus, T = mg √2 (1 + μs). … (ii) From (i) and (ii) m(1 + μs) = M (1 − μs) … (iii) or, m = (1 − μs) (1 + μs) M = 1 − 0. 28 1 + 0. 28 × 2 kg = 9 8 kg. When maximum possible value of m is supplied, the directions of friction are reversed because m has the tendency to slip down and 2 kg block to slip up. Thus, the maximum value of m can be obtained from (iii) by putting μ = − 0. 28. Thus, the maximum value of m is m = 1 + 0. 28 1 − 0. 28 × 2 kg = 32 9 kg. (b) If m = 9/8 kg and the system is gently pushed, kinetic friction will operate. Thus, f = μk ⋅ Mg √2 and f ′ = μk mg √2 , where μk = 0. 25. If the acceleration is a, Newton’s second law for M gives (figure 6-W16). Mg sin 45° − T − f = Ma or, Mg √2 − T − μk Mg √2 = Ma. … (iv) Applying Newton’s second law m (figure 6-W17), T − mg sin 45° − f ′ = ma or, T − mg √2 − μk mg √2 = ma. … (v) Adding (iv) and (v) Mg √2 (1 − μk) − mg √2 (1 + μk) = (M + m) a or, a = M(1 − μk) − m (1 + μk) √2 (M + m) g = 2 × 0. 75 − 9/8 × 1. 25 √2 (2 + 9/8) g = 0. 31 m/s 2 . QUESTIONS FOR SHORT ANSWER 1. For most of the surfaces used in daily life, the friction coefficient is less than 1. Is it always necessary that the friction coefficient is less than 1 ? 2. Why is it easier to push a heavy block from behind than to press it on the top and push ? 3. What is the average friction force when a person has a usual 1 km walk ? 4. Why is it difficult to walk on solid ice ? 5. Can you accelerate a car on a frictionless horizontal road by putting more petrol in the engine ? Can you stop a car going on a frictionless horizontal road by applying brakes ? 6. Spring fitted doors close by themselves when released. You want to keep the door open for a long time, say for an hour. If you put a half kg stone in front of the open door, it does not help. The stone slides with the door and the door gets closed. However, if you sandwitch a   Figure 6-W16  N Figure 6-W17 Friction 95

20 g piece of wood in the small gap between the door and the floor, the door stays open. Explain why a much lighter piece of wood is able to keep the door open while the heavy stone fails. 7. A classroom demonstration of Newton’s first law is as follows : A glass is covered with a plastic card and a coin is placed on the card. The card is given a quick strike and the coin falls in the glass. (a) Should the friction coefficient between the card and the coin be small or large ? (b) Should the coin be light or heavy ? (c) Why does the experiment fail if the card is gently pushed ? 8. Can a tug of war be ever won on a frictionless surface ? 9. Why do tyres have a better grip of the road while going on a level road than while going on an incline ? 10. You are standing with your bag in your hands, on the ice in the middle of a pond. The ice is so slippery that it can offer no friction. How can you come out of the ice ? 11. When two surfaces are polished, the friction coefficient between them decreases. But the friction coefficient increases and becomes very large if the surfaces are made highly smooth. Explain. OBJECTIVE I 1. In a situation the contact force by a rough horizontal surface on a body placed on it has constant magnitude. If the angle between this force and the vertical is decreased, the frictional force between the surface and the body will (a) increase (b) decrease (c) remain the same (d) may increase or decrease. 2. While walking on ice, one should take small steps to avoid slipping. This is because smaller steps ensure (a) larger friction (b) smaller friction (c) larger normal force (d) smaller normal force. 3. A body of mass M is kept on a rough horizontal surface (friction coefficient = µ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on A is F, where (a) F = Mg (b) F = µ Mg (c) Mg ≤ F ≤ Mg √ 1 + µ 2 (d) Mg ≥ F ≥ Mg √ 1 − µ 2 . 4. A scooter starting from rest moves with a constant acceleration for a time ∆t1, then with a constant velocity for the next ∆t2 and finally with a constant deceleration for the next ∆t3 to come to rest. A 500 N man sitting on the scooter behind the driver manages to stay at rest with respect to the scooter without touching any other part. The force exerted by the seat on the man is (a) 500 N throughout the journey (b) less than 500 N throughout the journey (c) more than 500 N throughout the journey (d) > 500 N for time ∆t1 and ∆t3 and 500 N for ∆t2 . 5. Consider the situation shown in figure (6-Q1). The wall is smooth but the surfaces of A and B in contact are rough. The friction on B due to A in equilibrium (a) is upward (b) is downward (c) is zero (d) the system cannot remain in equilibrium. 6. Suppose all the surfaces in the previous problem are rough. The direction of friction on B due to A (a) is upward (b) is downward (c) is zero (d) depends on the masses of A and B. 7. Two cars of unequal masses use similar tyres. If they are moving at the same initial speed, the minimum stopping distance (a) is smaller for the heavier car (b) is smaller for the lighter car (c) is same for both cars (d) depends on the volume of the car. 8. In order to stop a car in shortest distance on a horizontal road, one should (a) apply the brakes very hard so that the wheels stop rotating (b) apply the brakes hard enough to just prevent slipping (c) pump the brakes (press and release) (d) shut the engine off and not apply brakes. 9. A block A kept on an inclined surface just begins to slide if the inclination is 30°. The block is replaced by another block B and it is found that it just begins to slide if the inclination is 40°. (a) mass of A > mass of B (b) mass of A < mass of B (c) mass of A = mass of B (d) all the three are possible. 10. A boy of mass M is applying a horizontal force to slide a box of mass M ′ on a rough horizontal surface. The coefficient of friction between the shoes of the boy and the floor is µ and that between the box and the floor is µ′. In which of the following cases it is certainly not possible to slide the box ? (a) µ < µ′ , M < M ′ (b) µ > µ′ , M < M ′ (c) µ < µ′ , M > M ′ (d) µ > µ′ , M > M ′. A B F Figure 6-Q1 96 Concepts of Physics

OBJECTIVE II 1. Let F, FN and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero, (a) F > FN (b) F > f (c) FN > f (d) FN − f < F < FN + f. 2. The contact force exerted by a body A on another body B is equal to the normal force between the bodies. We conclude that (a) the surfaces must be frictionless (b) the force of friction between the bodies is zero (c) the magnitude of normal force equals that of friction (d) the bodies may be rough but they don’t slip on each other. 3. Mark the correct statements about the friction between two bodies. (a) Static friction is always greater than the kinetic friction. (b) Coefficient of static friction is always greater than the coefficient of kinetic friction. (c) Limiting friction is always greater than the kinetic friction. (d) Limiting friction is never less than static friction. 4. A block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them. (a) The graph is a straight line of slope 45°. (b) The graph is a straight line parallel to the F-axis. (c) The graph is a straight line of slope 45° for small F and a straight line parallel to the F-axis for large F. (d) There is a small kink on the graph. 5. Consider a vehicle going on a horizontal road towards east. Neglect any force by the air. The frictional forces on the vehicle by the road (a) is towards east if the vehicle is accelerating (b) is zero if the vehicle is moving with a uniform velocity (c) must be towards east (d) must be towards west. EXERCISES 1. A body slipping on a rough horizontal plane moves with a deceleration of 4. 0 m/s 2. What is the coefficient of kinetic friction between the block and the plane ? 2. A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0. 10, how far will it travel before coming to rest ? 3. A block of mass m is kept on a horizontal table. If the static friction coefficient is µ, find the frictional force acting on the block. 4. A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two. 5. Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest ? The mass of the block is 4 kg. 6. A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0. 2. 7. Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline. 8. In a children-park an inclined plane is constructed with an angle of incline 45° in the middle part (figure 6-E1). Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0. 6 and g = 10 m/s 2 . 9. A body starts slipping down an incline and moves half meter in half second. How long will it take to move the next half meter ? 10. The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that, if λ be the angle of friction and µ the coefficient of static friction, λ ≤ tan− 1 µ. 11. Consider the situation shown in figure (6-E2). Calculate (a) the acceleration of the 1. 0 kg blocks, (b) the tension in the string connecting the 1. 0 kg blocks and (c) the tension in the string attached to 0. 50 kg. 45° Figure 6-E1 0.5 kg 1.0 kg 1.0 kg Figure 6-E2 Friction 97

12. If the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0. 5 m/s 2, find the friction coefficients at the two contacts with the blocks. 13. The friction coefficient between the table and the block shown in figure (6-E4) is 0. 2. Find the tensions in the two strings. 14. The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m. 15. The friction coefficient between an athelete’s shoes and the ground is 0. 90. Suppose a superman wears these shoes and races for 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time that he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop ? 16. A car is going at a speed of 21. 6 km/hr when it encounters a 12. 8 m long slope of angle 30° (figure 6-E5). The friction coefficient between the road and the tyre is 1/2√3. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take g = 10 m/s 2 . 17. A car starts from rest on a half kilometer long bridge. The coefficient of friction between the tyre and the road is 1. 0. Show that one cannot drive through the bridge in less than 10 s. 18. Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2. 0 kg and the incline is µ1, and that between the block of mass 4. 0 kg and the incline is µ2. Calculate the acceleration of the 2. 0 kg block if (a) µ1 = 0. 20 and µ2 = 0. 30, (b) µ1 = 0. 30 and µ2 = 0. 20. Take g = 10 m/s 2 . 19. Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is µ. Find the acceleration of the system and the force by the rod on one of the blocks. 20. A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is µ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block ? In which direction should this force act ? 21. The friction coefficient between the board and the floor shown in figure (6-E7) is µ. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor. 22. A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0. 20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take g = 10 m/s 2 . 23. Find the accelerations a1 , a2 , a3 of the three blocks shown in figure (6-E8) if a horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take g = 10 m/s 2 . 24. The friction coefficient between the two blocks shown in figure (6-E9) is µ but the floor is smooth. (a) What maximum horizontal force F can be applied without disturbing the equilibrium of the system ? (b) Suppose the horizontal force applied is double of that found in part (a). Find the accelerations of the two masses. 2 kg 4 kg 30° 2 1 Figure 6-E3 15 kg 5 kg 5 kg Figure 6-E4 30° Figure 6-E5 30° 4 kg 2 kg Figure 6-E6 M m Figure 6-E7 a a a 1 2 3 1 2 3 2 kg 3 kg 7 kg Figure 6-E8 98 Concepts of Physics

25. Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b). 26. Consider the situation shown in figure (6-E9). Suppose a small electric field E exists in the space in the vertically upward direction and the upper block carries a positive charge Q on its top surface. The friction coefficient between the two blocks is but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium ? [Hint : The force on a charge Q by the electric field E is F QE in the direction of E.] 27. A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient of friction between the block and the table is . The table does not move on the floor. Find the total frictional force applied by the floor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table ? 28. Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the two blocks is 1 and that between the bigger block and the ground is 2. 29. A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of static friction being 0. 5. Another horizontal force of 15 N, is applied on the block in a direction parallel to the wall. Will the block move ? If yes, in which direction ? If no, find the frictional force exerted by the wall on the block. 30. A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back (figure 6-E11). Assume that the friction coefficient between his body and the walls is 0. 8 and that limiting friction acts at all the contacts. (a) Show that the person pushes the two walls with equal force. (b) Find the normal force exerted by either wall on the person. Take g 10 m/s 2 . 31. Figure (6-E12) shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is and that between the block is /2. Find the time elapsed before the smaller blocks separates from the bigger block. ANSWERS OBJECTIVE I 1. (b) 2. (b) 3. (c) 4. (d) 5. (d) 6. (a) 7. (c) 8. (b) 9. (d) 10. (a) OBJECTIVE II 1. (a), (b), (d) 2. (b), (d) 3. (b), (c), (d) 4. (c), (d) 5. (a), (b) EXERCISES 1. 0. 4 2. 50 m 3. zero 4. 0. 11 5. 10 m 6. (a) 13 N (b) zero   Figure 6-E9  Figure 6-E10 Figure 6-E11 m M Figure 6-E12 Friction 99

7. 17. 5 N 8. 2√2 m/s 2 9. 0. 21 s 11. (a) 0. 4 m/s 2 (b) 2. 4 N (c) 4. 8 N 12. µ1 = 0. 75, µ2 = 0. 06 13. 96 N in the left string and 68 N in the right 14. 16° 15. (a) 10 3 s (b) 10 3 s 18. 2. 7 m/s 2, 2. 4 m/s 2 19. a = g (sin θ − µ cos θ), zero 20. µ mg √ 1 + µ 2 at an angle tan − 1 µ with the horizontal 21. µ (M + m) g 1 + µ 22. (a) upper block 4 m/s 2, lower block 1 m/s 2 (b) both blocks 2 m/s 2 23. (a) a1 = 3 m/s 2 , a2 = a3 = 0. 4 m/s 2 (b) a1 = a2 = a3 = 5 6 m/s 2 (c) same as (b) 24. (a) 2 µ mg (b) 2 µ mg M + m in opposite directions 25. (a) 2 µ m (g − a) (b) 2 µ m (g − a) m + M 26. 2µ (mg − QE) 27. µ mg 28. [2 m − µ2 (M + m)] g M + m [5 + 2 (µ1 − µ2)] 29. it will move at an angle of 53° with the 15 N force 30. (b) 250 N 31. √ 4 M l (M + m) µ g 100 Concepts of Physics

7.1 ANGULAR VARIABLES Suppose a particle P is moving in a circle of radius r (figure 7.1). Let O be the centre of the circle. Let O be the origin and OX the X-axis. The position of the particle P at a given instant may be described by the angle θ between OP and OX. We call θ the angular position of the particle. As the particle moves on the circle, its angular position θ changes. Suppose the particle goes to a nearby point P′ in time ∆t so that θ increases to θ + ∆θ. The rate of change of angular position is called angular velocity. Thus, the angular velocity is ω = lim ∆t → 0 ∆θ ∆t = dθ dt ⋅ The rate of change of angular velocity is called angular acceleration. Thus, the angular acceleration is α = dω dt = d 2 θ dt 2 ⋅ If the angular acceleration α is constant, we have θ = ω0 t + 1 2 α t 2 … (7.1) ω = ω0 + α t … (7.2) and ω 2 = ω0 2 + 2α θ … (7.3) where ω0 and ω are the angular velocities at t = 0 and at time t and θ is the angular position at time t. The linear distance PP′ travelled by the particle in time ∆t is ∆s = r ∆θ or, ∆s ∆t = r ∆θ ∆t or, v = r ω … (7.4) where v is the linear speed of the particle. Differentiating equation (7.4) with respect to time, the rate of change of speed is at = dv dt = r dω dt or, at = r α. … (7.5) Remember that at = dv dt is the rate of change of speed and is not the rate of the change of velocity. It is, therefore, not equal to the net acceleration. We shall show that at is the component of acceleration along the tangent and hence we have used the suffix t. It is called the tangential acceleration. Example 7.1 A particle moves in a circle of radius 20 cm with a linear speed of 10 m/s. Find the angular velocity. Solution : The angular velocity is ω = v r = 10 m/s 20 cm = 50 rad/s. Example 7.2 A particle travels in a circle of radius 20 cm at a speed that uniformly increases. If the speed changes from 5. 0 m/s to 6. 0 m/s in 2. 0 s, find the angular acceleration. Solution : The tangential acceleration is given by at = dv dt = v2 − v1 t2 − t1 = 6. 0 − 5. 0 2. 0 m/s 2 = 0. 5 m/s 2 . The angular acceleration is α = at /r = 0. 5 m/s 2 20 cm = 2. 5 rad/s 2 . P P O X Figure 7.1 CHAPTER 7 CIRCULAR MOTION

7.2 UNIT VECTORS ALONG THE RADIUS AND THE TANGENT Consider a particle moving in a circle. Suppose the particle is at a point P in the circle at a given instant (figure 7.2). Take the centre of the circle to be the origin, a line OX as the X-axis and a perpendicular radius OY as the Y-axis. The angular position of the particle at this instant is θ. Draw a unit vector PA → = er → along the outward radius and a unit vector PB → = et → along the tangent in the direction of increasing θ. We call er → the radial unit vector and et → the tangential unit vector. Draw PX ′ parallel to the X-axis and PY ′ parallel to the Y-axis. From the figure, PA → = i → PA cosθ + j → PA sinθ or, er → = i → cosθ + j → sinθ, … (7.6) where i → and j → are the unit vectors along the X and Y axes respectively. Similarly, PB → = − i → PB sinθ + j → PB cosθ or, et → = − i → sinθ + j → cosθ. … (7.7) 7.3 ACCELERATION IN CIRCULAR MOTION Consider the situation shown in figure (7.2). It is clear from the figure that the position vector of the particle at time t is r → = OP → = OP er → = r (i → cosθ + j → sinθ). … (i) Differentiating equation (i) with respect to time, the velocity of the particle at time t is v →= dr → dt = d dt [r (i → cosθ + j → sinθ)] = r    i →    − sinθ dθ dt    + j →    cosθ dθ dt       = r ω[− i → sinθ + j → cosθ]. … (ii) The term r ω is the speed of the particle at time t (equation 7.4) and the vector in the square bracket is the unit vector e → t along the tangent. Thus, the velocity of the particle at any instant is along the tangent to the circle and its magnitude is v = r ω. The acceleration of the particle at time t is a →= dv → dt ⋅ From (ii), a →= r    ω d dt [− i → sinθ + j → cosθ] + dω dt [− i → sinθ + j → cosθ]    = ωr    − i → cosθ dθ dt − j → sinθ dθ dt    + r dω dt et → = − ω 2 r [i → cosθ + j → sinθ] + r dω dt et → = − ω 2 r er → + dv dt et → , … (7.8) where er → and et → are the unit vectors along the radial and tangential directions respectively and v is the speed of the particle at time t. We have used r dω dt = d dt (r ω) = dv dt ⋅ Uniform Circular Motion If the particle moves in the circle with a uniform speed, we call it a uniform circular motion. In this case, dv dt = 0 and equation (7.8) gives a →= − ω 2 r er → . Thus, the acceleration of the particle is in the direction of − e → r, that is, towards the centre. The magnitude of the acceleration is ar = ω 2 r = v 2 r 2 r = v 2 r ⋅ … (7.9) Thus, if a particle moves in a circle of radius r with a constant speed v, its acceleration is v 2 /r directed towards the centre. This acceleration is called centripetal acceleration. Note that the speed remains constant, the direction continuously changes and hence the “velocity” changes and there is an acceleration during the motion. Example 7.3 Find the magnitude of the linear acceleration of a particle moving in a circle of radius 10 cm with uniform speed completing the circle in 4 s. Solution : The distance covered in completing the circle is 2π r = 2π × 10 cm. The linear speed is v = 2π r/t P e e O X X Y Y A B r t Figure 7.2 102 Concepts of Physics

= 2π × 10 cm 4 s = 5π cm/s. The linear acceleration is a = v 2 r = (5π cm/s) 2 10 cm = 2. 5 π 2 cm/s 2 . This acceleration is directed towards the centre of the circle. Nonuniform Circular Motion If the speed of the particle moving in a circle is not constant, the acceleration has both the radial and the tangential components. According to equation (7.8), the radial and the tangential accelerations are and ar = − ω 2 r = − v 2 /r at = dv dt      ⋅ … (7.10) Thus, the component of the acceleration towards the centre is ω 2 r = v 2 /r and the component along the tangent (along the direction of motion) is dv/dt. The magnitude of the acceleration is a = √ ar 2 + at 2 = √    v 2 r    2 +    dv dt    2 ⋅ The direction of this resultant acceleration makes an angle α with the radius (figure 7.3) where tanα =    dv dt    /    v 2 r    ⋅ Example 7.4 A particle moves in a circle of radius 20 cm. Its linear speed is given by v = 2 t, where t is in second and v in metre/second. Find the radial and tangential acceleration at t = 3 s. Solution : The linear speed at t = 3 s is v = 2 t = 6 m/s. The radial acceleration at t = 3 s is ar = v 2 /r = 36 m2 /s 2 0. 20 m = 180 m/s 2 . The tangential acceleration is at = dv dt = d (2t) dt = 2 m/s 2 . 7.4 DYNAMICS OF CIRCULAR MOTION If a particle moves in a circle as seen from an inertial frame, a resultant nonzero force must act on the particle. That is because a particle moving in a circle is accelerated and acceleration can be produced in an inertial frame only if a resultant force acts on it. If the speed of the particle remains constant, the acceleration of the particle is towards the centre and its magnitude is v 2 /r. Here v is the speed of the particle and r is the radius of the circle. The resultant force must act towards the centre and its magnitude F must satisfy a = F m or, v 2 r = F m or, F = mv 2 r ⋅ … (7.11) Since this resultant force is directed towards the centre, it is called centripetal force. Thus, a centripetal force of magnitude mv 2/r is needed to keep the particle in uniform circular motion. It should be clearly understood that “centripetal force” is another word for “force towards the centre”. This force must originate from some external source such as gravitation, tension, friction, coulomb force, etc. Centripetal force is not a new kind of force, just as an “upward force” or a “downward force” is not a new kind of force. Example 7.5 A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical side walls, of radius 25 cm. If the block takes 2. 0 s to complete one round, find the normal contact force by the side wall of the groove. Solution : The speed of the block is v = 2π × (25 cm) 2. 0 s = 0. 785 m/s . The acceleration of the block is a = v 2 r = (0. 785 m/s) 2 0. 25 m = 2. 5 m/s 2 towards the centre. The only force in this direction is the normal contact force due to the side walls. Thus, from Newton’s second law, this force is N = ma = (0. 100 kg) (2. 5 m/s 2) = 0. 25 N. 2 a O v r dv dt Figure 7.3 Circular Motion 103

7.5 CIRCULAR TURNINGS AND BANKING OF ROADS When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which will produce the required acceleration. If the vehicle goes in a horizontal circular path, this resultant force is also horizontal. Consider the situation as shown in figure (7.4). A vehicle of mass M moving at a speed v is making a turn on the circular path of radius r. The external forces acting on the vehicle are (i) weight Mg (ii) Normal contact force N and (iii) friction fs. If the road is horizontal, the normal force N is vertically upward. The only horizontal force that can act towards the centre is the friction fs. This is static friction and is self adjustable. The tyres get a tendency to skid outward and the frictional force which opposes this skidding acts towards the centre. Thus, for a safe turn we must have v 2 r fs M or, fs Mv 2 r However, there is a limit to the magnitude of the frictional force. If s is the coefficient of static friction between the tyres and the road, the magnitude of friction fs cannot exceed sN. For vertical equilibrium N Mg, so that fs s Mg. Thus, for a safe turn Mv 2 r s Mg or, s v 2 rg (7.12) Friction is not always reliable at circular turns if high speeds and sharp turns are involved. To avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is somewhat lifted up as compared to the inner part (figure 7.5). The surface of the road makes an angle with the horizontal throughout the turn. The normal force N makes an angle with the vertical. At the correct speed, the horizontal component of N is sufficient to produce the acceleration towards the centre and the self adjustable frictional force keeps its value zero. Applying Newton’s second law along the radius and the first law in the vertical direction, N sin Mv 2 r and N cos Mg. These equations give tan v 2 rg (7.13) The angle depends on the speed of the vehicle as well as on the radius of the turn. Roads are banked for the average expected speed of the vehicles. If the speed of a particular vehicle is a little less or a little more than the correct speed, the self adjustable static friction operates between the tyres and the road and the vehicle does not skid or slip. If the speed is too different from that given by equation (7.13), even the maximum friction cannot prevent a skid or a slip. Example 7.6 The road at a circular turn of radius 10 m is banked by an angle of 10. With what speed should a vehicle move on the turn so that the normal contact force is able to provide the necessary centripetal force ? Solution : If v is the correct speed, tan v 2 rg Figure 7.4 W W cos sin Figure 7.5 104 Concepts of Physics

or, v rg tan 10 m 9 . 8 m/s 2 tan 10 4. 2 m/s. 7.6 CENTRIFUGAL FORCE We discussed in chapter 5 that Newton’s laws of motion are not valid if one is working from a noninertial frame. If the frame translates with respect to an inertial frame with an acceleration a0 , one must assume the existence of a pseudo force ma0 , acting on a particle of mass m. Once this pseudo force is included, one can use Newton’s laws in their usual form. What pseudo force is needed if the frame of reference rotates at a constant angular velocity with respect to an inertial frame ? Suppose the observer is sitting in a closed cabin which is made to rotate about the vertical Z-axis at a uniform angular velocity (figure 7.6). The X and Y axes are fixed in the cabin. Consider a heavy box of mass m kept on the floor at a distance r from the Z-axis. Suppose the floor and the box are rough and the box does not slip on the floor as the cabin rotates. The box is at rest with respect to the cabin and hence is rotating with respect to the ground at an angular velocity . Let us first analyse the motion of the box from the ground frame. In this frame (which is inertial) the box is moving in a circle of radius r. It, therefore, has an acceleration v 2 /r 2 r towards the centre. The resultant force on the box must be towards the centre and its magnitude must be m2 r. The forces on the box are (a) weight mg (b) normal force N by the floor (c) friction f by the floor. Figure (7.6b) shows the free body diagram for the box. Since the resultant is towards the centre and its magnitude is m2 r, we should have f m2 r. The floor exerts a force of static friction f m2 r towards the origin. Now consider the same box when observed from the frame of the rotating cabin. The observer there finds that the box is at rest. If he or she applies Newton’s laws, the resultant force on the box should be zero. The weight and the normal contact force balance each other but the frictional force f m2 r acts on the box towards the origin. To make the resultant zero, a pseudo force must be assumed which acts on the box away from the centre (radially outward) and has a magnitude m2 r. This pseudo force is called the centrifugal force. The analysis from the rotating frame is as follows : The forces on the box are (a) weight mg (b) normal force N (c) friction f (d) centrifugal force m2 r. The free body diagram is shown in figure (7.6c). As the box is at rest, Newton’s first law gives f m2 r. Note that we get the same equation for friction as we got from the ground frame. But we had to apply Newton’s second law from the ground frame and Newton’s first law from the rotating frame. Let us now summarise our discussion. Suppose we are working from a frame of reference that is rotating at a constant angular velocity with respect to an inertial frame. If we analyse the dynamics of a particle of mass m kept at a distance r from the axis of rotation, we have to assume that a force m2 r acts radially outward on the particle. Only then we can apply Newton’s laws of motion in the rotating frame. This radially outward pseudo force is called the centrifugal force. In fact, centrifugal force is a sufficient pseudo force, only if we are analysing the particles at rest in a uniformly rotating frame. If we analyse the motion of a particle that moves in the rotating frame, we may have to assume other pseudo forces, together with the centrifugal force. Such forces are called the coriolis forces. The coriolis force is perpendicular to the velocity of the particle and also perpendicular to the axis of rotation of the frame. Once again, we emphasise that all these pseudo forces, centrifugal or coriolis, are needed only if the working frame is rotating. If we work from an inertial frame, there is no need to apply any pseudo force. It is a common misconception among the beginners that centrifugal force acts on a particle because the f mg f r mg Z (a) (b) (c) m r 2 Figure 7.6 Circular Motion 105

particle goes on a circle. Centrifugal force acts (or is assumed to act) because we describe the particle from a rotating frame which is noninertial and still use Newton’s laws. 7.7 EFFECT OF EARTH’S ROTATION ON APPARENT WEIGHT The earth rotates about its axis at an angular speed of one revolution per 24 hours. The line joining the north and the south poles is the axis of rotation. Every point on the earth moves in a circle. A point at equator moves in a circle of radius equal to the radius of the earth and the centre of the circle is same as the centre of the earth. For any other point on the earth, the circle of rotation is smaller than this. Consider a place P on the earth (figure 7.7). Drop a perpendicular PC from P to the axis SN. The place P rotates in a circle with the centre at C. The radius of this circle is CP. The angle between the axis SN and the radius OP through P is called the colatitude of the place P. We have CP OP sin or, r R sin where R is the radius of the earth. If we work from the frame of reference of the earth, we shall have to assume the existence of the pseudo forces. In particular, a centrifugal force m2 r has to be assumed on any particle of mass m placed at P. Here is the angular speed of the earth. If we discuss the equilibrium of bodies at rest in the earth’s frame, no other pseudo force is needed. Consider a heavy particle of mass m suspended through a string from the ceiling of a laboratory at colatitude (figure 7.8). Looking from the earth’s frame the particle is in equilibrium and the forces on it are (a) gravitational attraction mg towards the centre of the earth, i.e., vertically downward, (b) centrifugal force m2 r towards CP and (c) the tension in the string T along the string. As the particle is in equilibrium (in the frame of earth), the three forces on the particle should add up to zero. The resultant of mg and m2 r mg 2 m2 r 2 2mg m2 r cos90 m g 2 4 R 2 sin2 2g2 R sin2 mg where g g 2 2 R sin2 2g 2 R. (7.14) Also, the direction of this resultant makes an angle with the vertical OP, where tan m2 r sin90 mg m2 r cos90 2 R sin cos g 2 R sin2 (7.15) As the three forces acting on the particle must add up to zero, the force of tension must be equal and opposite to the resultant of the rest two. Thus, the magnitude of the tension in the string must be mg and the direction of the string should make an angle with the true vertical. The direction of g is the apparent vertical direction, because a plumb line stays in this direction only. The walls of the buildings are constructed by making them parallel to g and not to g. The water surface placed at rest is perpendicular to g. The magnitude of g is also different from g. As 2g > 2 R, it is clear from equation (7.14) that g < g. One way of measuring the weight of a body is to suspend it by a string and find the tension in the string. The tension itself is taken as a measure of the weight. As T mg, the weight so observed is less than the true weight mg. This is known as the apparent weight. Similarly, if a person stands on the platform of a weighing machine, the platform exerts a normal      Figure 7.7         Figure 7.8 106 Concepts of Physics

force N which is equal to mg′. The reading of the machine responds to the force exerted on it and hence the weight recorded is the apparent weight mg′. At equator, θ = 90° and equation (7.14) gives g′ = √ g  2 − 2gω 2 R + ω 4 R 2 = g − ω 2 R or, mg′ = mg − mω 2 R. … (7.16) This can be obtained in a more straightforward way. At the equator, mω 2 R is directly opposite to mg and the resultant is simply mg – mω 2 R. Also, this resultant is towards the centre of the earth so that at the equator the plumb line stands along the true vertical. At poles, θ = 0 and equation (7.14) gives g′ = g and equation (7.15) shows that α = 0. Thus, there is no apparent change in g at the poles. This is because the poles themselves do not rotate and hence the effect of earth’s rotation is not felt there. Example 7.7 A body weighs 98 N on a spring balance at the north pole. What will be its weight recorded on the same scale if it is shifted to the equator ? Use g = GM/R 2 = 9. 8 m/s 2 and the radius of the earth R = 6400 km. Solution : At poles, the apparent weight is same as the true weight. Thus, 98 N = mg = m(9. 8 m/s 2 ) or, m = 10 kg. At the equator, the apparent weight is mg′ = mg − mω 2 R. The radius of the earth is 6400 km and the angular speed is ω = 2π rad 24 × 60 × 60 s = 7. 27 × 10 − 5 rad/s. Thus, mg′ = 98 N − (10 kg) (7. 27 × 10 − 5 s − 1) 2 (6400 km) = 97. 66 N. Worked Out Examples 1. A car has to move on a level turn of radius 45 m. If the coefficient of static friction between the tyre and the road is µs = 2.0, find the maximum speed the car can take without skidding. Solution : Let the mass of the car be M. The forces on the car are (a) weight Mg downward (b) normal force N by the road upward (c) friction fs by the road towards the centre. The car is going on a horizontal circle of radius R, so it is accelerating. The acceleration is towards the centre and its magnitude is v 2 /R, where v is the speed. For vertical direction, acceleration = 0. Resolving the forces in vertical and horizontal directions and applying Newton’s laws, we have N = mg and fs = Mv 2 /R. As we are looking for the maximum speed for no skidding, it is a case of limiting friction and hence fs = µsN = µsMg. So, we have µsMg = Mv 2 /R or, v 2 = µs gR. Putting the values, v = √ 2 × 10 m/s  2 × 45 m = 30 m/s = 108 km/hr. 2. A circular track of radius 600 m is to be designed for cars at an average speed of 180 km/hr. What should be the angle of banking of the track ? Solution : Let the angle of banking be θ. The forces on the car are (figure 7-W1) (a) weight of the car Mg downward and (b) normal force N . For proper banking, static frictional force is not needed. For vertical direction the acceleration is zero. So, N cosθ = Mg. … (i) For horizontal direction, the acceleration is v 2 /r towards the centre, so that N sinθ = Mv 2 /r. … (ii) N Figure 7-W1 Circular Motion 107

From (i) and (ii), tan v 2 /rg. Putting the values, tan 180 km/hr 2 600 m 10 m/s 2 0. 4167 or, 22. 6. 3. A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r. Find (a) the speed of the particle and (b) the tension in the string. Such a system is called a conical pendulum. Solution : The situation is shown in figure (7-W2). The angle made by the string with the vertical is given by sin r/L. (i) The forces on the particle are (a) the tension T along the string and (b) the weight mg vertically downward. The particle is moving in a circle with a constant speed v. Thus, the radial acceleration towards the centre has magnitude v 2 /r. Resolving the forces along the radial direction and applying Newton’s second law, T sin m v 2 /r. (ii) As there is no acceleration in vertical direction, we have from Newton’s first law, T cos mg. (iii) Dividing (ii) by (iii), tan v 2 rg or, v rg tan . And from (iii), T mg cos Using (i), v rg L 2 r 2 1/4 and T mgL L 2 r 2 1/2 4. One end of a massless spring of spring constant 100 N/m and natural length 0. 5 m is fixed and the other end is connected to a particle of mass 0. 5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring. Solution : The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude v 2 /r. The horizontal force on the particle is due to the spring and equals kl, where l is the elongation and k is the spring constant. Thus, kl mv 2 /r m 2 r m 2 l0 l. Here is the angular velocity, l0 is the natural length (0. 5 m) and l0 + l is the total length of the spring which is also the radius of the circle along which the particle moves. Thus, k m 2 l m 2 l0 or, l m 2 l0 k m 2 Putting the values, l 0. 5 4 0. 5 100 0. 5 4 m 1 100 m 1 cm. 5. A simple pendulum is constructed by attaching a bob of mass m to a string of length L fixed at its upper end. The bob oscillates in a vertical circle. It is found that the speed of the bob is v when the string makes an angle with the vertical. Find the tension in the string at this instant. Solution : The forces acting on the bob are (figure 7-W3) (a) the tension T (b) the weight mg. As the bob moves in a vertical circle with centre at O, the radial acceleration is v 2 /L towards O. Taking the components along this radius and applying Newton’s second law, we get, T mg cos mv 2 /L or, T m g cos v 2 /L. 6. A cylindrical bucket filled with water is whirled around in a vertical circle of radius r. What can be the minimum speed at the top of the path if water does not fall out from the bucket ? If it continues with this speed, what normal contact force the bucket exerts on water at the lowest point of the path ?    Figure 7-W2 Figure 7-W3 108 Concepts of Physics

Solution : Consider water as the system. At the top of the circle its acceleration towards the centre is vertically downward with magnitude v 2 /r. The forces on water are (figure 7-W4) (a) weight Mg downward and (b) normal force by the bucket, also downward. So, from Newton’s second law Mg N Mv 2 /r. For water not to fall out from the bucket, N 0. Hence, Mv 2 /r Mg or, v 2 rg. The minimum speed at the top must be rg . If the bucket continues on the circle with this minimum speed rg , the forces at the bottom of the path are (a) weight Mg downward and (b) normal contact force N by the bucket upward. The acceleration is towards the centre which is vertically upward, so N Mg Mv 2 /r or, N Mg v 2 /r 2 Mg. 7. A fighter plane is pulling out for a dive at a speed of 900 km/hr. Assuming its path to be a vertical circle of radius 2000 m and its mass to be 16000 kg, find the force exerted by the air on it at the lowest point. Take g 9. 8 m/s 2 . Solution : At the lowest point in the path the acceleration is vertically upward (towards the centre) and its magnitude is v 2 /r. The forces on the plane are (a) weight Mg downward and (b) force F by the air upward. Hence, Newton’s second law of motion gives F Mg Mv 2 /r or, F Mg v 2 /r. Here v 900 km/hr 9 10 5 3600 m/s 250 m/s or, F 16000 9. 8 62500 2000 N 6. 56 10 5 N upward. 8. Figure (7-W5) shows a rod of length 20 cm pivoted near an end and which is made to rotate in a horizontal plane with a constant angular speed. A ball of mass m is suspended by a string also of length 20 cm from the other end of the rod. If the angle made by the string with the vertical is 30, find the angular speed of the rotation. Take g 10 m/s 2 . Solution : Let the angular speed be . As is clear from the figure, the ball moves in a horizontal circle of radius L L sin, where L 20 cm. Its acceleration is, therefore, 2 L L sin towards the centre. The forces on the bob are (figure 7-W5) (a) the tension T along the string and (b) the weight mg. Resolving the forces along the radius and applying Newton’s second law, T sin m 2 L 1 sin. (i) Applying Newton’s first law in the vertical direction, T cos mg. (ii) Dividing (i) by (ii), tan 2 L1 sin g or, 2 g tan L1 sin 10 m/s 2 1/3 0. 20 m 1 1/2 or, 4. 4 rad/s. 9. Two blocks each of mass M are connected to the ends of a light frame as shown in figure (7-W6). The frame is rotated about the vertical line of symmetry. The rod breaks if the tension in it exceeds T0 . Find the maximum frequency with which the frame may be rotated without breaking the rod. Figure 7-W4    Figure 7-W5 M M Figure 7-W6 Circular Motion 109

Solution : Consider one of the blocks. If the frequency of revolution is f, the angular velocity is ω = 2π f. The acceleration towards the centre is v 2 /l = ω 2 l = 4π 2 f 2 l. The only horizontal force on the block is the tension of the rod. At the point of breaking, this force is T0. So from Newton’s second law, T0 = M ⋅ 4π 2 f 2 l or, f = 1 2π    T0 Ml    1/2 ⋅ 10. In a rotor, a hollow vertical cylindrical structure rotates about its axis and a person rests against the inner wall. At a particular speed of the rotor, the floor below the person is removed and the person hangs resting against the wall without any floor. If the radius of the rotor is 2 m and the coefficient of static friction between the wall and the person is 0. 2, find the minimum speed at which the floor may be removed. Take g = 10 m/s 2 . Solution : The situation is shown in figure (7-W7). When the floor is removed, the forces on the person are (a) weight mg downward (b) normal force N due to the wall, towards the centre (c) frictional force fs , parallel to the wall, upward. The person is moving in a circle with a uniform speed, so its acceleration is v 2 /r towards the centre. Newton’s law for the horizontal direction (2nd law) and for the vertical direction (1st law) give N = mv 2 /r … (i) and fs = mg. … (ii) For the minimum speed when the floor may be removed, the friction is limiting one and so equals µs N . This gives µs N = mg or, µsmv 2 r = mg [using (i)] or, v = √rg µs = √ 2 m × 10 m/s 2 0. 2 = 10 m/s. 11. A hemispherical bowl of radius R is set rotating about its axis of symmetry which is kept vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface. If the surface of the bowl is smooth, and the angle made by the radius through the block with the vertical is θ, find the angular speed at which the bowl is rotating. Solution : Suppose the angular speed of rotation of the bowl is ω. The block also moves with this angular speed. The forces on the block are (figure 7-W8) (a) the normal force N and (b) the weight mg. The block moves in a horizontal circle with the centre at C, so that the radius is PC = OP sinθ = R sinθ. Its acceleration is, therefore, ω 2 R sinθ. Resolving the forces along PC and applying Newton’s second law, N sinθ = mω 2 R sinθ or, N = mω 2 R. … (i) As there is no vertical acceleration, N cosθ = mg. … (ii) Dividing (i) by (ii), 1 cosθ = ω 2 R g ⋅ or, ω = √g R cosθ ⋅ 12. A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v. Find the tension in the ring. Solution : Consider a small part ACB of the ring that subtends an angle ∆θ at the centre as shown in figure (7-W9). Let the tension in the ring be T. The forces on this small part ACB are (a) tension T by the part of the ring left to A, (b) tension T by the part of the ring right to B, mg fs N Figure 7-W7 P mg C O N Figure 7-W8 O A C B T T Figure 7-W9 110 Concepts of Physics

(c) weight mg and (d) normal force N by the table. The tension at A acts along the tangent at A and the tension at B acts along the tangent at B. As the small part ACB moves in a circle of radius R at a constant speed v, its acceleration is towards the centre (along CO) and has a magnitude mv 2 /R. Resolving the forces along the radius CO, T cos 90 2 T cos 90 2 m v 2 R or, 2T sin 2 m v 2 R (i) The length of the part ACB is R. As the total mass of the ring is m, the mass of the part ACB will be m m 2R R m 2 Putting m in (i), 2T sin 2 m 2 v 2 R or, T mv 2 2 R /2 sin/2 As is very small, /2 sin/2 1 and T mv 2 2 R 13. A table with smooth horizontal surface is turning at an angular speed about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the particle as its distance from the centre becomes L. Solution : The situation is shown in figure (7-W10). Let us work from the frame of reference of the table. Let us take the origin at the centre of rotation O and the X-axis along the groove (figure 7-W10). The Y-axis is along the line perpendicular to OX coplanar with the surface of the table and the Z-axis is along the vertical. Suppose at time t the particle in the groove is at a distance x from the origin and is moving along the X-axis with a speed v. The forces acting on the particle (including the pseudo forces that we must assume because we have taken our frame on the table which is rotating and is noninertial) are (a) weight mg vertically downward, (b) normal contact force N 1 vertically upward by the bottom surface of the groove, (c) normal contact force N 2 parallel to the Y-axis by the side walls of the groove, (d) centrifugal force m 2 x along the X-axis, and (e) coriolis force along Y-axis (coriolis force is perpendicular to the velocity of the particle and the axis of rotation.) As the particle can only move in the groove, its acceleration is along the X-axis. The only force along the X-axis is the centrifugal force m 2 x. All the other forces are perpendicular to the X-axis and have no components along the X-axis. Thus, the acceleration along the X-axis is a F m m 2 x m 2 x or, dv dt 2 x or, dv dx dx dt 2 x or, dv dx v 2 x or, v dv 2 x dx or, 0 v vdv a L 2 x dx or, 1 2 v 2 0 v 1 2 2 x 2 a L or, v 2 2 1 2 2 L2 a 2 or, v L2 a 2 . QUESTIONS FOR SHORT ANSWER 1. You are driving a motorcycle on a horizontal road. It is moving with a uniform velocity. Is it possible to accelerate the motorcyle without putting higher petrol input rate into the engine ? O x X Y Z m x2 Figure 7-W10 Circular Motion 111

2. Some washing machines have cloth driers. It contains a drum in which wet clothes are kept. As the drum rotates, the water particles get separated from the cloth. The general description of this action is that “the centrifugal force throws the water particles away from the drum”. Comment on this statement from the viewpoint of an observer rotating with the drum and the observer who is washing the clothes. 3. A small coin is placed on a record rotating at 331 3 rev/minute. The coin does not slip on the record. Where does it get the required centripetal force from ? 4. A bird while flying takes a left turn, where does it get the centripetal force from ? 5. Is it necessary to express all angles in radian while using the equation 0 t ? 6. After a good meal at a party you wash your hands and find that you have forgotten to bring your handkerchief. You shake your hands vigorously to remove the water as much as you can. Why is water removed in this process ? 7. A smooth block loosely fits in a circular tube placed on a horizontal surface. The block moves in a uniform circular motion along the tube (figure 7-Q1). Which wall (inner or outer) will exert a nonzero normal contact force on the block ? 8. Consider the circular motion of the earth around the sun. Which of the following statements is more appropriate ? (a) Gravitational attraction of the sun on the earth is equal to the centripetal force. (b) Gravitational attraction of the sun on the earth is the centripetal force. 9. A car driver going at some speed v suddenly finds a wide wall at a distance r. Should he apply brakes or turn the car in a circle of radius r to avoid hitting the wall ? 10. A heavy mass m is hanging from a string in equilibrium without breaking it. When this same mass is set into oscillation, the string breaks. Explain. OBJECTIVE I 1. When a particle moves in a circle with a uniform speed (a) its velocity and acceleration are both constant (b) its velocity is constant but the acceleration changes (c) its acceleration is constant but the velocity changes (d) its velocity and acceleration both change. 2. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the circle in equal time, the ratio of their angular speeds 1 /2 is (a) m1 /m2 (b) r1 /r2 (c) m1r1 /m2 r2 (d) 1. 3. A car moves at a constant speed on a road as shown in figure (7-Q2). The normal force by the road on the car is NA and NB when it is at the points A and B respectively. (a) NA NB (b) NA > NB (c) NA < NB (d) insufficient information to decide the relation of NA and NB . 4. A particle of mass m is observed from an inertial frame of reference and is found to move in a circle of radius r with a uniform speed v. The centrifugal force on it is (a) mv 2 r towards the centre (b) mv 2 r away from the centre (c) mv 2 r along the tangent through the particle (d) zero. 5. A particle of mass m rotates about the Z-axis in a circle of radius a with a uniform angular speed . It is viewed from a frame rotating about the Z-axis with a uniform angular speed 0. The centrifugal force on the particle is (a) m 2 a (b) m 0 2 a (c) m 0 2 2 a (d) m 0 a. 6. A particle is kept fixed on a turntable rotating uniformly. As seen from the ground the particle goes in a circle, its speed is 20 cm/s and acceleration is 20 cm/s 2 . The particle is now shifted to a new position to make the radius half of the original value. The new values of the speed and acceleration will be (a) 10 cm/s, 10 cm/s 2 (b) 10 cm/s, 80 cm/s 2 (c) 40 cm/s, 10 cm/s 2 (d) 40 cm/s, 40 cm/s 2 . 7. Water in a bucket is whirled in a vertical circle with a string attached to it. The water does not fall down even when the bucket is inverted at the top of its path. We conclude that in this position Figure 7-Q1  Figure 7-Q2 112 Concepts of Physics

(a) mg = mv 2 r (b) mg is greater than mv 2 r (c) mg is not greater than mv 2 r (d) mg is not less than mv 2 r ⋅ 8. A stone of mass m tied to a string of length l is rotated in a circle with the other end of the string as the centre. The speed of the stone is v. If the string breaks, the stone will move (a) towards the centre (b) away from the centre (c) along a tangent (d) will stop. 9. A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of (a) 1 cm (b) 2 cm (c) 4 cm (d) 8 cm. 10. A motorcyle is going on an overbridge of radius R. The driver maintains a constant speed. As the motorcycle is ascending on the overbridge, the normal force on it (a) increases (b) decreases (c) remains the same (d) fluctuates. 11. Three identical cars A, B and C are moving at the same speed on three bridges. The car A goes on a plane bridge, B on a bridge convex upward and C goes on a bridge concave upward. Let FA , FB and FC be the normal forces exerted by the cars on the bridges when they are at the middle of bridges. (a) FA is maximum of the three forces. (b) FB is maximum of the three forces. (c) FC is maximum of the three forces. (d) FA = FB = FC . 12. A train A runs from east to west and another train B of the same mass runs from west to east at the same speed along the equator. A presses the track with a force F1 and B presses the track with a force F2 . (a) F1 > F2 . (b) F1 < F2 . (c) F1 = F2 . (d) the information is insufficient to find the relation between F1 and F2 . 13. If the earth stops rotating, the apparent value of g on its surface will (a) increase everywhere (b) decrease everywhere (c) remain the same everywhere (d) increase at some places and remain the same at some other places. 14. A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T1 and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted ends. (a) T1 > T2 . (b) T2 > T1 . (c) T1 = T2 . (d) The relation between T1 and T2 depends on whether the rod rotates clockwise or anticlockwise. 15. A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and lands at some distance. Let R be the maximum height of the car from the top of the cliff. The tension in the string when the car is in air is (a) mg (b) mg − mv 2 R (c) mg + mv 2 R (d) zero. 16. Let θ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is m, the tension in the string is mg cosθ (a) always (b) never (c) at the extreme positions (d) at the mean position. OBJECTIVE II 1. An object follows a curved path. The following quantities may remain constant during the motion (a) speed (b) velocity (c) acceleration (d) magnitude of acceleration. 2. Assume that the earth goes round the sun in a circular orbit with a constant speed of 30 km/s. (a) The average velocity of the earth from 1st Jan, 90 to 30th June, 90 is zero. (b) The average acceleration during the above period is 60 km/s 2 . (c) The average speed from 1st Jan, 90 to 31st Dec, 90 is zero. (d) The instantaneous acceleration of the earth points towards the sun. 3. The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. Its (a) velocity remains constant (b) speed remains constant (c) acceleration remains constant (d) tangential acceleration remains constant. 4. A particle is going in a spiral path as shown in figure (7-Q3) with constant speed. (a) The velocity of the particle is constant. (b) The acceleration of the particle is constant. Figure 7-Q3 Circular Motion 113

(c) The magnitude of acceleration is constant. (d) The magnitude of acceleration is decreasing continuously. 5. A car of mass M is moving on a horizontal circular path of radius r. At an instant its speed is v and is increasing at a rate a. (a) The acceleration of the car is towards the centre of the path. (b) The magnitude of the frictional force on the car is greater than mv 2 r ⋅ (c) The friction coefficient between the ground and the car is not less than a/g. (d) The friction coefficient between the ground and the car is µ = tan– 1 v 2 rg ⋅ 6. A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible. (a) The car cannot make a turn without skidding. (b) If the car turns at a speed less than 40 km/hr, it will slip down. (c) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is equal to mv 2 r ⋅ (d) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than mv 2 r ⋅ 7. A person applies a constant force F → on a particle of mass m and finds that the particle moves in a circle of radius r with a uniform speed v as seen from an inertial frame of reference. (a) This is not possible. (b) There are other forces on the particle. (c) The resultant of the other forces is mv 2 r towards the centre. (d) The resultant of the other forces varies in magnitude as well as in direction. EXERCISES 1. Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and the moon = 3. 85 × 10 5 km and the time taken by the moon to complete one revolution around the earth = 27. 3 days. 2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth’s rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis. 3. A particle moves in a circle of radius 1. 0 cm at a speed given by v = 2. 0 t where v is in cm/s and t in seconds. (a) Find the radial acceleration of the particle at t = 1 s. (b) Find the tangential acceleration at t = 1 s. (c) Find the magnitude of the acceleration at t = 1 s. 4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible ? 5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking ? 6. A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking ? 7. If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that a scooter going at 18 km/hr does not skid ? 8. A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used ? 9. In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron goes round the proton in a circle of radius 5. 3 × 10 – 11 m. Find the speed of the electron in the ground state. Mass of the electron = 9. 1 × 10 – 31 kg and charge of the electron = 1. 6 × 10 – 19 C. 10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed the stone can have at the highest point of the circle. 11. A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed ? Who exerts this force on the particle ? How much force does the particle exert on the blade along its surface ? 12. A mosquito is sitting on an L.P. record disc rotating on a turn table at 331 3 revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than π 2 /81. Take g = 10 m/s 2 . 13. A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by the string of the pendulum with the vertical if this angle does not change during the turn. Take g = 10 m/s 2 . 114 Concepts of Physics

14. The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1. 4 m/s at the lowest point in its path. Find the tension in the string at this instant. 15. Suppose the bob of the previous problem has a speed of 1. 4 m/s when the string makes an angle of 0. 20 radian with the vertical. Find the tension at this instant. You can use cosθ ≈ 1 − θ 2 /2 and sinθ ≈ θ for small θ. 16. Suppose the amplitude of a simple pendulum having a bob of mass m is θ0 . Find the tension in the string when the bob is at its extreme position. 17. A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight ? (b) If the speed of earth’s rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case ? 18. A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0. 4, what are the possible speeds of a vehicle so that it neither slips down nor skids up ? 19. A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular arc of radius R and has a total length L. Suppose the motorcycle starts from the highest point. (a) What can its maximum velocity be for which the contact with the road is not broken at the highest point ? (b) If the motorcycle goes at speed 1/√2 times the maximum found in part (a), where will it lose the contact with the road ? (c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge ? 20. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv dt = a. The friction coefficient between the road and the tyre is µ. Find the speed at which the car will skid. 21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is µ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip ? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip ? 22. A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track. (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed ? Take g = 10 m/s 2 . 23. In a children’s park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (figure 7-E2). Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3. 0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids. 24. A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is µ. Find the range of the angular speed for which the block will not slip. 25. A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle ? This radius is called the radius of curvature of the curve at the point. 26. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ/2 with the horizontal ? 27. A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is µ. The block is given an initial speed v0. As a function of the speed v write (a) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the A B C D E Figure 7-E1 Figure 7-E2 Circular Motion 115

tangential acceleration    dv dt = v dv ds    to obtain the speed of the block after one revolution. 28. A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in a circular path of radius R (figure 7-E3). A smooth groove AB of length L( << R) is made on the surface of the table. The groove makes an angle θ with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B. 29. A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road (figure 7-E4). A small block of mass 100 g is kept on the seat which rests against the plate. The friction coefficient between the block and the plate is µ = 0. 58. (a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate. 30. A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R (figure 7-E5). A smooth pulley of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the strings along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string. ANSWERS OBJECTIVE I 1. (d) 2. (d) 3. (c) 4. (d) 5. (b) 6. (a) 7. (c) 8. (c) 9. (a) 10. (a) 11. (c) 12. (a) 13. (d) 14. (a) 15. (d) 16. (c) OBJECTIVE II 1. (a), (d) 2. (d) 3. (b), (d) 4. (c) 5. (b), (c) 6. (b), (d) 7. (b), (d) EXERCISES 1. 2. 73 × 10 − 3 m/s 2 2. 0. 0336 m/s 2 3. (a) 4. 0 cm/s 2 (b) 2. 0 cm/s 2 (c) √20 cm/s 2 4. 500 N 5. tan – 1 (1/3) 6. tan – 1 (1/4) 7. 0. 25 8. 17 m/s 9. 2. 2 × 10 6 m/s 10. √Rg 11. 14. 8 N, 14. 8 N 13. 45° 14. 1. 2 N A B O R Figure 7-E3 O Figure 7-E4 m m 1 2 Figure 7-E5 116 Concepts of Physics

15. 1. 16 N 16. mg cos0 17. (a) 3. 5 10 3 (b) 2. 0 hour 18. Between 14. 7 km/h and 54 km/hr 19. (a) Rg, (b) a distance R/3 along the bridge from the highest point, (c) gRcosL/2 R 20. 2 g 2 a 2 R 2 1/4 21. (a) g/L (b) g L 2 2 1/4 22. (a) 975 N, 1025 N (b) 0, 707 N, 0 (c) 682 N, 732 N (d) 1. 037 23. 10 2 N 24. gsin cos R sincos sin 1/2 to gsin cos R sincos sin 1/2 25. u 2 cos 2 g 26. u 2 cos 2 g cos 3 /2 27. (a) mv 2 R (b) mv 2 R (c) v 2 R (d) v0e 2 28. 2 L 2 R cos 29. (a) 0. 2 N (b) 30 30. 2 R 3 , 4 3 m 2 R Circular Motion 117

8.1 KINETIC ENERGY A dancing, running man is said to be more energetic compared to a sleeping snoring man. In physics, a moving particle is said to have more energy than an identical particle at rest. Quantitatively the energy of the moving particle (over and above its energy at rest) is defined by Kv 1 2 mv 2 1 2 mv v (8.1) and is called the kinetic energy of the particle. The kinetic energy of a system of particles is the sum of the kinetic energies of all its constituent particles, i.e., K i 1 2 mi vi 2 The kinetic energy of a particle or a system of particles can increase or decrease or remain constant as time passes. If no force is applied on the particle, its velocity v remains constant and hence the kinetic energy remains the same. A force is necessary to change the kinetic energy of a particle. If the resultant force acting on a particle is perpendicular to its velocity, the speed of the particle does not change and hence the kinetic energy does not change. Kinetic energy changes only when the speed changes and that happens only when the resultant force has a tangential component. When a particle falls near the earth’s surface, the force of gravity is parallel to its velocity. Its kinetic energy increases as time passes. On the other hand, a particle projected upward has the force opposite to the velocity and its kinetic energy decreases. From the definition of kinetic energy dK dt d dt 1 2 mv 2 mv dv dt Ft v, where Ft is the resultant tangential force. If the resultant force F makes an angle with the velocity, Ft F cos and dK dt Fv cos F v F dr dt or, dK F dr . (8.2) 8.2 WORK AND WORK-ENERGY THEOREM The quantity F dr F dr cos is called the work done by the force F on the particle during the small displacement dr . The work done on the particle by a force F acting on it during a finite displacement is obtained by W F dr F cos dr, (8.3) where the integration is to be performed along the path of the particle. If F is the resultant force on the particle we can use equation (8.2) to get W F dr dK K2 K1 . Thus, the work done on a particle by the resultant force is equal to the change in its kinetic energy. This is called the work-energy theorem. Let F 1 , F 2 , F 3 , be the individual forces acting on a particle. The resultant force is F F 1 F 2 F 3 , and the work done by the resultant force on the particle is W F dr F 1 F 2 F 3 ....... dr F 1 dr F 2 dr F 3 dr , where F 1 dr is the work done on the particle by F 1 and so on. Thus, the work done by the resultant force is equal to the sum of the work done by the individual forces. Note that the work done on a particle by an individual force is not equal to the change in its More energetic balls Less energetic balls Less energetic man More energetic man Figure 8.1 CHAPTER 8 WORK AND ENERGY

kinetic energy; the sum of the work done by all the forces acting on the particle (which is equal to the work done by the resultant force) is equal to the change in its kinetic energy. The rate of doing work is called the power delivered. The work done by a force F → in a small displacement dr → is dW = F → ⋅ dr → . Thus, the power delivered by the force is P = dW dt = F → ⋅ dr → dt = F → ⋅ v → . The SI unit of power is joule/second and is written as “watt”. A commonly used unit of power is horsepower which is equal to 746 W. 8.3 CALCULATION OF WORK DONE The work done by a force on a particle during a displacement has been defined as W = ∫ F → ⋅ dr → . Constant Force Suppose, the force is constant (in direction and magnitude) during the displacement. Then W = ∫ F → ⋅ dr → = F → ⋅ ∫ dr → = F → ⋅ r → , where r → is the total displacement of the particle during which the work is calculated. If θ be the angle between the constant force F → and the displacement r → , the work is W = Fr cosθ. … (8.4) In particular, if the displacement is along the force, as is the case with a freely and vertically falling particle, θ = 0 and W = Fr. The force of gravity (mg→ ) is constant in magnitude and direction if the particle moves near the surface of the earth. Suppose a particle moves from A to B along some curve and that AB → makes an angle θ with the vertical as shown in figure (8.2). The work done by the force of gravity during the transit from A to B is W = mg (AB) cosθ = mgh, where h is the height descended by the particle. If a particle ascends a height h, the work done by the force of gravity is – mgh. If the particle goes from the point A to the point B along some other curve, the work done by the force of gravity is again mgh. We see that the work done by a constant force in going from A to B depends only on the positions of A and B and not on the actual path taken. In case of gravity, the work is weight mg times the height descended. If a particle starts from A and reaches to the same point A after some time, the work done by gravity during this round trip is zero, as the height descended is zero. We shall encounter other forces having this property. Spring Force Consider the situation shown in figure (8.3). One end of a spring is attached to a fixed vertical support and the other end to a block which can move on a horizontal table. Let x = 0 denote the position of the block when the spring is in its natural length. We shall calculate the work done on the block by the spring-force as the block moves from x = 0 to x = x1. The force on the block is k times the elongation of the spring. But the elongation changes as the block moves and so does the force. We cannot take F → out of the integration ∫ F → ⋅ dr → . We have to write the work done during a small interval in which the block moves from x to x + dx. The force in this interval is kx and the displacement is dx. The force and the displacement are opposite in direction. So, F → ⋅ dr → = − F dx = − kx dx during this interval. The total work done as the block is displaced from x = 0 to x = x1 is W = ∫ 0 x1 − kx dx =    − 1 2 k x 2   0 x1 = − 1 2 k x1 2 . If the block moves from x = x1 to x = x2 , the limits of integration are x1 and x2 and the work done is W =    1 2 k x1 2 − 1 2 k x2 2   ⋅ … (8.5) Note that if the block is displaced from x1 to x2 and brought back to x = x1 , the work done by the spring-force is zero. The work done during the return journey is negative of the work during the onward journey. The net work done by the spring-force in a round trip is zero. A B mg h Figure 8.2 x = 0 x = x1 A A Figure 8.3 Work and Energy 119

Three positions of a spring are shown in figure (8.4). In (i) the spring is in its natural length, in (ii) it is compressed by an amount x and in (iii) it is elongated by an amount x. Work done by the spring-force on the block in various situations is shown in the following table. Table 8.1 Initial state of the spring Final state of the spring x1 x2 W Natural Compressed 0 – x − 1 2 k x 2 Natural Elongated 0 x − 1 2 k x 2 Elongated Natural x 0 1 2 k x 2 Compressed Natural − x 0 1 2 k x 2 Elongated Compressed x − x 0 Compressed Elongated – x x 0 Force Perpendicular to Velocity Suppose F → ⊥ v → for all the time. Then F → ⋅ dr → = F → ⋅ v →dt is zero in any small interval and the work done by this force is zero. For example, if a particle is fastened to the end of a string and is whirled in a circular path, the tension is always perpendicular to the velocity of the particle and hence the work done by the tension is zero in circular motion. Example 8.1 A spring of spring constant 50 N/m is compressed from its natural position through 1 cm. Find the work done by the spring-force on the agency compressing the spring. Solution : The magnitude of the work is 1 2 kx 2 = 1 2 × (50 N/m) × (1 cm) 2 = (25 N/m) × (1 × 10 − 2 m) 2 = 2. 5 × 10 − 3 J. As the compressed spring will push the agency, the force will be opposite to the displacement of the point of application and the work will be negative. Thus, the work done by the spring-force is –2. 5 mJ. The following three cases occur quite frequently : (a) The force is perpendicular to the velocity at all the instants. The work done by the force is then zero. (b) The force is constant (both in magnitude and direction). The work done by the force is W = Fd cosθ, where F and d are magnitudes of the force and the displacement and θ is the angle between them. The amount of work done depends only on the end positions and not on the intermediate path. The work in a round trip is zero. Force of gravity on the bodies near the earth’s surface is an example. The work done due to the force of gravity on a particle of mass m is mgh, where h is the vertical height ‘descended’ by the particle. (c) The force is F = − kx as is the case with an elastic spring. The magnitude of the work done by the force during a displacement x from or to its natural position (x = 0) is 1 2 kx 2 . The work may be + 1 2 kx 2 or − 1 2 kx 2 depending on whether the force and the displacement are along the same or opposite directions. Example 8.2 A particle of mass 20 g is thrown vertically upwards with a speed of 10 m/s. Find the work done by the force of gravity during the time the particle goes up. Solution : Suppose the particle reaches a maximum height h. As the velocity at the highest point is zero, we have 0 = u 2 − 2gh or, h = u 2 2g ⋅ The work done by the force of gravity is − mgh = − mg u 2 2g = − 1 2 mu 2 = − 1 2 (0. 02 kg) × (10 m/s) 2 = − 1. 0 J. 8.4 WORK-ENERGY THEOREM FOR A SYSTEM OF PARTICLES So far we have considered the work done on a single particle. The total work done on a particle equals the change in its kinetic energy. In other words, to change the kinetic energy of a particle we have to apply a force on it and the force must do work on it. Next, consider a system containing more than one particle and suppose the particles exert forces on each other. As a simple example, take a system of two x x (i) (ii) (iii) Figure 8.4 120 Concepts of Physics

charged particles as shown in figure (8.5) attracting each other (such as a positive and a negative charge). Because of mutual attraction, the particles are accelerated towards each other and the kinetic energy of the system increases. We have not applied any external force on the system, yet the kinetic energy has changed. Let us examine this in more detail. The particle B exerts a force F AB on A. As A moves towards B, this force does work. The work done by this force is equal to the increase in the kinetic energy of A. Similarly, A exerts a force F BA on B. This force does work on B and this work is equal to the increase in the kinetic energy of B. The work by F AB + the work by F BA is equal to the increase in the total kinetic energy of the two particles. Note that F AB F BA , so that F AB F BA 0. But the work by F AB + the work by F BA 0. The two forces are opposite in direction but the displacements are also opposite. Thus, the work done by both the forces are positive and are added. The total work done on different particles of the system by the internal forces may not be zero. The change in the kinetic energy of a system is equal to the work done on the system by the external as well as the internal forces. 8.5 POTENTIAL ENERGY Consider the example of the two charged particles A and B taken in the previous section. Suppose at some instant t1 the particles are at positions A, B and are going away from each other with speeds v1 and v2 (figure 8.6). The kinetic energy of the system is K1 . We call the positions of the particles at time t1 as configuration-1. The particle B attracts A and hence the speed v1 decreases as time passes. Similarly, the speed v2 of B decreases. Thus, the kinetic energy of the two-particle system decreases as time passes. Suppose at a time t2, the particles are at A and B, the speeds have changed to v1 and v2 and the kinetic energy becomes K2. We call the positions of the particles at time t2 as configuration-2. The kinetic energy of the system is decreased by K1 K2 . However, if you wait for some more time, the particles return to the original positions A and B, i.e., in configuration-1. At this time, say t3 , the particles move towards each other with speeds v1 and v2 . Their kinetic energy is again K1 . When the particles were in configuration-1 the kinetic energy was K1 . When they reached configuration-2 it decreased to K2 . The kinetic energy has decreased but is not lost for ever. We just have to wait. When the particles return to configuration-1 at time t3 , the kinetic energy again becomes K1 . It seems meaningful and reasonable if we think of yet another kind of energy which depends on the configuration. We call this as the potential energy of the system. Some kinetic energy was converted into potential energy when the system passed from configuration-1 to configuration-2. As the system returns to configuration-1, this potential energy is converted back into kinetic energy. The sum of the kinetic energy and the potential energy remains constant. How do we precisely define the potential energy of a system ? Before defining potential energy, let us discuss the idea of conservative and nonconservative forces. 8.6 CONSERVATIVE AND NONCONSERVATIVE FORCES Let us consider the following two examples. (1) Suppose a block of mass m rests on a rough horizontal table (figure 8.7). It is dragged horizontally towards right through a distance l and then back to its initial position. Let be the friction coefficient between the block and the table. Let us calculate the work done by friction during the round trip. The normal force between the table and the block is N mg and hence the force of friction is mg. When        Figure 8.5               Figure 8.6 Figure 8.7 Work and Energy 121

the block moves towards right, friction on it is towards left and the work by friction is (− µmgl) . When the block moves towards left, friction on it is towards right and the work is again (− µmgl) . Hence, the total work done by the force of friction in the round trip is (− 2µmgl) . (2) Suppose a block connected by a spring is kept on a rough table as shown in figure (8.8). The block is pulled aside and then released. It moves towards the centre A and has some velocity v0 as it passes through the centre. It goes to the other side of A and then comes back. This time it passes through the centre with somewhat smaller velocity v1 . Compare these two cases in which the block is at A, once going towards left and then towards right. In both the cases the system (table + block + spring) has the same configuration. The spring has the same length. The block is at the same point on the table and the table of course is fixed to the ground. The kinetic energy in the second case is less than the kinetic energy in the first case. This loss in the kinetic energy is a real loss. Every time the block passes through the mean position A, the kinetic energy of the system is smaller and in due course, the block stops on the table. We hold friction as the culprit, because in absence of friction the system regains its kinetic energy as it returns to its original configuration. Remember, work done by friction in a round trip is negative and not zero [example (1) above]. We divide the forces in two categories (a) conservative forces and (b) nonconservative forces. If the work done by a force during a round trip of a system is always zero, the force is said to be conservative. Otherwise, it is called nonconservative. Conservative force can also be defined as follows : If the work done by a force depends only on the initial and final states and not on the path taken, it is called a conservative force. Thus, the force of gravity, Coulomb force and the force of spring are conservative forces, as the work done by these forces are zero in a round trip. The force of friction is nonconservative because the work done by the friction is not zero in a round trip. 8.7 DEFINITION OF POTENTIAL ENERGY AND CONSERVATION OF MECHANICAL ENERGY We define the change in potential energy of a system corresponding to a conservative internal force as Uf − Ui = − W = − ∫ i f F → ⋅ dr → where W is the work done by the internal force on the system as the system passes from the initial configuration i to the final configuration f. We don’t (or can’t) define potential energy corresponding to a nonconservative internal force. Suppose only conservative internal forces operate between the parts of the system and the potential energy U is defined corresponding to these forces. There are either no external forces or the work done by them is zero. We have Uf − Ui = − W = − (Kf − Ki) or, Uf + Kf = Ui + Ki . … (8.6) The sum of the kinetic energy and the potential energy is called the total mechanical energy. We see from equation (8.6) that the total mechanical energy of a system remains constant if the internal forces are conservative and the external forces do no work. This is called the principle of conservation of energy. The total mechanical energy K + U is not constant if nonconservative forces, such as friction, act between the parts of the system. We can’t apply the principle of conservation of energy in presence of nonconservative forces. The work-energy theorem is still valid even in the presence of nonconservative forces. Note that only a change in potential energy is defined above. We are free to choose the zero potential energy in any configuration just as we are free to choose the origin in space anywhere we like. If nonconservative internal forces operate within the system, or external forces do work on the system, the mechanical energy changes as the configuration changes. According to the work-energy theorem, the work done by all the forces equals the change in the kinetic energy. Thus, Wc + Wnc + Wext = Kf − Ki where the three terms on the left denote the work done by the conservative internal forces, nonconservative internal forces and the external forces. As Wc = − (Uf − Ui) , we get Wnc + Wext = (Kf + Uf) − (Ki + Ui) = Ef − Ei … (8.7) v = 0 v0 1v A A A Figure 8.8 122 Concepts of Physics

where E = K + U is the total mechanical energy. If the internal forces are conservative but external forces also act on the system and they do work, Wnc = 0 and from (8.7), Wext = Ef − Ei . … (8.8) The work done by the external forces equals the change in the mechanical energy of the system. Let us summarise the concepts developed so far in this chapter. (1) Work done on a particle is equal to the change in its kinetic energy. (2) Work done on a system by all the (external and internal) forces is equal to the change in its kinetic energy. (3) A force is called conservative if the work done by it during a round trip of a system is always zero. The force of gravitation, Coulomb force, force by a spring etc. are conservative. If the work done by it during a round trip is not zero, the force is nonconservative. Friction is an example of nonconservative force. (4) The change in the potential energy of a system corresponding to conservative internal forces is equal to negative of the work done by these forces. (5) If no external forces act (or the work done by them is zero) and the internal forces are conservative, the mechanical energy of the system remains constant. This is known as the principle of conservation of mechanical energy. (6) If some of the internal forces are nonconservative, the mechanical energy of the system is not constant. (7) If the internal forces are conservative, the work done by the external forces is equal to the change in mechanical energy. Example 8.3 Two charged particles A and B repel each other by a force k/r 2 , where k is a constant and r is the separation between them. The particle A is clamped to a fixed point in the lab and the particle B which has a mass m, is released from rest with an initial separation r0 from A. Find the change in the potential energy of the two-particle system as the separation increases to a large value. What will be the speed of the particle B in this situation ? Solution : The situation is shown in figure (8.9). Take A + B as the system. The only external force acting on the system is that needed to hold A fixed. (You can imagine the experiment being conducted in a gravity free region or the particles may be kept and allowed to move on a smooth horizontal surface, so that the normal force balances the force of gravity). This force does no work on the system because it acts on the charge A which does not move. Thus, the external forces do no work and internal forces are conservative. The total mechanical energy must, therefore, remain constant. There are two internal forces; FAB acting on A and FBA acting on B. The force FAB does no work because it acts on A which does not move. The work done by FBA as the particle B is taken away is, W = ∫ F → ⋅ dr → = ∫ r0 ∞ k r 2 dr = k r0 ⋅ … (i) The change in the potential energy of the system is Uf − Ui = − W = − k r0 ⋅ As the total mechanical energy is conserved, Kf + Uf = Ki + Ui or, Kf = Ki − (Uf − Ui) or, 1 2 mv 2 = k r0 or, v = √2k mr0 ⋅ 8.8 CHANGE IN THE POTENTIAL ENERGY IN A RIGID-BODY-MOTION If the separation between the particles do not change during motion, such as in the case of the motion of a rigid body, the internal forces do no work. This is a consequence of Newton’s third law. As an example, consider a system of two particles A and B. Suppose, the particles move in such a way that the line AB translates parallel to itself. The displacement dr → A of the particle A is equal to the displacement dr → B of the particle B in any short time interval. The net work done by the internal forces F → AB and F → BA is W = ∫ (F → AB ⋅ dr → A + F → BA ⋅ dr → B) = ∫ (F → AB + F → BA) ⋅ dr → A = 0. Thus, the work done by F → AB and F → BA add up to zero. Even if AB does not translate parallel to itself but rotates, the result is true. The internal forces acting between the particles of a rigid body do no work in its motion and we need not consider the potential energy corresponding to these forces. The potential energy of a system changes only when the separations between the parts of the system change. In other words, the potential energy depends only on the separation between the interacting particles. A B AB BA F = k/r 2 F = k/r 2 External force Figure 8.9 Work and Energy 123

8.9 GRAVITATIONAL POTENTIAL ENERGY Consider a block of mass m kept near the surface of the earth and suppose it is raised through a height h. Consider “the earth + the block” as the system. The gravitational force between the earth and the block is conservative and we can define a potential energy corresponding to this force. The earth is very heavy as compared to the block and so one can neglect its acceleration. Thus, we take our reference frame attached to the earth, it will still be very nearly an inertial frame. The work done by the gravitational force due to the block on the earth is zero in this frame. The force mg on the block does work (–mgh) if the block ascends through a height h and hence the potential energy is increased by mgh. Thus, if a block of mass m ascends a height h above the earth’s surface (h << radius of earth), the potential energy of the “earth + block” system increases by mgh. If the block descends by a height h, the potential energy decreases by mgh. Since the earth almost remains fixed, it is customary to call the potential energy of the earth-block system as the potential energy of the block only. We then say that the gravitational potential energy of the “block” is increased by an amount mgh when it is raised through a hieght h above the earth’s surface. We have been talking in terms of the changes in gravitational potential energy. We can choose any position of the block and call the gravitational potential energy to be zero in this position. The potential energy at a height h above this position is mgh. The position of the zero potential energy is chosen according to the convenience of the problem. Example 8.4 A block of mass m slides along a frictionless surface as shown in the figure (8.10). If it is released from rest at A, what is its speed at B ? Solution : Take the block + the earth as the system. Only the block moves, so only the work done on the block will contribute to the gravitational potential energy. As it descends through a height h between A and B, the potential energy decreases by mgh. The normal contact force N on the block by the surface does no work as it is perpendicular to its velocity. No external force does any work on the system. Hence, increase in kinetic energy decrease in potential energy or, 1 2 mv 2 mgh or, v 2gh. Example 8.5 A pendulum bob has a speed 3 m/s while passing through its lowest position. What is its speed when it makes an angle of 60 with the vertical ? The length of the pendulum is 0. 5 m. Take g 10 m/s 2 . Solution : Take the bob + earth as the system. The external force acting on the system is that due to the string. But this force is always perpendicular to the velocity of the bob and so the work done by this force is zero. Hence, the total mechanical energy will remain constant. As is clear from figure (8.11), the height ascended by the bob at an angular displacement is l l cos l 1 cos. The increase in the potential energy is mgl 1 cos. This should be equal to the decrease in the kinetic energy of the system. Again, as the earth does not move in the lab frame, this is the decrease in the kinetic energy of the bob. If the speed at an angular displacement is v1, the decrease in kinetic energy is 1 2 mv0 2 1 2 mv1 2 , where v0 is the speed of the block at the lowest position. Thus, 1 2 mv0 2 1 2 mv1 2 mgl 1 cos or, v1 v 0 2 2gl 1 cos 9 m 2 /s 2 2 10 m/s 2 0. 5 m 1 1 2 2 m/s. 8.10 POTENTIAL ENERGY OF A COMPRESSED OR EXTENDED SPRING Consider a massless spring of natural length l, one end of which is fastened to a wall (figure 8.12). The other end is attached to a block which is slowly pulled     Figure 8.10 Figure 8.11 124 Concepts of Physics

on a smooth horizontal surface to extend the spring. Take the spring as the system. When it is elongated by a distance x, the tension in it is kx, where k is its spring constant. It pulls the wall towards right and the block towards left by forces of magnitude kx. The forces exerted on the spring are (i) kx towards left by the wall and (ii) kx towards right by the block. How much work has been done on the spring by these two external forces ? The force by the wall does no work as the point of application is fixed. The force by the block does work ∫ 0 x kx dx = 1 2 kx 2 . The work is positive as the force is towards right and the particles of the spring, on which this force is acting, also move towards right. Thus, the total external work done on the spring is 1 2 kx 2 , when the spring is elongated by an amount x from its natural length. The same is the external work done on the spring if it is compressed by a distance x. We have seen (equation 8.8) that the external work done on a system is equal to the change in its total mechanical energy. The spring is assumed to be massless and hence its kinetic energy remains zero all the time. Thus, its potential energy has increased by 1 2 kx 2 . We conclude that a stretched or compressed spring has a potential energy 1 2 kx 2 larger than its potential energy at its natural length. The potential energy of the spring corresponds to the internal forces between the particles of the spring when it is stretched or compressed. It is called elastic potential energy or the strain energy of the spring. Again, the calculation gives only the change in the elastic potential energy of the spring and we are free to choose any length of the spring and call the potential energy zero at that length. It is customary to choose the potential energy of a spring in its natural length to be zero. With this choice the potential energy of a spring is 1 2 kx 2 , where x is the elongation or the compression of the spring. Example 8.6 A block of mass m, attached to a spring of spring constant k, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. If it has a speed v when the spring is at its natural length, how far will it move on the table before coming to an instantaneous rest ? Solution : Consider the block + the spring as the system. The external forces acting on the system are (a) the force of gravity, (b) the normal force by the table and (c) the force by the wall. None of these do any work on this system and hence the total mechanical energy is conserved. If the block moves a distance x before comming to rest, we have, 1 2 mv 2 = 1 2 kx 2 or, x = v √ m/k . Example 8.7 A block of mass m is suspended through a spring of spring constant k and is in equilibrium. A sharp blow gives the block an initial downward velocity v. How far below the equilibrium position, the block comes to an instantaneous rest ? Solution : Let us consider the block + the spring + the earth as the system. The system has gravitational potential energy corresponding to the force between the block and the earth as well as the elastic potential energy corresponding to the spring-force. The total mechanical energy includes kinetic energy, gravitational potential energy and elastic potential energy. When the block is in equilibrium, it is acted upon by two forces, (a) the force of gravity mg and (b) the tension in the spring T = kx, where x is the elongation. For equilibrium, mg = kx, so that the spring is stretched by a length x = mg/k. The potential energy of the spring in this position is 1 2 k (mg/k) 2 = m 2 g 2 2k ⋅ Take the gravitational potential energy to be zero in this position. The total mechanical energy of the system just after the blow is 1 2 mv 2 + m 2 g 2 2k ⋅ The only external force on this system is that due to the ceiling which does no work. Hence, the mechanical kx kx k L x Figure 8.12 l 0 x = mg/k h Figure 8.13 Work and Energy 125

energy of this system remains constant. If the block descends through a height h before coming to an instantaneous rest, the elastic potential energy becomes 1 2 k (mg/k + h) 2 and the gravitational potential energy – mgh. The kinetic energy is zero in this state. Thus, we have 1 2 mv 2 + m 2 g 2 2k = 1 2 k (mg/k + h) 2 − mgh. Solving this we get, h = v √ m/k . Compare this with the result obtained in Example (8.6). If we neglect gravity and consider the length of the spring in equilibrium position as the natural length, the answer is same. This simplification is often used while dealing with vertical springs. 8.11 DIFFERENT FORMS OF ENERGY : MASS ENERGY EQUIVALENCE The kinetic energy and the potential energy of a system, taken together, form mechanical energy. Energy can exist in many other forms. In measuring kinetic energy of an extended body, we use the speed of the body as a whole. Even if we keep the body at rest, the particles in it are continuously moving inside the body. These particles also exert forces on each other and there is a potential energy corresponding to these forces. The total energy corresponding to the internal motion of molecules and their interaction, is called internal energy or thermal energy of the body. Light and sound are other forms of energy. When a source emits light or sound, it loses energy. Chemical energy is significant if there are chemical reactions. Einstein’s special theory of relativity shows that a material particle itself is a form of energy. Thus, about 8. 18 × 10 – 14 J of energy may be converted to form an electron and equal amount of energy may be obtained by destroying an electron. The ralation between the mass of a particle m and its equivalent energy E is given as E = mc 2 , where c = 3 × 10 8 m/s is the speed of light in vacuum. When all forms of energy are taken into account, we arrive at the generalised law of conservation of energy. Energy can never be created or destroyed, it can only be changed from one form into another. Worked Out Examples 1. A porter lifts a suitcase weighing 20 kg from the platform and puts it on his head 2. 0 m above the platform. Calculate the work done by the porter on the suitcase. Solution : The kinetic energy of the suitcase was zero when it was at the platform and it again became zero when it was put on the head. The change in kinetic energy is zero and hence the total work done on the suitcase is zero. Two forces act on the suitcase, one due to gravity and the other due to the porter. Thus, the work done by the porter is negative of the work done by gravity. As the suitcase is lifted up, the work done by gravity is W = − mgh = − (20 kg) (9. 8 m/s 2 ) (2 m) = − 392 J The work done by the porter is 392 J ≈ 390 J. 2. An elevator weighing 500 kg is to be lifted up at a constant velocity of 0. 20 m/s. What would be the minimum horsepower of the motor to be used ? Solution : As the elevator is going up with a uniform velocity, the total work done on it is zero in any time interval. The work done by the motor is, therefore, equal to the work done by the force of gravity in that interval (in magnitude). The rate of doing work, i.e., the power delivered is P = F v = mgv = (500 kg) (9. 8 m/s2 ) (0. 2 m/s) = 980 W Assuming no loss against friction etc., in the motor, the minimum horsepower of the motor is P = 980 W = 980 746 hp = 1. 3 hp. 3. A block of mass 2. 0 kg is pulled up on a smooth incline of angle 30° with the horizontal. If the block moves with an acceleration of 1. 0 m/s 2 , find the power delivered by the pulling force at a time 4. 0 s after the motion starts. What is the average power delivered during the 4. 0 s after the motion starts ? Solution : The forces acting on the block are shown in figure (8-W1). Resolving the forces parallel to the incline, we get F − mg sinθ = ma or, F = mg sinθ + ma = (2. 0 kg) [(9. 8 m/s 2 ) (1/2) + 1. 0 m/s 2 ] = 11. 8 N. 126 Concepts of Physics

The velocity at t = 4. 0 s is v = at = (1. 0 m/s 2 ) (4. 0 s) = 4. 0 m/s. The power delivered by the force at t = 4. 0 s is P = F → ⋅ v →= (11. 8 N) (4. 0 m/s) ≈ 47 W. The displacement during the first four seconds is x = 1 2 at 2 = 1 2 (1. 0 m/s 2 ) (16 s 2 ) = 8. 0 m. The work done in these four seconds is, therefore, W = F → ⋅ d →= (11. 8 N) (8. 0 m) = 94. 4 J. The average power delivered = 94. 4 J 4. 0 s = 23. 6 W ≈ 24 W. 4. A force F = (10 + 0. 50 x) acts on a particle in the x direction, where F is in newton and x in meter. Find the work done by this force during a displacement from x = 0 to x = 2. 0 m. Solution : As the force is variable, we shall find the work done in a small displacement x to x + dx and then integrate it to find the total work. The work done in this small displacement is dW = F → ⋅ dx →= (10 + 0. 5 x) dx. Thus, W = ∫ 0 2⋅0 (10 + 0. 50 x) dx =    10 x + 0. 50 x 2 2    0 2⋅0 = 21 J. 5. A body dropped from a height H reaches the ground with a speed of 1. 2 √gH. Calculate the work done by air-friction. Solution : The forces acting on the body are the force of gravity and the air-friction. By work-energy theorem, the total work done on the body is W = 1 2 m(1. 2 √gH) 2 − 0 = 0. 72 mgH. The work done by the force of gravity is mgH. Hence, the work done by the air-friction is 0. 72 mgH − mgH = − 0. 28 mgH. 6. A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with the horizontal. The friction coefficient between the block and the surface is µ. If the block travels at a uniform velocity, find the work done by this applied force during a displacement d of the block. Solution : Forces on the block are (i) its weight Mg, (ii) the normal force N, (iii) the applied force F and (iv) the kinetic friction µ N . The forces are shown in figure (8-W2). As the block moves with a uniform velocity, the forces add up to zero. Taking horizontal and vertical components, F cosθ = µ N and F sinθ + N = Mg. Eliminating N from these equations, F cosθ = µ (Mg − F sinθ) or, F = µ Mg cosθ + µ sinθ ⋅ The work done by this force during a displacement d is W = F d cosθ = µ Mgd cosθ cosθ + µ sinθ ⋅ 7. Two cylindrical vessels of equal cross-sectional area A contain water upto heights h1 and h2 . The vessels are interconnected so that the levels in them become equal. Calculate the work done by the force of gravity during the process. The density of water is ρ. Solution : Since the total volume of the water is constant, the height in each vessel after interconnection will be (h1 + h2)/2 . The level in the left vessel shown in the figure, drops from A to C and that in the right vessel rises from B to D. Effectively, the water in the part AC has dropped down to DB. The mass of this volume of water is m = ρ A    h1 − h1 + h2 2    30° mg F N Figure 8-W1 F Mg N N Figure 8-W2 A C D B h1 h2 2 h + h 1 2 Figure 8-W3 Work and Energy 127

= ρ A    h1 − h2 2    ⋅ The height descended by this water is AC = (h1 − h2)/2. The work done by the force of gravity during this process is, therefore, = ρ A    h1 − h2 2    2 g. 8. What minimum horizontal speed should be given to the bob of a simple pendulum of length l so that it describes a complete circle ? Solution : Suppose the bob is given a horizontal speed v0 at the bottom and it describes a complete vertical circle. Let its speed at the highest point be v. Taking the gravitational potential energy to be zero at the bottom, the conservation of energy gives, 1 2 mv0 2 = 1 2 mv 2 + 2mgl or, mv 2 = mv0 2 − 4 mgl. … (i) The forces acting on the bob at the highest point are mg due to the gravity and T due to the tension in the string. The resultant force towards the centre is, therefore, mg + T. As the bob is moving in a circle, its acceleration towards the centre is v 2 /l. Applying Newton’s second law and using (i), mg + T = mv 2 l = 1 l (mv0 2 − 4mgl) or, mv0 2 = 5 mgl + T l. Now, for v0 to be minimum, T should be minimum. As the minimum value of T can be zero, for minimum speed, mv0 2 = 5 mgl or, v0 = √ 5 gl. 9. A uniform chain of length l and mass m overhangs a smooth table with its two third part lying on the table. Find the kinetic energy of the chain as it completely slips off the table. Solution : Let us take the zero of potential energy at the table. Consider a part dx of the chain at a depth x below the surface of the table. The mass of this part is dm = m/l dx and hence its potential energy is − (m/l dx)gx. The potential energy of the l/3 of the chain that overhangs is U1 = ∫ 0 l/3 − m l gx dx = −    m l g    x 2 2       0 l/3 = − 1 18 mgl. This is also the potential energy of the full chain in the initial position because the part lying on the table has zero potential energy. The potential energy of the chain when it completely slips off the table is U2 = ∫ 0 l − m l gx dx = − 1 2 mgl. The loss in potential energy =    − 1 18 mgl   −    − 1 2 mgl   = 4 9 mgl. This should be equal to the gain in the kinetic energy. But the initial kinetic enegry is zero. Hence, the kinetic energy of the chain as it completely slips off the table is 4 9 mgl. 10. A block of mass m is pushed against a spring of spring constant k fixed at one end to a wall. The block can slide on a frictionless table as shown in figure (8-W6). The natural length of the spring is L0 and it is compressed to half its natural length when the block is released. Find the velocity of the block as a function of its distance x from the wall. Solution : When the block is released, the spring pushes it towards right. The velocity of the block increases till the spring acquires its natural length. Thereafter, the block loses contact with the spring and moves with constant velocity. Initially, the compression of the spring is L0 /2. When the distance of the block from the wall becomes x, where v T mg 2l v0 Figure 8-W4 dx x Figure 8-W5 m L /2 v x k 0 Figure 8-W6 128 Concepts of Physics

x < L0 , the compression is (L0 − x). Using the principle of conservation of energy, 1 2 k    L0 2    2 = 1 2 k (L0 − x) 2 + 1 2 mv 2 . Solving this, v = √ k m    L0 2 4 − (L0 − x) 2   1/2 ⋅ When the spring acquires its natural length, x = L0 and v = √ k m L0 2 ⋅ Thereafter, the block continues with this velocity. 11. A particle is placed at the point A of a frictionless track ABC as shown in figure (8-W7). It is pushed slightly towards right. Find its speed when it reaches the point B. Take g = 10 m/s 2 . Solution : Let us take the gravitational potential energy to be zero at the horizontal surface shown in the figure. The potential energies of the particle at A and B are UA = Mg (1 m) and UB = Mg (0. 5 m). The kinetic energy at the point A is zero. As the track is frictionless, no energy is lost. The normal force on the particle does no work. Applying the principle of conservation of energy, UA + KA = UB + KB or, Mg(1 m) = Mg(0. 5 m) + 1 2 MvB 2 or, 1 2 vB 2 = g(1 m − 0. 5 m) = (10 m/s 2 ) × 0. 5 m = 5 m 2 /s 2 or, vB = √10 m/s. 12. Figure (8-W8) shows a smooth curved track terminating in a smooth horizontal part. A spring of spring constant 400 N/m is attached at one end to a wedge fixed rigidly with the horizontal part. A 40 g mass is released from rest at a height of 4. 9 m on the curved track. Find the maximum compression of the spring. Solution : At the instant of maximum compression the speed of the 40 g mass reduces to zero. Taking the gravitational potential energy to be zero at the horizontal part, the conservation of energy shows, mgh = 1 2 kx 2 where m = 0. 04 kg, h = 4. 9 m, k = 400 N/m and x is the maximum compression. Thus, x = √2mgh k = √ 2 × (0. 04 kg) × (9. 8 m/s 2 ) × (4. 9 m) (400 N/m) = 9. 8 cm. 13. Figure (8-W9) shows a loop-the-loop track of radius R. A car (without engine) starts from a platform at a distance h above the top of the loop and goes around the loop without falling off the track. Find the minimum value of h for a successful looping. Neglect friction. Solution : Suppose the speed of the car at the topmost point of the loop is v. Taking the gravitational potential energy to be zero at the platform and assuming that the car starts with a negligible speed, the conservation of energy shows, 0 = − mgh + 1 2 mv 2 or, mv 2 = 2 mgh, … (i) where m is the mass of the car. The car moving in a circle must have radial acceleration v 2 /R at this instant. The forces on the car are, mg due to gravity and N due to the contact with the track. Both these forces are in radial direction at the top of the loop. Thus, from Newton’s Law mg + N = mv 2 R A B C 1m 0.5 m Figure 8-W7 4.9 m Figure 8-W8 h R Figure 8-W9 Work and Energy 129

or, mg + N = 2 mgh/R. For h to be minimum, N should assume the minimum value which can be zero. Thus, 2 mg hmin R = mg or, hmin = R/2. 14. A heavy particle is suspended by a string of length l. The particle is given a horizontal velocity v0 . The string becomes slack at some angle and the particle proceeds on a parabola. Find the value of v0 if the particle passes through the point of suspension. Solution : Suppose the string becomes slack when the particle reaches the point P (figure 8-W10). Suppose the string OP makes an angle θ with the upward vertical. The only force acting on the particle at P is its weight mg. The radial component of the force is mg cosθ. As the particle moves on the circle upto P, mg cosθ = m    v 2 l    or, v 2 = gl cosθ … (i) where v is its speed at P. Using conservation of energy, 1 2 mv0 2 = 1 2 mv 2 + mgl (1 + cosθ) or, v 2 = v0 2 − 2gl(1 + cosθ). … (ii) From (i) and (ii), v0 2 − 2 g l (1 + cosθ) = g l cosθ or, v0 2 = gl (2 + 3 cosθ). … (iii) Now onwards the particle goes in a parabola under the action of gravity. As it passes through the point of suspension O, the equations for horizontal and vertical motions give, l sinθ = (v cosθ) t and − l cosθ = (v sinθ) t − 1 2 gt 2 or, − l cosθ = (v sinθ)    l sinθ v cosθ    − 1 2 g    l sinθ v cosθ    2 or, − cos 2 θ = sin 2 θ − 1 2 g l sin 2 θ v 2 cosθ or, − cos 2 θ = 1 − cos 2 θ − 1 2 gl sin 2 θ gl cos 2 θ [From (i)] or, 1 = 1 2 tan 2 θ or, tan θ = √2. From (iii), v0 = [gl (2 + √3)] 1/2 . QUESTIONS FOR SHORT ANSWER 1. When you lift a box from the floor and put it on an almirah the potential energy of the box increases, but there is no change in its kinetic energy. Is it a violation of conservation of energy ? 2. A particle is released from the top of an incline of height h. Does the kinetic energy of the particle at the bottom of the incline depend on the angle of incline ? Do you need any more information to answer this question in Yes or No ? 3. Can the work by kinetic friction on an object be positive ? Zero ? 4. Can static friction do nonzero work on an object ? If yes, give an example. If no, give reason. 5. Can normal force do a nonzero work on an object. If yes, give an example. If no, give reason. 6. Can kinetic energy of a system be increased without applying any external force on the system ? 7. Is work-energy theorem valid in noninertial frames ? 8. A heavy box is kept on a smooth inclined plane and is pushed up by a force F acting parallel to the plane. Does the work done by the force F as the box goes from A to B depend on how fast the box was moving at A and B ? Does the work by the force of gravity depend on this ? 9. One person says that the potential energy of a particular book kept in an almirah is 20 J and the other says it is 30 J. Is one of them necessarily wrong ? 10. A book is lifted from the floor and is kept in an almirah. One person says that the potential energy of the book is increased by 20 J and the other says it is increased by 30 J. Is one of them necessarily wrong ? 11. In one of the exercises to strengthen the wrist and fingers, a person squeezes and releases a soft rubber ball. Is the work done on the ball positive, negative or zero during compression ? During expansion ? 12. In tug of war, the team that exerts a larger tangential force on the ground wins. Consider the period in which a team is dragging the opposite team by applying a larger tangential force on the ground. List which of the O P Figure 8-W10 130 Concepts of Physics

following works are positive, which are negative and which are zero ? (a) work by the winning team on the losing team (b) work by the losing team on the winning team (c) work by the ground on the winning team (d) work by the ground on the losing team (e) total external work on the two teams. 13. When an apple falls from a tree what happens to its gravitational potential energy just as it reaches the ground ? After it strikes the ground ? 14. When you push your bicycle up on an incline the potential energy of the bicyle and yourself increases. Where does this energy come from ? 15. The magnetic force on a charged particle is always perpendicular to its velocity. Can the magnetic force change the velocity of the particle ? Speed of the particle ? 16. A ball is given a speed v on a rough horizontal surface. The ball travels through a distance l on the surface and stops. (a) What are the initial and final kinetic energies of the ball ? (b) What is the work done by the kinetic friction ? 17. Consider the situation of the previous question from a frame moving with a speed v0 parallel to the initial velocity of the block. (a) What are the initial and final kinetic energies ? (b) What is the work done by the kinetic friction ? OBJECTIVE I 1. A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown (a) vertically downward (b) vertically upward (c) horizontally (d) the speed does not depend on the initial direction. 2. Two springs A and B(kA = 2kB) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is (a) E/2 (b) 2E (c) E (d) E/4. 3. Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is (a) 1 2 kx 2 (b) − 1 2 kx 2 (c) 1 4 kx 2 (d) − 1 4 kx 2 . 4. The negative of the work done by the conservative internal forces on a system equals the change in (a) total energy (b) kinetic energy (c) potential energy (d) none of these. 5. The work done by the external forces on a system equals the change in (a) total energy (b) kinetic energy (c) potential energy (d) none of these. 6. The work done by all the forces (external and internal) on a system equals the change in (a) total energy (b) kinetic energy (c) potential energy (d) none of these. 7. ————— of a two particle system depends only on the separation between the two particles. The most appropriate choice for the blank space in the above sentence is (a) Kinetic energy (b) Total mechanical energy (c) Potential energy (d) Total energy. 8. A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be (a) zero (b) mgvt cos 2 θ (c) mgvt sin 2θ (d) mgvt sin 2θ. 9. A block of mass m slides down a smooth vertical circular track. During the motion, the block is in (a) vertical equilibrium (b) horizontal equilibrium (c) radial equilibrium (d) none of these. 10. A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is (a) √gl (b) √2gl (c) √3gl (d) √5gl. OBJECTIVE II 1. A heavy stone is thrown from a cliff of height h in a given direction. The speed with which it hits the ground (a) must depend on the speed of projection (b) must be larger than the speed of projection (c) must be independent of the speed of projection (d) may be smaller than the speed of projection. 2. The total work done on a particle is equal to the change in its kinetic energy (a) always Work and Energy 131

(b) only if the forces acting on it are conservative (c) only if gravitational force alone acts on it (d) only if elastic force alone acts on it. 3. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that (a) its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a circular path. 4. Consider two observers moving with respect to each other at a speed v along a straight line. They observe a block of mass m moving a distance l on a rough surface. The following quantities will be same as observed by the two observers (a) kinetic energy of the block at time t (b) work done by friction (c) total work done on the block (d) acceleration of the block. 5. You lift a suitcase from the floor and keep it on a table. The work done by you on the suitcase does not depend on (a) the path taken by the suitcase (b) the time taken by you in doing so (c) the weight of the suitcase (d) your weight. 6. No work is done by a force on an object if (a) the force is always perpendicular to its velocity (b) the force is always perpendicular to its acceleration (c) the object is stationary but the point of application of the force moves on the object (d) the object moves in such a way that the point of application of the force remains fixed. 7. A particle of mass m is attached to a light string of length l, the other end of which is fixed. Initially the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle. (a) The string becomes slack when the particle reaches its highest point. (b) The velocity of the particle becomes zero at the highest point. (c) The kinetic energy of the ball in initial position was 1 2 mv 2 mgl. (d) The particle again passes through the initial position. 8. The kinetic energy of a particle continuously increases with time. (a) The resultant force on the particle must be parallel to the velocity at all instants. (b) The resultant force on the particle must be at an angle less than 90 all the time. (c) Its height above the ground level must continuously decrease. (d) The magnitude of its linear momentum is increasing continuously. 9. One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is 1 2 kx 2 . The possible cases are (a) the spring was initially compressed by a distance x and was finally in its natural length (b) it was initially stretched by a distance x and finally was in its natural length (c) it was initially in its natural length and finally in a compressed position (d) it was initially in its natural length and finally in a stretched position. 10. A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F. The kinetic energy of the block increases by 20 J in 1 s. (a) The tension in the string is Mg. (b) The tension in the string is F. (c) The work done by the tension on the block is 20 J in the above 1 s. (d) The work done by the force of gravity is –20 J in the above 1 s. EXERCISES 1. The mass of cyclist together with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6. 0 km/h to 12 km/h. 2. A block of mass 2. 00 kg moving at a speed of 10. 0 m/s accelerates at 3. 00 m/s 2 for 5. 00 s. Compute its final kinetic energy. 3. A box is pushed through 4. 0 m across a floor offering 100 N resistance. How much work is done by the resisting force ? 4. A block of mass 5. 0 kg slides down an incline of inclination 30 and length 10 m. Find the work done by the force of gravity. 5. A constant force of 2. 50 N accelerates a stationary particle of mass 15 g through a displacement of 2. 50 m. Find the work done and the average power delivered. 6. A particle moves from a point r 1 2 m i 3 m j to another point r 2 3 m i 2 m j during which a certain force F 5 N i 5 N j acts on it. Find the work done by the force on the particle during the displacement. 7. A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he covers a distance of 40 m with an acceleration of 0. 5 m/s 2, find the work done by the man on the block during the motion. 132 Concepts of Physics

8. A force F = a + bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d. 9. A block of mass 250 g slides down an incline of inclination 37° with a uniform speed. Find the work done against the friction as the block slides through 1. 0 m. 10. A block of mass m is kept over another block of mass M and the system rests on a horizontal surface (figure 8-E1). A constant horizontal force F acting on the lower block produces an acceleration F 2 (m + M) in the system, the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement d of the system. 11. A box weighing 2000 N is to be slowly slid through 20 m on a straight track having friction coefficient 0. 2 with the box. (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ which ensures him the minimum magnitude of the force. 12. A block of weight 100 N is slowly slid up on a smooth incline of inclination 37° by a person. Calculate the work done by the person in moving the block through a distance of 2. 0 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction. 13. Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h. 14. Find the average force needed to accelerate a car weighing 500 kg from rest to 72 km/h in a distance of 25 m. 15. A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation v = a√x, where a is a constant. Find the total work done by all the forces during a displacement from x = 0 to x = d. 16. A block of mass 2. 0 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take g = 10 m/s 2 . 17. A block of mass 2. 0 kg is pushed down an inclined plane of inclination 37° with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of 10 m/s 2. If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take g = 10 m/s 2 . 18. A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 cm/s. If the friction coefficient between the table and the block is 0. 1, how far does the block move before coming to rest ? 19. Water falling from a 50 m high fall is to be used for generating electric energy. If 1. 8 × 10 5 kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit ? 20. A person is painting his house walls. He stands on a ladder with a bucket containing paint in one hand and a brush in other. Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had a mass of 6. 0 kg and was at a height of 2. 0 m at the time it slipped, how much gravitational potential energy is lost together with the paint ? 21. A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground. 22. The 200 m free style women’s swimming gold medal at Seol Olympic 1988 went to Heike Friendrich of East Germany when she set a new Olympic record of 1 minute and 57. 56 seconds. Assume that she covered most of the distance with a uniform speed and had to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water during the swim. 23. The US athlete Florence Griffith-Joyner won the 100 m sprint gold medal at Seol Olympic 1988 setting a new Olympic record of 10. 54 s. Assume that she achieved her maximum speed in a very short-time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith-Joyner at her full speed. (b) Assuming that the track, the wind etc. offered an average resistance of one tenth of her weight, calculate the work done by the resistance during the run. (c) What power GriffithJoyner had to exert to maintain uniform speed ? 24. A water pump lifts water from a level 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower the engine should have to do this. 25. An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3. 00 m/s. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw, what horsepower does he use ? 26. In a factory it is desired to lift 2000 kg of metal through a distance of 12 m in 1 minute. Find the minimum horsepower of the engine to be used. 27. A scooter company gives the following specifications about its product. F m M Figure 8-E1 Work and Energy 133

Weight of the scooter — 95 kg Maximum speed — 60 km/h Maximum engine power — 3. 5 hp Pick up time to get the maximum speed — 5 s Check the validity of these specifications. 28. A block of mass 30. 0 kg is being brought down by a chain. If the block acquires a speed of 40. 0 cm/s in dropping down 2. 00 m, find the work done by the chain during the process. 29. The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is 16. 0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest. 30. The two blocks in an Atwood machine have masses 2. 0 kg and 3. 0 kg. Find the work done by gravity during the fourth second after the system is released from rest. 31. Consider the situation shown in figure (8-E2). The system is released from rest and the block of mass 1. 0 kg is found to have a speed 0. 3 m/s after it has descended through a distance of 1 m. Find the coefficient of kinetic friction between the block and the table. 32. A block of mass 100 g is moved with a speed of 5. 0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process. 33. A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, calculate the work done against friction (negative of the work done by the friction). 34. A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3. 2 m high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline ? What will be the speed of the block when it reaches the ground, if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline ? Take g = 10 m/s 2 . 35. In a children’s park, there is a slide which has a total length of 10 m and a height of 8. 0 m (figure 8-E3). Vertical ladder are provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. Find (a) the work done by the boy on the ladder as he goes up, (b) the work done by the slide on the boy as he comes down, (c) the work done by the ladder on the boy as he goes up. Neglect any work done by forces inside the body of the boy. 36. Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground ? 37. A block weighing 10 N travels down a smooth curved track AB joined to a rough horizontal surface (figure 8-E5). The rough surface has a friction coefficient of 0. 20 with the block. If the block starts slipping on the track from a point 1. 0 m above the horizontal surface, how far will it move on the rough surface ? 38. A uniform chain of mass m and length l overhangs a table with its two third part on the table. Find the work to be done by a person to put the hanging part back on the table. 39. A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is μ. Find the work done by the friction during the period the chain slips off the table. 40. A block of mass 1 kg is placed at the point A of a rough track shown in figure (8-E6). If slightly pushed towards right, it stops at the point B of the track. Calculate the work done by the frictional force on the block during its transit from A to B.   Figure 8-E2 Figure 8-E3    Figure 8-E4    Figure 8-E5 134 Concepts of Physics

41. A block of mass 5. 0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of 2. 0 m/s. How high will it rise ? Take g = 10 m/s 2 . 42. A block of mass 250 g is kept on a vertical spring of spring constant 100 N/m fixed from below. The spring is now compressed to have a length 10 cm shorter than its natural length and the system is released from this position. How high does the block rise ? Take g = 10 m/s 2 . 43. Figure (8-E7) shows a spring fixed at the bottom end of an incline of inclination 37°. A small block of mass 2 kg starts slipping down the incline from a point 4. 8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 m up the incline. Find (a) the friction coefficient between the plane and the block and (b) the spring constant of the spring. Take g = 10 m/s 2 . 44. A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring. 45. Consider the situation shown in figure (8-E8). Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring. 46. A block of mass m is attached to two unstretched springs of spring constants k1 and k2 as shown in figure (8-E9). The block is displaced towards right through a distance x and is released. Find the speed of the block as it passes through the mean position shown. 47. A block of mass m sliding on a smooth horizontal surface with a velocity v → meets a long horizontal spring fixed at one end and having spring constant k as shown in figure (8-E10). Find the maximum compression of the spring. Will the velocity of the block be the same as v → when it comes back to the original position shown ? 48. A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5. 0 cm (figure 8-E11). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring ? 49. A small heavy block is attached to the lower end of a light rod of length l which can be rotated about its clamped upper end. What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle ? 50. Figure (8-E12) shows two blocks A and B, each having a mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block A can slide is smooth. The block A is attached 1.0 m 0.8 m A B Figure 8-E6 37° Figure 8-E7 k m Figure 8-E8 k1 2k m Figure 8-E9 v m k Figure 8-E10 Figure 8-E11 A m B m Figure 8-E12 Work and Energy 135

to a spring of spring constant 40 N/m whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g 10 m/s 2 . 51. One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height h (figure 8-E13). Initially, the spring makes an angle of 37 with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical. 52. Figure (8-E14) shows a light rod of length l rigidly attached to a small heavy block at one end and a hook at the other end. The system is released from rest with the rod in a horizontal position. There is a fixed smooth ring at a depth h below the initial position of the hook and the hook gets into the ring as it reaches there. What should be the minimum value of h so that the block moves in a complete circle about the ring ? 53. The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity 10 gl, where l is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of 60 with the upward vertical. 54. A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37 with the vertical and is then released. Find the tension in the string when the bob is at its lowest position. 55. Figure (8-E15) shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal. 56. The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of 3 gl. Find the angle rotated by the string before it becomes slack. 57. A heavy particle is suspended by a 1. 5 m long string. It is given a horizontal velocity of 57 m/s. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g 10 m/s 2 . 58. A simple pendulum of length L having a bob of mass m is deflected from its rest position by an angle and released (figure 8-E16). The string hits a peg which is fixed at a distance x below the point of suspension and the bob starts going in a circle centred at the peg. (a) Assuming that initially the bob has a height less than the peg, show that the maximum height reached by the bob equals its initial height. (b) If the pendulum is released with 90 and x L/2 find the maximum height reached by the bob above its lowest position before the string becomes slack. (c) Find the minimum value of x/L for which the bob goes in a complete circle about the peg when the pendulum is released from 90. 59. A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere. 60. A particle of mass m is kept on a fixed, smooth sphere of radius R at a position, where the radius through the particle makes an angle of 30 with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release ? (b) Find the distance travelled by the particle before it leaves contact with the sphere. 61. A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed v. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of v for which the particle does not slip on the sphere ? (c) Assuming the velocity v to be half the minimum calculated in part (b) find the angle made by the radius   Figure 8-E13 Figure 8-E14    Figure 8-E15   Figure 8-E16 136 Concepts of Physics

through the particle with the vertical when it leaves the sphere. 62. Figure (8-E17) shows a smooth track which consists of a straight inclined part of length l joining smoothly with the circular part. A particle of mass m is projected up the incline from its bottom. (a) Find the minimum projection-speed v0 for which the particle reaches the top of the track. (b) Assuming that the projection-speed is 2v0 and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is only slightly greater than v0, where will the block lose contact with the track ? 63. A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle . (c) Find the tangential acceleration dv dt of the chain when the chain starts sliding down. 64. A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle it slides. ANSWERS OBJECTIVE I 1. (d) 2. (b) 3. (d) 4. (c) 5. (a) 6. (b) 7. (c) 8. (c) 9. (d) 10. (c) OBJECTIVE II 1. (a), (b) 2. (a) 3. (c), (d) 4. (d) 5. (a), (b), (d) 6. (a), (c), (d) 7. (a), (d) 8. (b), (d) 9. (a), (b) 10. (b) EXERCISES 1. 375 J 2. 625 J 3. 400 J 4. 245 J 5. 6. 25 J, 36. 1 W 6. zero 7. 40 J 8. a 1 2 bd d 9. 1. 5 J 10. (a) F 2 M m g (b) mF 2 M m (c) mFd 2M m 11. (a) 40000 J 5 tan (b) 7690 J 12. (a) 120 J (b) 120 J 13. 4000 N 14. 4000 N 15. ma 2 d/2 16. (b) 24 J (c) 16 J 17. (a) 100 J (b) 60 J (c) 60 J 18. 002 J, 82 cm 19. 122 20. 118 J 21. 58 m/s 22. 270 N 23. (a) 2250 J (b) 4900 J (c) 465 W 24. 6. 6 10 – 2 hp 25. 3. 84 J, 5. 14 10 – 3 hp 26. 5. 3 hp 27. Seems to be somewhat overclaimed. 28. 586 J 29. 19. 6 J 30. 67 J 31. 0. 12 32. 1. 45 J 33. 20300 J 34. (a) 6. 4 J (b) 6. 4 J (c) 8. 0 m/s (d) 8. 0 m/s 35. (a) zero (b) – 600 J (c) 1600 J 36. At a horizontal distance of 1 m from the end of the track. 37. 5. 0 m Figure 8-E17 Work and Energy 137

38. mgl/18 39. − 2 μ MgL/9 40. – 2 J 41. 20 cm 42. 20 cm 43. (a) 0. 5 (b) 1000 N/m 44. 3 mv 2 4 x 2 45. 2 mg/k 46. √⎯⎯⎯k1 + k2 m x 47. v √⎯⎯⎯⎯ m/k⎯ , No 48. At a horizontal distance of 1 m from the free end of the spring. 49. 2 √⎯⎯gl 50. 1. 5 m/s 51. h 4 √⎯⎯⎯⎯ k/m⎯ 52. l 53. (a) 8 mg (b) 5 mg (c) 6. 5 mg 54. 1. 4 N 55. √⎯⎯⎯3 mg R k 56. cos – 1 (− 1/3) 57. (a) 53° (b) 3. 0 m/s (c) 1. 2 m 58. (b) 5L/6 above the lowest point (c) 0. 6 59. cos – 1(2/3) 60. √3 mg/2 (b) 0. 43 R 61. (a) mg − mv 2 R (b) √⎯⎯⎯Rg (c) cos – 1 (3/4) 62. (a) √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 2 g [R(1 − cosθ) + l sinθ] (b) 6 mg ⎛ ⎜ ⎝ 1 − cosθ + l R sinθ ⎞ ⎟ ⎠ (c) The radius through the particle makes an angle cos – 1(2/3) with the vertical. 63. (a) mR 2 g l sin (l/R) (b) mR 2 g l ⎡ ⎢ ⎣ sin ⎛ ⎜ ⎝ l R ⎞ ⎟ ⎠ + sinθ − sin⎛ ⎜ ⎝ θ + l R ⎞ ⎟ ⎠ ⎤ ⎥ ⎦ (c) Rg l [1 − cos(l/R)] 64. [2 R(a sinθ + g − g cosθ)] 1/2 138 Concepts of Physics

9.1 CENTRE OF MASS Suppose a spin bowler throws a cricket ball vertically upward. Being a spinner, his fingers turn while throwing the ball and the ball goes up spinning rapidly. Focus your attention to a particular point on the surface of the ball. How does it move in the space ? Because the ball is spinning as well as rising up, in general, the path of a particle at the surface is complicated, not confined to a straight line or to a plane. The centre of the ball, however, still goes on the vertical straight line and the spinner’s fingers could not make its path complicated. If he does not throw the ball vertically up, rather passes it to his fellow fielder, the centre of the ball goes in a parabola. All the points of the ball do not go in parabolic paths. If the ball is spinning, the paths of most of the particles of the ball are complicated. But the centre of the ball always goes in a parabola irrespective of how the ball is thrown. (In fact the presence of air makes the path of the centre slightly different from a parabola and bowlers utilise this deviation. This effect will be discussed in a later chapter. At present we neglect it.) The centre of the ball is a very special point which is called the centre of mass of the ball. Its motion is just like the motion of a single particle thrown. Definition of Centre of Mass Let us consider a collection of N particles (Figure 9.2). Let the mass of the ith particle be mi and its coordinates with reference to the chosen axes be xi , yi , zi . Write the product mi xi for each of the particles and add them to get mi xi . Similarly get i mi yi and i mi zi. Then find X 1 M i mi xi , Y 1 M i mi yi and Z 1 M i mi zi where M i mi is the total mass of the system. Locate the point with coordinates X, Y, Z. This point is called the centre of mass of the given collection of the particles. If the position vector of the i th particle is r i , the centre of mass is defined to have the position vector R CM 1 M i mi r i . (9.1) Taking x, y, z components of this equation, we get the coordinates of centre of mass as defined above X 1 M i mi xi , Y 1 M i mi yi , Z 1 M i mi zi (9.2)   Figure 9.1 m (x , y , z ) (X, Y, Z) i i i i Z 0 Y X Figure 9.2 CHAPTER 9 CENTRE OF MASS, LINEAR MOMENTUM, COLLISION