Pre-U STPM Physics Term 1 CC039242a

94 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Relation between Force and Potential Energy For a mechanical system, Total energy = Kinetic energy + Potential energy E = K + U ∆K + ∆U = 0 (∆E = 0 since E = constant) ∆K = –∆U Work done W = Force × (displacement in the direction of the force) = F ∆x ∆x = displacement Also W = ∆K Hence, F∆x = ∆K = –∆U F = – ∆U ∆x Force = – dU dx Example 3 The initial velocity of a body of mass 5.0 kg is 2.0 m s–1. The figure below shows the variation of the force F with displacement s acting on the body in the direction of its motion. F / N 0 8.0 3.0 20 s / m What is the velocity of the body when it is displaced by 20 m? Solution: By the work-energy theorem, Gain in kinetic energy = Work done by the force 1 2 mv 2 – 1 2 mu2 = Area under F-s graph 1 2 (5.0)v 2 = 1 2 (3.0 + 8.0)(20) + 1 2 (5.0)(2.0)2 Velocity, v = 6.9 m s–1

95 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Quick Check 2 1. When the brakes of a car of mass 1 500 kg were applied, 600 kJ of heat were produced before the car was brought to rest. What was the speed of the car just before the brakes were applied? A 0.89 m s–1 B 1.26 m s–1 C 28.3 m s–1 D 63.2 m s–1 2. An object of mass m passes a point P with a velocity u and slides up a frictionless incline to stop at point Q which is at a height h above P. Q P u h A second object of mass 2m passes P with a velocity of 2u. What is the height it will rise? A 1 2 h C 2h B h D 4h 3. A bullet is fired into a sand bag. The figure shows the variation of kinetic energy E with the distance of penetration d of the bullet. The retarding force on the bullet exerted by the sand is A 3.6 N C 40 N B 7.2 N D 80 N 4. A bullet is fired vertically upwards from the ground and rises to a height h before falling back to the ground. Neglecting air resistance, which of the following represents the variation of the kinetic energy E of the bullet with the height h of the bullet above the ground. A C B D 5. A body of mass 2.0 kg moves with an initial velocity of 4.0 m s–1 in acted by a force F which varies with the displacement x as shown in the figure. What is the velocity of the body when x = 6.0 m? A 3.0 m s–1 C 7.0 m s–1 B 5.0 m s–1 D 8.0 m s–1 6. A trolley is pushed with a constant force F on a horizontal surface of negligible friction. Which of the following graphs represent the variation of kinetic energy K with time t of the trolley? A C B D 0 24 0.3 d / m E / J

96 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power 7. A man on ski is pulled by a constant force of 1 000 N along a horizontal surface. The frictional force is 200 N. If the mass of the man is 100 kg, what is his velocity and the work done by the force when the displacement is 4.0 m? Velocity Work done A 8.0 m s–1 3 200 J B 8.0 m s–1 4 000 J C 8.9 m s–1 4 000 J D 8.9 m s–1 4 800 J 8. A block of mass m at P slides from rest down a plane inclined at an angle θ to the horizontal. The friction on the block is F. P After sliding through a distance d, the velocity v of the block is given by A 1 2 mv2 = Fd B 1 2 mv2 = mgd sin θ – Fd C 1 2 mv2 = Fd + mgd sin θ D 1 2 mv2 = (F – mg)d 9. The figure shows a mass of 4.0 kg and a mass of 5.0 kg connected by an inextensible string passing over a smooth pulley. The mass for 4.0 kg starts from rest and moves up the smooth plane inclined at 30° to the horizontal. What is the velocity of the masses when the 4.0 kg mass moves a distance of 2.0 m along the plane? A 3.62 m s–1 C 5.42 m s–1 B 4.67 m s–1 D 9.90 m s–1 10. The potential energy U of a body varies with its distance x from a point O according to the expression U = ax2 . The variation of the force F with distance x on the body is given by A F = –2ax C F = 2 3 ax3 B F = 2ax D F = – 2 3 ax3 11. The bob of a simple pendulum of length l is released from rest when the string makes an angle of 60° with the vertical. If g is the acceleration due to gravity, what is the speed of the bob when it passes the lowest position? A gl 2 C 3gl 2 B gl D 2gl 4.3 Power Students should be able to: • derive and use the formula P = Fv • use the concept of efficiency to solve problems Learning Outcomes 1. Power is the rate of doing work. Power = Work done Time taken Unit of power is watt (W) or J s–1. 2009/P1/Q4, 2013/P1/Q4, 2014/P1/Q4, 2015/P1/Q4

97 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power 2. If the force F on a body is constant, Work done, W= Fs s = displacement Power P = dW dt = d dt (Fs) = F ds dt ds dt = v P = Fv Example 4 The maximum power delivered by the engine of a car of mass 1 200 kg is 90 kW. (a) Assuming that the power delivered by the engine of the car is always at the maximum, calculate the time which the car takes to accelerate from rest to 30 m s–1. (b) If it is assumed that the engine only delivers the maximum power when its speed is 30 m s–1, and the force is constant, what is the time taken by the car to accelerate from rest? (c) Explain the difference in the answers for (a) and (b) above. Solution: (a) Since the power of the engine is constant, work done by engine = Pt where t = time taken. By the principle of conservation of energy. Work done = Gain in kinetic energy Pt = 1 2 mv2 Time taken t = 1 2 × 1 200 (30)2 90 × 103 = 6.0 s (b) If the power is maximum when the speed is 30 m s–1, using power = Fv Constant force delivered by engine, F = 90 000 30 = 3 000 N Using F = ma Acceleration of car, a = 3 000 1 200 = 2.5 m s–2 Using v = u + at Time taken to reach 30 m s–1, t = 30 – 0 2.5 = 12.0 s (c) Power = Fv In (a) power is always maximum and constant. Initially when v is small, F is greater. In (b) the power is only maximum when v is maximum. Hence the constant force is smaller and a longer time is required to achieve the final speed of 30 m s–1.

98 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Quick Check 3 1. A toy car which is powered by a motor moves with a constant velocity v along a straight line. The figure shows the variation of the energy E supplied by the motor with time t. a If the constant force provided by the motor is F, the velocity v is given by A a Fb C Fa b B b Fa D Fb a 2. The power of a toy car of mass 0.400 kg varies with its velocity as shown in the figure. What is the acceleration of the toy car? Power / W Velocity / m s–1 8 2 2.0 0 A 1.60 m s–2 C 10.0 m s–2 B 7.50 m s–2 D 12.0 m s–2 3. A car of mass 1 000 kg is moving with a constant velocity of 20 m s–1 along a horizontal road. The output power of the car is 20 kW. The output power is then suddenly increased to 60 kW. What is the acceleration of the car? A 0.20 m s–2 C 2.0 m s–2 B 0.30 m s–2 D 3.0 m s–2 4. What is the power required by a body of mass m to produce an acceleration of a when it is moving with a velocity v up a plane inclined at an angle θ to the horizontal? (Neglect friction) A mavg sin θ B mav sin θ + mgv C mav + mgv sin θ D (mav + mgv) sin θ 5. A car of mass m has an engine which delivers a constant power P. What is the minimum time it takes to accelerate from rest to a speed v? A mv P C mv 2 2P B P mv D 2P mv 2 6. A machine requires 200 J of energy to do 120 J of work. What is its efficiency? 7. The efficiency of a car engine is 25%. The total resistance to motion is 1 000 N when the car is moving at a constant velocity. If the energy in one litre of petrol is 40 MJ, how far can the car travel with 1.0 litre of petrol? 8. The power of an electric motor is 200 W. It is used to raise a load of 100 N at a constant speed 0.6 m s–1. What is the efficiency of the motor? (Take g = 10 m s–2) 9. (a) The total resistance to the motion of a cyclist on a level road is 25 N. If his speed is 12 m s–1, what is his output power? (b) The cyclist managed to reduce the resistance to 15 N by bending low over the handle bar. If his output power is as calculated in (a) above, what would be his speed? Important Formulae 1. Work done by a constant force F, W = F.s = Fs cos θ 2. Work done by a variable force, W = ∫ 0 s Fdx = Area under force-displacement graph 3. Relation between force F and potential energy U F = – dU dx 4. Power = Work done Time taken = Fv 5. Efficiency = Work done by machine Energy input × 100% = Output power Input power × 100%

99 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power STPM PRACTICE 4 1. A car of mass m and a van of mass 2m travel at the speed. They are brought to rest by the same braking force. What is the ratio of the distance travelled by the van to that of the car? A 1:2 C 2:1 B 1:4 D 4:1 2. The engine of a vehicle produces a constant thrust F when the vehicle accelerates from rest. The total resistive force is F’ when the speed is v. What is the power produced by the engine when the speed is v? A Fv C (F –F’)v B F’v D (F + F’)v 3. A light spring is compressed between two trolleys P and Q of masses 4.0 kg and 3.0 kg respectively as shown in the figure below. The trolleys are held together and then released. 4.0 kg P Q 3.0 kg When the spring returns to its original length, it drops, and the speed of trolley P is 1.5 m s-1. What was the energy stored in the compressed spring? A 4.5 J C 9.0 J B 10.5 J D 21.0 J 4. An aircraft has a weight of 2.94 × 106 N. At a certain instant during landing, its speed is 97.2 km h–1. If a total frictional force is 4.50 × 105 N, what distance does the aircraft travel before it comes to a stop? A 176 m B 243 m C 2380 m D 3150 m 5. A piece of plank with a rough surface is used to move a load through a height h as shown in the diagram. h Load Plank If a shorter plank is used, choose the correct statement about the work done. Against gravity Against friction A Remains the same Remains the same B Remains the same Decreases C Increases Remains the same D Increases Decreases 6. A car accelerates uniformly from rest. Which graph shows the variation of the power P of the car’s engine with velocity v? A C P 0 V P 0 V B D P 0 V P 0 V 7. A ball is thrown vertically upwards. Neglecting air resistance, which graph correctly shows how the kinetic energy K, potential energy U and total energy E of the ball vary the height h of the ball from the point of projection? A C E h KU E h KU B D E h K U E h K U

100 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power 8. A mass of 0.250 kg is dropped from a height of 0.60 m above the top of a helical spring as shown in the diagram. 0.250 kg 0.06 m The spring has a force constant of 560 N m–1. Air resistance is negligible. (a) Describe the energy transformation in the system after the mass is released. (b) Find the maximum compression of the spring. 9. (a) (i) Define work done by a force. (ii) A body is acted by a constant force F. Show that the work done by the force F after the body is displaced at a distance s in the direction of F equals to the gain in kinetic energy of the body. (iii) State the work-energy theorem. (b) Two bodies P and Q of masses 8.0 kg and 4.0 kg are connected by a light rope that passes over a smooth pulley as shown in the figure below. 8.0 kg P 3.0 m 4.0 kgQ P is released from a height of 3.0 m from the ground. Neglecting the mass of the pulley, calculate (i) the speed of P just before it hits the ground, (ii) the maximum height reached by Q. Discuss how the answer is affected if the mass of the trolley is not negligible. 10. A pump is used to pump water to a height of 9.0 m at a rate of 2.4 kg s–1. What is the minimum power of pump (a) if the speed at which water enters and leaves the pump is negligible? (b) if the speed of water entering the pump is negligible, but the speed at which it leaves the pump is 6.0 m s–1? 11. A constant force F acts on a body of mass m which is initially at rest on a smooth horizontal surface. Deduce an expression for the instantaneous power P delivered to the body is terms the time t and m. Hence, sketch a graph to show the variation of the power P with time t. 12. A body of mass m moves up a plane inclined at an angle θ to the horizontal. What is the power required by the body for it to have an acceleration a when its velocity is v. (Neglect friction). 13. A man of mass m climbs the wall of a building. He starts from rest and achieves a constant velocity after a very short time interval. After a time t, he has reached a height h. (a) What is his total gain in kinetic energy and potential energy? (b) What is the mean power developed? (c) The ratio power weight is used to compare the power of climbers of different weight. Calculate the value of this ratio for a climber of mass 46.0 kg when he climbs a height of 54 m in 8.0 minutes. 14. (a) Write an expression for the work done W by a variable force F. Identify any other symbol in your expression. (b) State the principle of conversation of energy. (c) A car of mass 1 000 kg moves with a velocity of 30 m s–1 along a straight road inclined at an angle of θ to the horizontal as shown in the figure. 30 m s–1 sin θ = 1 20 θ Calculate the power required by the car to move up along the road. (Neglect friction)

101 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power 15. (a) Define work and energy. (b) In a game, balls are thrown at wooden cubes in order to turn them over. Each cube is of mass 150 g and each of its sides is of length 10 cm as shown in figure (a). G 10 cm 10 cm Ball A G A (a) (b) In order to turn the block over, the centre of gravity G of the block must be raised so that G is vertically above the corner A, as shown in figure (b). For the block shown in figure (b), (i) calculate the vertical height through which the centre of gravity has been raised. (ii) calculate the gain in potential energy of the block. (c) The block is struck by a ball of mass 16 g travelling horizontal towards G as shown in figure (a). The collision is perfectly elastic and without sliding, the block turns about the corner A. The block is just able to reach the position in figure (b) if 30% of the kinetic energy of the ball is transferred to the block. Calculate (i) the kinetic energy of the ball just before it strikes the block. (ii) the speed at which the ball strikes the block. (iii) the speed at which the ball rebounce from the block. (iv) the change in momentum of the ball. (v) the average force on the block, assuming that the ball and the block are in contact for 0.020 s. 16. (a) Starting from the definition of work, deduce the change in (i) the gravitational potential energy of a mass m when it is raised through a height h. (ii) the change in kinetic energy of an object of mass m acting upon by a constant force F through a displacement s. Identify any additional symbols in your equations. (b) The total mass of a cyclist and his bicycle is 80 kg and travels with a constant speed of 18 m s–1 on a flat road at A as shown in the figure. He then travels down a slope to B. A B Calculate (i) the kinetic energy at A. (ii) the loss of potential energy. (iii) the speed at B, assuming that the loss potential energy is transformed into kinetic energy. (c) The cyclist travelling at a constant speed of 18 m s–1 on a level road provides a power of 320 W. Calculate the total resistive force. (d) The cyclist now travels at a higher constant speed. Explain why the cyclist needs to provide a greater power.

102 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power 1 11. (i) Work done = 1 2 Fx = 1 2 (10)(0.12) J = 0.60 J (ii) Work done = area under F-x graph = 1 2 (2)(0.04) + 1 2 (6 – 2)(0.08 – 0.04) + 1 2 (12 – 6) (0.12 - 0.08) J = 0.24 J 2. (i) Work done = area under F-s graph = 1 2 (8)(8) J = 32 J (ii) Work done = 1 2 (8)(8) + 1 2 (12 – 8)(–4) J = 24 J 2 1. C 2. D 3. D 4. B 5. B 6. B 7. B 8. B 9. A 10. A 11. B 3 1. B 2. B 3. C 4. C 5. C 6. 60% 7. 10 km 8. 30% output power = Fv 9. (a) Power = Fv = 300 W (b) v = 20 m s–1 STPM Practice 4 1. C : Work done by braking force = loss of K.E. Fs = 1 2 mv2 , s ∝ m 2. A : Power = Fv 3. B : Momentum is conserved, (4.0 kg)(1.5 m s-1) + (3.0 kg)v = 0 Velocity of Q, v = -2.0 m s-1 Energy of spring = total K.E. of trolleys = 1 2 (4.0)1.52 + 1 2 (3.0)2.02 J = 10.5 J 4. B : Fs = 1 2 mv2 s = 1 2 2 m = 243 m 5. A: Work done against gravity = mgh F R = mg cos L h mg θ µ = mg cos θ θ h Shorter plank θ Length of plank, L = h sin θ F = μ mg cos θ Work done against friction= (μ mg cos θ)( h sin θ ) = µmgh tan θ With a shorter plank, θ’ > θ, tan θ’ > tan θ Work done against friction decreases. 6. B: Power, P = Fv = (ma)v 7. C : E = U + K = constant U = mgh and K = E – mgh 8. (a) As the mass falls, its gravitational potential energy decreases and its kinetic energy increases. The gain in kinetic energy equals the loss gravitational potential energy. As the spring is compressed, the kinetic energy of the mass is transformed into elastic potential energy of the spring. At the maximum compression, elastic potential energy of the spring equals the loss of gravitational potential energy. (b) 1 2 kx2 = mgh x = 2mgh k = 2(0.250)(9.81)(0.60) 560 m = 7.25 cm 9. (a) (i) Work done by a force F when the displacement its point of application is s is given by work = F.s (ii) Let speed of body increases from u to v when the displacement is s, v2 = u2 + 2as, as = 1 2 (v2 – u2 ) Work done = Fs = (ma)s = 1 2 mv2 – 1 2mu2 = gain in K.E. (iii) Work energy theorem: Work done on a body by a force equals the gain in mechanical energy of the body. (b) (i) (8.0 + 4.0)a = 8.0g – 4.0g a = 4.0(9.81) 12.0 m s–1 = 3.27 m s–2 ANSWERS

103 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power v2 = u2 + 2as = 0 + 2(3.27)(3.0) v = 4.43 m s–1 Alternative K.E. of P and Q + P.E. gained by Q = loss of P.E. by P. 1 2 (8.0 + 4.0) v2 + (4.0)(9.81)(3.0) = (8.0)(9.81)(3.0) J v = 4.43 m s–1 (ii) After moving up 3.00 m, Q moves under gravity. v2 = u2 + 2as s = 0 – 4.432 2(–9.81) m = 1.00 m Maximum height reached = (3.00 + 1.00) m = 4.00 m If mass of pulley is not negligible, part of the loss of potential energy of P is transformed into kinetic energy of the pulley. Hence K.E. gained by Q is smaller. The maximum height reached is smaller. 10. (a) Power = mgh t = (2.4)(9.81)(9.0) W = 212 W (b) Power = mgh + 1 2 mv2 t = 212 W + 1 2 (2.4)(6.0)2 W = 255 W 11. F = ma, v = u + at, P = Fv 12. Power = Fv = mav + mgv sin θ 13. (a) 1 2 m ( h t ) 2 + mgh v = h t (b) Power = mh2 2t3 + mgh t (c) 0.113 m s–1 14. (a) W =  s 0 F dx (b) Refer to page 93 (c) P = (mg sin q)v = 14.7 kW 15. (a) Refer to page 89 and 92 (b) (i) [ 1 2 ( 2 × 10) – 5.0 cm] = 2.07cm (ii) ΔU = (mg)(0.0207) J = 3.046 × 10-2 J (c) (i) 1.015 × 10–1 J E = 100 30 × (b) (ii) (ii) 3.57 m s–1 (iii) –2.98 m s–1 1 2 mv2 = 0.70 E (iv) 0.105 N s ∆p = m (v – u) (v) 5.25 N 16. (a) (i) ∆U = Fh F = mg = mgh (ii) ∆K = Fs F = ma, v2 = u2 – 2as = 1 2 m (v2 – u2 ) (b) (i) K = 1 2 mu2 = 1.296 × 104 J (ii) ΔU = (mg)(5.0) J = 3.924 × 103 J (iii) 20.5 m s–1 1 2 mv2 – 1 2 mu2 = mgh (c) F = 17.8 N P = Fv (d) P = Fv P  v

CHAPTER Concept Map CIRCULAR MOTION 5 Bilingual Keywords Angular displacement: Sesaran sudut Angular velocity: Halaju sudut Centripetal force: Daya memusat Circular motion: Gerakan membulat Horizontal: Mengufuk Period: Kala Revolution: Putaran Tension: Tegangan Vertical: Tegak Circular Motion Angular Displacement and Angular Velocity ω = dθ dt v = rω Centripetal Acceleration a = v2 r = vω = rω2 Centripetal Force F = mv2 r = mvω = mrω2 • Motion in a horizontal circle • Motion in a vertical circle 104

Physics Term 1 STPM Chapter 5 Circular Motion 5.1 Angular Displacement and Angular Velocity Students should be able to: • express angular displacement in radians • define angular velocity and period • derive and use the formula v = rw Learning Outcomes 1. When a particles moves from A to B in a circle, the angle turned through θ (Figure 5.1) is known as the angular displacement. 2. The angular velocity of an object in circular motion is the rate of change of angular displacement θ. Angular velocity, ω = dθ dt . 3. The unit for angular velocity is radians per second. Angular velocity can also be stated in revolutions per second or revolutions per minute. 1 revolution per second = 2π radians per second 1 revolution per minute = 2π 60 radians per second 4. Figure 5.1 shows a particle going round a circle of radius r at a constant speed v. Suppose that the particle takes a time, t to move from A to B. During this time, the angular displacement is θ. Then, angular velocity, ω = θ t , angular displacement, θ = ωt. 5. The time taken for the particle to complete a circle is known as the period T of the circular motion. 6. In a complete circle, the angular displacement, θ = 2π radians. Hence, period, T = 2π ω 7. If s = length of the arc AB, the angular displacement θ in radian is given by θ = s r s = rθ Differentiating with respect to time t, ds dt = d dt (rθ) v = r dθ dt (dθ dt = ω) v = rω Example 1 The second hand of a watch is 1.2 cm long. (a) What is its angular velocity in radians per second? (b) What is the speed of the tip of the second hand? Solution: (a) Angular velocity, ω = θ t = 2π radians 60 seconds = 0.1047 rad s–1 B v v s r O θ A Figure 5.1 (b) Speed of the tip of the second arm, v = rω = 1.2 × 0.1047 = 0.1256 cm s–1 105 5

Physics Term 1 STPM Chapter 5 Circular Motion Quick Check 1 1. The minute hand of a large clock is 0.50 m. What is the mean of angular speed? A 8.75 × 10–4 rad s–1 B 1.75 × 10–3 rad s–1 C 3.50 × 10–3 rad s–1 D 1.05 × 10–1 rad s–1 2. On a tape recorder, the tape leaves a spool at constant speed v at a variable distance r from the centre. The angular velocity of the spool is Spool Spool Tape Tape v v A constant B proportional to r C proportional to r2 D proportional to 1 r 3. A disc is rotating about an axis through its centre and perpendicular to its plane. Points P and Q are at distance x and 2x from the centre of the circle. What is the ratio of linear speed of P linear speed of Q ? A 4 C 1 2 B 2 D 1 4 4. A measuring tape is wound on a roller rotated at a constant angular velocity. The radius of the roller increases at a steady rate. Radius v Tape Which of the graphs below correctly shows the variation of the speed v of the tape with time t? A C B D 5. A particle moves in a circle of radius 0.25 m at a constant speed. It takes 5.0 s to move from A to B. B A (a) What is the angular displacement in radians when the particle moves (i) from A to B, and (ii) from A back to A. (b) What is its angular velocity? (c) What is its speed? 6. The Earth takes 24 hours for a complete rotation about its axis. What is the angular velocity of the Earth? 106 5

Physics Term 1 STPM Chapter 5 Circular Motion 5.2 Centripetal Acceleration Students should be able to: • explain that uniform circular motion has an acceleration due to the change in direction of velocity • derive and use the formulae for centripetal acceleration a = v2 r and a = rw2 . Learning Outcomes A C B D P Q R |Δv| = v(Δθ) |v| = |vA| = |vB| (a) (b) Figure 5.2 1. Although the speed of a particle in uniform circular motion is constant, the particle is accelerated. This is because the direction of its velocity is constantly changing. This change in the velocity produces an acceleration. 2. Figure 5.2(a) shows a particle of mass m going round a circle of radius r with constant speed v. At A, the velocity of the particle vA is along the direction AC. After a very short interval of time Δt the particle is at B. The velocity of the particle at B is vB along the direction BD. The angular displacement Δθ is very small. In the figure, Δθ is shown bigger. Figure 5.2(b) is the vector diagram to show the change of velocity, Δv = vB – vA 3. The direction of the change of velocity Δv is QR. Hence, the acceleration a = Δv Δt is along QR. In triangle PQR, the angle Δθ is small, hence angle PQR ≈ 90°. As Δt → 0, Δθ → 0° and ∠PQR = 90°. Hence, the direction of acceleration is perpendicular to vA, that is towards the centre, O. 4. This acceleration is known as the centripetal acceleration. Magnitude of the centripetal acceleration a = Δv Δt Δv = v Δθ (Figure 5.2) = v Δθ Δt Δθ Δt = ω a = vω a = rω2 a = v2 r Quick Check 2 2008/P1/Q5, 2010/P1/Q6, 2013/P1/Q6, 2016/P1/Q4, 2017/P1/Q6 1. A body moves at a constant speed around a circle of radius 2.0 m in 0.50 s. What is its acceleration? A zero C 16 π2 m s–2 B 4 π2 m s–2 D 32 π2 rad s–2 2. The speed of rotation of a turntable changes from 45 revolutions per minute to 33 1 3 revolutions per minute. Find the ratio of the centripetal accelerations of a point on the rim of the turntable. 3. What is the centripetal acceleration of a particle at a distance of 20.0 cm from the axis of rotation of a centrifuge which rotates at 3 000 revolutions per minute? INFO Centripetal Acceleration 107 5

Physics Term 1 STPM Chapter 5 Circular Motion 5.3 Centripetal Force Students should be able to: • explain that uniform circular motion is due to the action of a resultant force that is always directed to the centre of the circle • use the formulae for centripetal force F = mv2 r and F = mrw2 • solve problems involving uniform horizontal circular motion for a point mass • solve problems involving vertical circular motions for a point mass (knowledge of tangential acceleration is not required) Learning Outcomes 1. A particle of mass m moving in a circle of radius r with constant speed v has acceleration, known as the centripetal acceleration of v2 r towards the centre of the circle. 2. The particle is able to move in a circle due to the resultant force towards the centre of the circle. This resultant force towards the centre of the circle is known as the centripetal force. 3. According to Newton’s second law of motion, centripetal force, F = ma (a is the centripetal acceleration) = m( v2 r ) (v = rω) = mrω2 = mvω Uniform Horizontal Circular Motion 2009/P1/Q5, 2009/P2/Q1 1. The simplest form of circular motion is motion of a body in a horizontal circle with uniform speed. O Figure 5.3 2. Figure 5.3 shows a particle of mass m at the end of an inelastic string being whirled round a horizontal circle with uniform speed v on top of a smooth table. 3. The centripetal force is provided by the tension T in the string. The acceleration of the particle is v 2 r towards the centre of the circle. Using F = ma Tension, T = mv 2 r 4. If the speed v is gradually increased, the tension in the string increases until the breaking tension is attained. 5. When the string breaks, no more resultant force acts on the particle. 6. According to Newton’s first law of motion, the particle continues its motion along a straight line with uniform speed at the direction tangential to the circle. Exam Tips Centripetal force F = mv 2 r = mrω2 acts towards centre of circle. 2013/P1/Q5, 2014/P1/Q6, 2015/P1/Q5,Q6, 2016/P1/Q4, 2017/P1/Q6 108 5

Physics Term 1 STPM Chapter 5 Circular Motion Example 2 With the aid of labelled diagrams, explain why a mass at the end of an inelastic light string cannot be whirled in a circle above the head with the string horizontal. Solution: O (a) (b) If the string is horizontal (figure (a)), there is no force to balance the weight, mg. Hence, the string must slant as shown in figure (b). The vertical component of the tension balances the weight. T cos θ = mg The horizontal component of the tension provides the centripetal force, T sin θ = mv 2 r . Example 3 The figure shows a U-tube with the distance between the limbs 25.0 cm. The U-tube is filled with a liquid. Calculate the difference in height of the liquid levels in the two limbs when (a) the tube moves along the x-axis with constant velocity 0.50 m s–1. (b) the tube moves along the x-axis with uniform acceleration 1.00 m s–2. (c) the tube is rotated about the y-axis at a speed of 30 revolutions per minute. Neglect effect of viscosity and surface tension. Solution: (a) When the velocity is constant, acceleration = 0, the liquid is in equilibrium. Hence, difference in height of liquid level = 0. (b) When the U-tube accelerates in the Ox direction, the liquid level at A is higher than at B (figure). 25 cm = 1.00 m s –2 A B F h a Resultant force on liquid in the horizontal direction, F = (hρg)A where ρ = density of liquid A = cross-sectional area of U-tube Using F = ma (hρg) A = (0.25 ρA) × 1.00 h = 0.25 9.81 = 0.0255 m = 2.55 cm A B 109 5

Physics Term 1 STPM Chapter 5 Circular Motion (c) When the U-tube is rotated about the y-axis, the direction of the acceleration a = rω2 is towards the axis of rotation. Hence, the liquid level at B is higher than at A. Angular velocity, ω = 30 rev. per minute = 30 × 2π 60 rad s–1 ω = π rad s–1 Resultant force, F = (hρg)A Using F = ma (hρg) A = (0.25 × ρA) (ω2 r) h = 0.25 × π2 9.81 × (0.25 2 ) = 0.0315 m = 3.15 cm Quick Check 3 1. Which of the following statements is correct for a particle moving in a horizontal circle with constant speed? A The angular velocity is constant but the linear velocity varies. B Both the linear momentum and kinetic energy is constant. C The speed is constant and the acceleration is zero. D The resultant force is zero and the kinetic energy varies. 2. A particle of mass m moves in a horizontal circle of radius r at constant speed v as shown in the figure. X Y O What is the change in linear momentum of the particle and the work done by the centripetal force, when it moves from X to Y? Change in linear Work done momentum A 0 mv2 B 0 πmv2 C 2mv 0 D 2mv 2mv2 3. When a particle moves in a horizontal circle with angular velocity ω1 , the centripetal force is F. When it moves in the same circle with angular velocity ω2 , the centripetal force is 2F. What is the value of ω2 ω1 ? A 0.5 B 1.4 C 2.0 D 4.0 4. A turntable rotates at a constant angular velocity. Which graphs best represents the relation between the centripetal force F on a particle placed at difference distance r from the centre of rotation? A C B D r = distance of centre of mass of liquid in horizontal section of tube = 0.25 2 m = 2 r A B h y F ω a ω 110 5

Physics Term 1 STPM Chapter 5 Circular Motion 5. A body of mass 0.5 kg rotates at a constant speed in a circle of radius 1.5 m. The period of the circular motion is 2.0 s. What is the magnitude of the resultant force on the body? A 2.36 N C 7.40 N B 3.00 N D 56.3 N 6. A stone tied to the end of a string is whirled in a horizontal circle on a smooth table with uniform speed. Which statement is correct? A A centrifugal force acts on the stone B There is no resultant force on the stone C The centrifugal force equals the tension in the string D The tension in the string provides the centripetal force. 2014/P1/Q6 7. A man of mass 72 kg stands on the equator. Calculate (a) his angular velocity due to the rotation of the Earth about its axis, (b) his linear speed, (c) his centripetal acceleration, (d) the centripetal force on the man. (Radius of the Earth = 6.4 × 106 m) 8. A turntable rotates at a constant speed of 2 revolutions per second. A coin of mass 50 g stays stationary on the turntable if its distance from the axis of rotation does not exceed 5.0 cm. Explain why. Calculate the limiting static friction between the coin and the turntable. Conical Pendulum mg T l r θ ω O mg T Resultant of T and mg mg T θ Resultant force (a) (b) (c) Figure 5.4 1. Figure 5.4(a) shows a particle at the end of string moving in a horizontal circle and forming a conical pendulum. 2. The forces acting on the particle of mass m are • T, the tension in the string, and • mg, the weight of the particle. 3. Since the particle is in circular motion about the point O, the centripetal acceleration a = v 2 r towards O Using F = ma Horizontal component of tension, T sin θ = mv 2 r = mrω2 = m(l sin θ) ω2 T = mlω2 111 5

Physics Term 1 STPM Chapter 5 Circular Motion Vertical component of tension, T cos θ = mg (mlω2 ) cos θ = mg cos θ = g lω2 θ = cos–1 ( g lω2) Example 4 (a) A pendulum bob is whirled in a horizontal circle with constant angular velocity ω as shown in the figure. Find in terms of m, ω, l and g (i) the tension T in the string. (ii) the angle θ. (b) A student suggests a method to determine g by measuring the period t of rotation of the bob for various values of the angle θ. (i) Deduce an expression for t in terms of l, g and θ. (ii) Suggest a suitable graph that would enable you to determine the value of g. (iii) Discuss critically whether this is a suitable method to determine the value of g. Solution: (a) (i) The horizontal component of T, (T sin θ) provides the centripetal force. Using F= ma a = rω2 r = l sin θ T sin θ = mrω2 = m(l sin θ) ω2 T = mlω2 (ii) Vertical component of T, T cos θ = mg (mlω2 ) cos θ = mg cos θ = g lω2 θ = cos–1 ( g lω2) (b) (i) From cos θ = g lω2 ω = g l cos θ Period, t = 2π ω = 2π l cos θ g (ii) Period t = 2π l cos θ g t2 = 4π2 l cos θ g mg T θ ω mg T r θ ω 112 5

Physics Term 1 STPM Chapter 5 Circular Motion Comparing with the equation of a straight line y = mx, a suitable graph is t 2 against cos θ. Gradient of the graph, k = 4π2 l g Acceleration due to gravity, g = 4π2 l k (iii) This is not a suitable method to determine g because • It is difficult to measure accurately the angle θ. • It is difficult to fix θ at a constant value when measuring the period t. Example 5 An aircraft is flying at a constant speed of 180 m s-1 in a horizontal circle of radius 20 km. A plumb line, attached to the roof of the cabin is inclined at an angle φ to the vertical while the aircraft is turning. (a) What is the centripetal acceleration of the aircraft. (b) Draw a labelled diagram to show the directions of the forces and their resultant on the bob. Indicate the centre of the circle. (c) Calculate the value of the angle φ. (d) Show by means of a simple sketch of the cross-section of the aircraft and its cabin, how the plumb line is oriented with respect to the aircraft. Solution: (a) Centripetal acceleration = v 2 r = 1802 20 × 103 = 1.62 m s–2 (c) T sin φ = mv 2 r T cos φ = mg tan φ = v 2 rg = 1802 (20 × 103 ) × 9.81 φ = 9°23′ (b) • Centre of circle Tension T Weight mg Resultant force (d) • Centre of circle String Pendulum bob Gradient = k 0 cos θ t 2 113 5

Physics Term 1 STPM Chapter 5 Circular Motion Quick Check 4 3. A mass at the end of a string is set in motion so that it describes a circle in a horizontal plane. Which of the following diagrams shows correctly the resultant force F on the mass? A C B D 4. A particle P attached to the end of a string moves in a horizontal circle on the outer surface of a cone as shown in the figure. The string is inclined at an angle θ to the vertical. Half the apex angle of the cone is α. P If the tension in the string is T and the force on P by the surface of the cone is R, which of the following gives the centripetal force? A T sin θ + R cos α C T cos θ – R sin α B T sin θ – R cos α D T cos θ + R sin α 1. An aircraft is travelling at constant speed in a horizontal circle, centre Q. Each diagram below shows a tailview of the aircraft. The dotted line representing the wings and the circle representing the centre of gravity of the aircraft. Which of the following diagrams correctly shows the forces acting on the aircraft? A C Q Q B D Q Q 2. A pendulum hangs from the roof of a car in front of a passenger. The car travels along a circular arc bending to the left. Which of the following diagrams shows the position of the pendulum as seen by the passenger and the directions of the forces on the bob? A C B D 114 5

Physics Term 1 STPM Chapter 5 Circular Motion 5. The figure shows a mass m at the end of string moving in a horizontal circle of radius r with a speed v. The string is inclined at an angle θ to the vertical. The tension in the string is T. O Which of the following is the centripetal force? A T sin θ B T cos θ C T sin θ + mv 2 r D T sin θ – mv 2 r 6. The figure shows an aircraft flying in a horizontal circle about the point O. O (a) Obtain an expression for the centripetal force on the aircraft in terms of the weight W and the lift L. (b) Then, deduce the acceleration of the aircraft. Identify any additional symbol in your expression. 7. Use Newton’s Laws of motion to explain each of the following for a body moving in a circle with uniform speed. (a) Although the speed is uniform, the body accelerates. (b) Although the radius is constant, the body has an acceleration towards the centre of the circle. (c) Explain how the centripetal acceleration is produced for a conical pendulum. (d) A pendulum bob of mass 100 g hangs from the end of an inelastic string of length 50.0 cm. It moves in a horizontal circle of radius 40.0 cm, so that the centre of the circle is directly below the point of support of the string. (i) Draw a sketch to show the two forces on the pendulum bob. (ii) Find the resultant force on the bob. (iii) What is the period of revolution of the bob? 8. (a) Use Newton’s Laws of motion to explain why a body moving with uniform speed in a circle must experience a resultant force. What is the direction of this force? (b) An aircraft of mass 2.0 × 104 kg is moving at a constant speed of 0.20 km s–1 in a horizontal circle of radius 1.5 km. (i) What is the angular velocity of the aircraft? (ii) Find the magnitude and direction of the resultant force on the aircraft. (iii) Explain why a passenger in the aircraft experiences a resultant force. What is the direction of this force? 9. A sphere X moves in a horizontal circle on the inner surface of an inverted cone as shown in the figure below. The apex angle of the cone is 30o . X (a) Name and draw the directions of the forces on the sphere if the inner surface of the cone is (i) smooth, (ii) rough. Show also the direction of the resultant force on the sphere. (b) A sphere rotates on the smooth inner surface of the cone in a circular motion with an angular velocity is 5.0 revolutions per second. Calculate the radius of the circular motion. 115 5

Physics Term 1 STPM Chapter 5 Circular Motion Vertical Circular Motion 1. When an object at the end of a string is whirled in a vertical circle, the speed of the particle varies as it moves in circular motion. 2. Suppose v is the instantaneous velocity of the body when the string makes an angle θ with the vertical. (Figure 5.4) 3. The centripetal acceleration a = v 2 r towards O Using F = ma T – mg cos θ = mv 2 r Tension T = mv 2 r + mg cos θ When θ = 0, the object is at the lowest point, v is maximum and the tension T is maximum. Tmax = mv 2 r + mg When θ = π 2 rad, or 3π 2 , the string is horizontal, T = mv 2 r cos θ = 0 When θ = π rad, the object is at the highest point, the velocity v is minimum, and the tension is minimum. Tmax = mv 2 r – mg cos θ = –1 4. For the object to reach the highest point of the vertical circle, the string must be taut when the particle is at the highest point. That is Tmin  0 mv 2 r – mg  0 Centripetal force, mv 2 r  mg, weight v  gr The minimum velocity of the object at the highest point is gr . 5. Figure 5.5 shows a pail of water whirling in a vertical circle. The water in the pail does not drop because of the reaction of the pail on the water, R  0 mv 2 r – mg  0 (mg = W) mv 2 r  W Centripetal force  Weight 6. Similarly, a person on a roller coaster at the top of the corkscrew does not drop down because the centripetal force on the person is greater than his weight. O Figure 5.5 Figure 5.4 116 5

Physics Term 1 STPM Chapter 5 Circular Motion Example 6 A stone of mass m is attached to a string of length l, which will break if the tension in it exceeds Tmax. The stone is whirled in a vertical circle. (a) Draw diagrams to show the forces acting on the stone when it is (i) at the top, (ii) at the bottom, of the circle. (b) In what position the string is most likely to break? (c) What is the angular velocity in terms of m, l, Tmax and g when the string breaks? Solution: (a) Tension T Weight W O Tension T Weight W O (i) (ii) (b) The string is most likely to break when the stone is at the lowest point of the circle. (c) When the stone is at the lowest point, using F = ma a = lω2 T′ – mg = mlω2 W = mg When the string breaks, T ′ = Tmax Tmax – mg = mlω2 Angular velocity, ω = Tmax – mg ml Example 7 The figure shows a carriage in an amusement park ‘loop-the-loop’ in a vertical circle. The radius of the loop is 8.0 m and the total mass of the carriage and passenger is 120 kg. At the highest point, the carriage is travelling at 15 m s–1. Assume that friction may be neglected. (a) Calculate, (i) the centripetal acceleration. (ii) the force that the track exerts on the carriage when the carriage is at the highest point. 15 m s–1 8.0 m Carriage entering loop Carriage leaving loop O 117 5

Physics Term 1 STPM Chapter 5 Circular Motion (b) The carriage moves round and descends to the bottom of the loop. Calculate (i) the loss in potential energy when the carriage and passengers move from the top to the bottom. (ii) the speed of the carriage when leaving the loop. (c) The speed of the carriage when entering the loop must be above certain minimum value. Explain why. Solution: (a) (i) Centripetal acceleration, a = v 2 r = 152 8.0 = 28.1 m s–2 (ii) Suppose R = Force of the track on the carriage Using F = ma R + mg = mv 2 r R = 120(28.1 – 9.81) = 2.19 × 103 N (b) (i) Loss in potential energy = mgh = 120 × 9.81 × 16.0 J = 1.88 × 104 J (ii) Using the principle of conservation of energy Gain in kinetic energy = Loss in potential energy 1 2 mv1 2 – 1 2 mv2 = mgh v1 2 = v2 + 2gh = 152 + 2 × 9.81 × 16.0 v1 = 18.5 m s–1 (c) If the speed of the carriage enters the loop is below a certain value, the carriage would not be able to reach the highest point. Quick Check 5 R mg 1. The drum of a washing machine rotates about a horizontal axis with constant angular velocity w. The radius of the drum is r. When a piece of laundry of weight W is at the highest point P of the circular motion the force exerted by the drum on the laundry is R1 . The force is R2 when the laundry is at the lowest point Q. P Rotating drum Q R1 – R2 is A 2W C W – mrw2 B mrw2 D W + mrw2 118 5

Physics Term 1 STPM Chapter 5 Circular Motion 2. The bob of a simple pendulum is displaced to the point X and then released. When it reaches Y, the string breaks. X Y Which of the following shows the path of the bob as it falls to the ground? A Y C Y B Y D Y 3. A car of mass m moves at a constant speed v passes over a humpback bridge of radius of curvature r. The car remains in constant contact with the road. The reaction that the road exerts on the car when the car is at the top of the bridge is A mv 2 r C mg – mv 2 r B mg + mv 2 r D mv 2 r – mg 4. A particle is suspended at the end of an inextensible string of length l which is attached to a point A. It is projected from B with a velocity v perpendicular to AB which is just sufficient for it to reach the point C. C A B (a) Find the speed of the particle at C when the string is just taut. (b) Find the speed v which the particle is projected from B. 5. A pendulum bob of mass m is hung from an inextensible string of length L. The bob is pulled to the side until the string is horizontal and then released. Find (a) the speed of the bob, (b) its acceleration, (c) the tension in the string, when the bob is at the lowest point. 6. (a) A body moves at constant angular velocity ω in a circle of radius a. State its acceleration. (b) In a ride at an amusement park, two persons, each of mass 80 kg, sit in cages which travel at constant speed in a vertical circle of radius 8.0 m as shown in the figure. Each revolution takes 4.2 s. When a cage is at the top of the circle (position A), the person in it is upside down. For the person in cage A, calculate the magnitudes of (i) the angular velocity. (ii) the linear speed. (iii) t h e c e n t r i p e t a l acceleration. (c) (i) Draw a vector diagram to show the directions of the following forces acting on the person in cage A: The weight W of the person. The force F exerted by the cage on the person. (ii) Draw the corresponding diagram for the person in cage B. (iii) What is the value of the resultant of these forces at A and B? (iv) Explain why the person remain on the floor of the cage at the top of the circle. (v) State the position of the cage when the force it exerts on the person is maximum. Calculate the magnitude of this force. (d) Draw a vector diagram showing W, F and their resultant when the line joining the cages is horizontal. Numerical values are not required but the force vectors should be drawn so that they have approximately correct relative sizes. A B 119 5

Physics Term 1 STPM Chapter 5 Circular Motion Important Formulae 1. Angular velocity, w = dq dt , tangential velocity, v = rw 2. Centripetal acceleration, a = v 2 r = vw = rw2 3. Centripetal force, F = mv 2 r = mvw = mrw2 STPM PRACTICE 5 1. A car travels at a constant speed along a winding road as shown in the figure. The magnitude of its acceleration is greatest when the car is at D C B A Car 2. A coin is placed on a turntable at a distance of 0.20 m from the axis of rotation. The coefficient of static friction between the coin and turntable is 0.60. The speed of rotation of the turntable is increased. The coin slips on the turntable when the angular speed exceeds A 29 rad s–1 B 5.4 rad s–1 C 1.1 rad s–1 D 1.2 rad s–1 3. A particle is in uniform circular motion of radius r. The period of the circular motion is T. The centripetal force on the particle is proportional to A rT2 C r2 T B r T D r T2 4. The positions of two chidlren with the same mass riding on a merry-go-round are shown in the diagram below. Child 2 Child 1 If the merry-go-round rotates at constant angular speed, A both children will have the same linear speed. B child 1 will have a greater angular speed than child 2. C both children will have angular and centripetal accelerations. D child 1 will have a smaller centripetal acceleration than child 2. 5. A particle of mass 25 g is tied to an elastic string of natural length 0.40 m and has a force constant of 50 N m–1. The particle is whirled in horizontal circle on a smooth table. What is the angular speed of the particle when the radius of the circle is 0.45 m? A 0.47 rad s–1 C 22 rad s–1 B 15 rad s–1 D 25 rad s–1 6. A bus of mass m moves around a curved track that is banked at an angle θ as shown in the figure. The centre of the circular track is O. The normal reaction of the track on the bus is R, and F is the total lateral frictional force between the tyres of the bus and the track. 120 5

Physics Term 1 STPM Chapter 5 Circular Motion R F O mg θ Which expression represents the centripetal force on the bus? A F + mg sin θ B F + mg cos θ C F cos θ + R sin θ D F sin θ + R cos θ 7. A car travels round a bend which is banked at an angle of 20°. The radius of curvature of the bend is 75.0 m. What is the speed of the car so that the lateral friction between the tyres of the car is balanced by the component of the weight of the car? A 16 m s–1 C 25 m s–1 B 20 m s–1 D 27 m s–1 8. A body of mass 5.0 kg rotates at a constant speed in a horizontal circle of radius 2.0 m. The time taken for a complete revolution is 4.0 s. What is the resultant force on the body? A 5 8 π2 N C 5π2 N B 5 2 π2 N D 10π2 N 9. An aircraft is flying in a horizontal circle of radius 80 km at a constant speed of 300 m s–1. What is the ratio of the lift on the aircraft to its weight? A 0.11 C 1.01 B 0.99 D 8.80 10. A particle moves with constant speed v in a circle of radius r. What is the magnitude of its tangential acceleration and centripetal acceleration? Tangential Centripetal acceleration acceleration A 0 rv2 B 0 v 2 r C rv2 0 D v 2 r 0 11. A stone of mass 0.50 kg at the end of a string swings in a vertical circle of radius 0.20 m. Its speed at the top of the circle is 3.0 m s–1. 3.0 m s–1 String m = 0.50 kg r = 0.20 m T What is the tension T in the string? A 4.9 N C 22.5 N B 17.6 N D 27.4 N 12. A corner on a racing track has a radius of curvature of 125 m. What is the minimum coefficient of static friction between the tyres of a car and the track so that a car does not skid when taking the bend at 108 km h–1? 13. (a) The diagram shows a particle moving with a constant speed v in a circle of radius r. At time t = 0, the particle is at A. After a short time interval ∆t, the particle is at B and the angular displacement is ∆θ. A B V r (i) Draw a vector diagram to show the change in velocity ∆v. (ii) Show that ∆θ = v∆t r . (iii) Hence show that the acceleration is a = v 2 r . Why is the acceleration known as centripetal acceleration? (b) A mass of 50g is attached to the end of a string of length 0.80 m, and the other end of the string is fixed to a disc of radius 0.12 m. When the disc rotates about a vertical axis with an angular velocity ω, the string makes an angle of 40° with the vertical as shown in the diagram. 121 5

Physics Term 1 STPM Chapter 5 Circular Motion 40º 0.80 m ω Rotating disc 0.12 m Calculate (i) tension in the string, (ii) angular velocity, ω of the disc. 14. A mass m hangs from the end of an inelastic string of length l. The mass is launched with a speed u from the lowest point A in order to reach the point P and complete the circular motion. P O A (a) State the condition for the mass to reach the highest point P. (b) Deduce the minimum speed of the mass at P in terms of u. (c) Discuss the motion of the mass if it is launched from A with a speed which is less than u. 15. (a) Define acceleration. Explain why an object moving with uniform speed in a circle is accelerated. (b) A wheel of radius r rotates with constant angular velocity ω, followed by constant angular acceleration ω . . Deduce expressions in terms of r, ω, ω . and any other quantities for (i) the tangential acceleration, and (ii) the radial acceleration, for each stage of the motion for a point on the rim of the wheel. State whether each of the acceleration is constant. (c) A mass is attached at the end of a string which hangs from roof of a car. Describe and explain the position of the mass as seen by a passenger in the car when (i) the car travels along a straight horizontal road with an acceleration which increases uniformly. (ii) the car travels with constant speed of 60 km h–1 in a horizontal circular track of radius 50 m. 1 1. B 2. D 3. C 4. D 5. (a) (i) θ = π 2 rad (ii) θ = 2π rad (b) ω = π/2 5.0 rad s–1 = 0.314 rad s–1 (c) v = rω = 7.85 × 10–2 m s–1 6. ω = 2π 24 × (60 × 60) rad s–1 = 2.26 × 104 rad s–1 2 1. D : a = rω2 = (2.0)( 2π 0.50) 2 m s-2 2. a = rω2 a2 a1 = (33 1 3 ) 2 452 = 0.549 m s-2 3. a = rω2 = (0.20)( 3000(2π) 60 ) 2 = 1.97 × 104 m s–2 ANSWERS 122 5

Physics Term 1 STPM Chapter 5 Circular Motion 3 1. A 2. C 3. B 4. B 5. C 6. D 7. (a) ω = θ t = 7.27 × 10–5 rad s–1 (b) v = rω = 465 m s–1 (c) a = vω = 3.39 × 10–3 m s–2 (d) F = ma = 2.44 N 8. r > 5.0 cm, mrω2 > limiting static friction F = mrω2 = 0.40 N 4 1. C 2. D 3. B 4. B 5. A 6. (a) F = L2 – W2 (b) a = F m = g [ L2 W2 – 1] 1 —2 7. (a) First law : Direction of velocity changes. Hence, a resultant force acts. Second law : Force produces an acceleration. (b) Second law : Direction of acceleration is the same as the direction of the resultant force. Resultant force is towards the centre of circle. (c) Horizontal component of the tension in the string provides the centripetal force. (d) (i) mg T θ (ii) Resultant force = 1.308 N toward centre of circle (iii) Period T = 1.099 s 8. (a) Refer to page 108 (b) (i) 0.133 rad s–1 ω = v r (ii) 5.31 × 105 N F = mrω2 (iii) Centripetal force enables passenger to move in a circle with the aircraft. Towards centre of the circles. 9. (a) (i) F W N X (ii) F W N X P W : weight N : normal reaction P : friction F : resultant force Centripetal force F L W (b) r = 3.71 cm W = mg N 15° F = mr tan 15o = mg mrω2 5 1. A R1 – W = mrω2 , R2 + W = mrω2 2. C 3. C 4. (a) T + mg = mv2 r When T = mv2 l – mg = 0 v = gl (b) v = 5gl 1 2 mv2 – 1 2 mu2 = mg(2l) 5. (a) v = 2gl Use 1 2 mv2 = mgh (b) a = 2g a = v2 r (c) T = 3mg T = mg + mv2 r 6. (a) Acceleration = aω2 (b) (i) 1.50 rad s–1 ω = 2π T (ii) 12.0 m s–1 v = rω (iii) 18.0 m s–2 a = rω2 (c) W F W F (i) (ii) (iii) 1 440 N, 1 440 N (iv) mrω2 > mg (v) At B, Fmax = 2 220 N (d) Resultant F W STPM Practice 5 1. C : Centripetal acceleration = v2 r . At C, r smallest. 2. B : Coin slips when friction, µmg < mrω2 ω > µg r = (0.60)(9.81) 0.20 = 5.4 rad s–1 3. D : F = mrω2 = mr( 2p T )2 = 4p2 mr T2 4. D : Centripetal acceleration = rω2 123 5

Physics Term 1 STPM Chapter 5 Circular Motion 5. B: Extension, x = (0.45 – 0.40) m = 0.05 m F = m(0.65)ω2 = kx ω2 = (50)(0.05) (0.025)(0.45) ω =15 rad s–1 6. C: Centripetal force = sum of components of forces towards O. 7. A: Since lateral frictional force balances the component of the weight, there is no resultant force along the incline. 20° mg R1 R2 (R1 + R2 ) cos 20° = mg (R1 + R2 ) sin 20° = mv2 r tan 20° = v2 rg v = (75.0)(9.81) tan 20° m s–1 = 16 m s–1 8. B : F = mrω2 = (5.0)(2.0)(( 2p 4.0) 2 ) N = 5p2 2 9. C : L W W L F L2 = W2 + F2 , W = mg, F = mv2 r L2 W2 = 1 + F2 W2 = 1 + (mv2 /r mg ) 2 L W = (1 + v4 r2 g2 )1/2 = (1 + 3004 (80 × 1003 )2 (9.81)2 ) 1/2 = 1.01 10. B : Tangential acceleration = 0 because speed is constant. 11. B : F = ma, T + mg = mv2 r T = (0.50)(3.0)2 0.20 – (0.50)(9.81) N = 17.6 N 12. Car does not skid if µmg > mv2 r µ > v2 rg = (108 × 103 /3600)2 (125)(9.81) = 0.73 13. (a) (i) V V V (ii) AB = v∆t ∆θ = AB r = v∆t r (iii) ∆v = v∆θ = v( v∆t r ) a = ∆v ∆t = v2 r Reason: direction of acceleration is towards the centre of the circle. (b) (i) r T mg 40º T cos 40° = mg T = (0.050)(9.81) cos 40° N = 0.64 N (ii) T sin 40°= mrω2 ω = (0.64)sin 40° (0.050)(0.12 + 0.80 sin 40°) rad s–1 = 3.6 rad s–1 14. (a) Tension T  0 (b) v = u 5 T = mv2 r – mg = 0 1 2 mv2 + mg(2l) = 1 2 mu2 (c) The string becomes slack. The mass leaves the circle and travels as a projectile until the string becomes taut again. 15. (a) Acceleration : rate of change of velocity. Direction of velocity constantly changing. (b) ω = constant: (i) 0 (ii) rω2 (constant) ω . constant (i) rω (constant) (ii) r (ω + ω. t)2 (not constant) (c) (i) The angle θ which the plumb line makes with the vertical keeps increasing. (ii) mg T θ a θ = 29.5° 124 5

CHAPTER Concept Map GRAVITATION 6 INTRODUCTION 1. Gravitational force is one of the fundamental forces. 2. Gravitational force is the force of attraction between bodies in the universe. 3. Gravitation force is not only experienced on the Earth, but also acts between the planets and the Sun, and between the Earth and the Moon. Bilingual Keywords Escape velocity: Halaju lepas Field strength: Kekuatan medan Geosynchronous: Geo segerak Gravitation: Kegravitian Satellite: Satelit Universal: Alam semesta Weightlessness: Ketiadaberatan Gravitational Field g = GM r 2 GM = g0R2 Satellite • r v v = GM r T 2 ∝ r 3 • Weightlessness Newton’s Law of Gravitation F = G m1m2 r 2 Escape Velocity v = 2GM R Gravitational Potential V = – GM r U = – GMm r g = – dV dr Gravitation 125

Physics Term 1 STPM Chapter 6 Gravitation 6.1 Newton’s Law of Universal Gravitation Students should be able to: • state Newton’s law of universal gravitation and use tha formula F = GMm r2 Learning Outcomes 1. The force between two bodies of mass m1 and m2 is attractive in nature. The gravitational attraction F is directly proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance between them. The gravitational force acts along the line joining the centres of mass of the two bodies. m1 m2 F r Figure 6.1 Gravitational attraction, F = Gm1m2 r2 where G is a constant known as universal gravitational constant, G = 6.67 × 10–11 kg–1 m3 s–2. 2. The gravitational attraction between two bodies form an action-reaction pair as in Newton’s third law of motion. Example 1 A rocket travels from the Earth to the Moon. Calculate the distance of the rocket from the Earth where it is under zero gravitational force. Consider only the gravitational attraction of the Earth and Moon. (Mass of Earth = 6.0 × 1024 kg, mass of Moon = 7.4 × 1022 kg. Distance of Moon from the Earth = 3.8 × 108 m). Solution: ME MM F1 F2 m x Moon Rocket Earth Suppose x = distance of rocket from Earth where the gravitational force on it is zero. Gravitational force on rocket by Earth, F1 = gravitational force on rocket by Moon, F2 GmME x2 = GmMM (3.8 × 108 – x)2 ME = mass of Earth ( x 3.8 × 108 – x ) 2 = ME MM = 6.0 × 1024 7.4 × 1022 MM = mass of Moon x 3.8 × 108 – x = 9 m = mass of rocket x = 3.42 × 108 m 2013/P1/Q19 126 6

Physics Term 1 STPM Chapter 6 Gravitation Quick Check 1 C d m m + M D d ( m m + M ) 3. The distance of the Earth from the Sun is 1.5 × 1011 m. The mass of the Sun is (3.24 × 105 ) ME where ME = mass of the Earth. Find the distance from the Earth along the line joining the Earth to the Sun where the gravitational attraction of the Sun is equal that of the Earth. 6.2 Gravitational Field Students should be able to: • explain the meaning of gravitational field • define gravitational field strength as force of gravity per unit mass • use the equation g = GM r 2 for a gravitational field Learning Outcomes 1. A gravitational field is a region where gravitational force acts on a body. For example, a body in the Earth’s gravitational field is attracted to the Earth. 2. The gravitational field strength, g at a point in a gravitational field is the gravitational pull per unit mass on a body at that point. 3. Gravitational field strength, g is a vector quantity and its SI unit is N kg–1. 4. Figure 6.2 shows a mass m at the point P in the Earth’s gravitational field at a distance r (r  R) from the Earth of radius R and mass M. The gravitational pull F of the Earth on the mass at P is F = – GMm r 2 The negative sign shows the direction of F is opposite to the positive direction. Hence, the gravitational field strength g = F m g = GM r 2 P g Earth Positive direction m R M r Figure 6.2 Gravitational field strength, g Exam Tips g0R2 = GM is an important expression which is used frequently. 1. Which of the following is the SI unit for the gravitational constant G? A m s–2 C m3 kg–1 s–2 B N m–2 kg–2 D J m kg–1 2. Two point masses m and M are separated by a distance d. A third point mass along the line joining m and M experiences no gravitational force. What is the distance of the third point mass from m? A d( M m ) B d m M 2014/P1/Q7 127 6

Physics Term 1 STPM Chapter 6 Gravitation 5. The weight of a body is the gravitational force on the body. Weight = gravitational force (mass) × (acceleration due to gravity) = (mass) × (gravitational field strength) Hence acceleration due to gravity = gravitational field strength That explains why the symbol, g for the acceleration due to gravity is also used for gravitational field strength. 6. If the acceleration due to gravity on the surface of the Earth is g0, then g0 = – GM R2 (r = R, radius of the Earth) Magnitude of g0 = GM R2 g0R2 = GM 7. This expression can be used to calculate the mass M of the Earth. Mass of the Earth, M = g0R2 G (g0 = 9.81 m s–2, R = 6.40 × 106 m) = (9.81)(6.40 × 106 )2 6.67 × 10–11 kg = 6.02 × 1024 kg 8. Mean density of the Earth, ρ = M V (V = 4 3 πR2 ) = (g0R2 )/G (4/3)πR3 = 3g0 4πGR = 3(9.81) 4π(6.67 × 10–11)(6.40 × 106 ) kg m–3 = 5.49 × 103 kg m–3 Example 2 The gravitational field strength on a planet of diameter d is g. What is the gravitational field strength on another planet of diameter 2d and having the same mean density? Solution: Using g = GM R2 , M = 4 3 πR3 ρ = G 4 3 πR3 ρ R2 = 4 3 πGρR g ∝ R R = d 2 ∝ d For the second planet, diameter = 2d g1 = 2g 128 6

Physics Term 1 STPM Chapter 6 Gravitation Example 3 The mass of the Earth is 81 times the mass of the Moon, and the radius of the Earth is 3.7 times of the Moon. If the acceleration of free fall on the Earth is 9.81 m s–2, what is its value of the Moon? Solution: Using g = GM R2 on the Earth .............................. ① and g1 = GM1 R1 2 on the Moon .............................. ② g1 g = M1 M . R2 R1 2 M1 M = 1 81, R R1 = 3.7 g1 = 1 81 . (3.7)2 × 9.81 On the Moon, g1 = 1.66 m s–2 Example 4 A planet of mass M and radius r rotates about its axis with an angular velocity large enough for substances on its equator to be just able to stay on its surface. Find in terms of M, r and G the period of rotation of the planet. Solution: If m = mass on the equator. When the planet rotates, the mass m rotates with the same angular velocity. Since the mass m is just able to stay on the surface on the planet, then the centripetal force mrω2 = GMm r 2 , the gravitational force. ω = ( GM r 3 ) 1 —2 Period, T = 2π ω = 2π r3 GM Quick Check 2 ω M m r 1. A body is released close to the surface of a planet of mass M and radius R. It falls freely under the gravitational attraction of the planet. What is its acceleration due to gravity? A MG R C R MG B MG R2 D R2 MG 2. What is the mean density of a planet which is spherical with radius r? (g = acceleration due to gravity on the planet) A 4πrG 3g C 4πg 3rG B 4πrg 3G D 3g 4πrG 129 6

Physics Term 1 STPM Chapter 6 Gravitation 3. A star of mass M and radius R rotates rapidly so that material on its equator is just able to remain on its surface. If G is the gravitational constant, the angular velocity of the star is A GM R3 C R3 GM B GM R D G R 4. The acceleration due to the gravity on the surface of the Moon is 1 6 that on the surface of the Earth. The mean density of the Moon is 3 5 of the Earth. What is the value of the ratio of the radius of the Moon to the radius of the Earth? A 0.53 B 0.31 C 0.28 D 0.10 5. If the values of the acceleration due to gravity on the surface of two planets are the same, which of the following quantities is the same for both the planets? A Mass C Mass Radius2 B Mass Radius D Mass Radius3 6. A star rotates at 1.0 revolution per second about its axis. The radius of the star is 20 km and the material on its surface is only just able to remain on its surface. What is the mass of the star? 7. Suppose that the Earth is a sphere of radius 6 × 106 m, and the acceleration of free fall on its surface is 10 m s–2, estimate the mass of the Earth. 8. The Earth of mass M1 revolves around the Sun of mass M2 in a circular orbit of radius R1. The Moon of mass m revolves around the Earth in a circular orbit of radius R2. Deduce an expression for the acceleration of the Earth towards the Sun (a) during an eclipse of the Sun (b) during an eclipse of the Moon in terms of M1, M2, R1 and R2. (c) What is the different in the accelerations in (a) and (b) above? Variation of g with Altitude 1. For a mass m on the surface of the Earth, mg0 = GMm R2 M = mass of Earth g0 = GM R2 R = radius of Earth 2. When the mass m is at a distance r from the centre of the Earth (r > R), mg = GMm r2 g = GM r2 GM = g0R2 = gR2 r 2 = ( R2 r 2 ) g0 g ∝ 1 r 2 ∝ 1 (R + h)2 r = R + h h = altitude O M R r h m Earth Figure 6.3 130 6

Physics Term 1 STPM Chapter 6 Gravitation Example 5 What is the acceleration due to gravity at a height 2R from the surface of the Earth? (R = radius of Earth) Solution: g ∝ 1 r 2 On the Earth’s surface, 9.81 ∝ 1 R2 .............................. a r = R At a height h = 2R, r = (R + 2R) = 3R g1 ∝ 1 (3R)2 .............................. b b a : g1 9.81 = R2 9R2 g1 = 1 9 × 9.81 = 1.09 m s–2 Variation of g with Depth below The Earth’s Surface 1. On the surface of the Earth, g0= GM R2 , M = 4 3 πR3 ρ = 4 3 πGRρ .............................. a 2. When the mass m is below the surface of the Earth, r < R, the gravitational pull is due to the sphere of radius r. Hence, g = 4 3 π Grρ .............................. b b a : g g0 = r R g = ( r R ) g0 g ∝ r (r < R) 3. The variation of g with distance r from the centre of the Earth can be summarised by the graph shown in Figure 6.5. 0 Acceleration due to gravity, g g0 g r R r g 1 r 2 Figure 6.5 O M’ R r M Figure 6.4 131 6

Physics Term 1 STPM Chapter 6 Gravitation Quick Check 3 1. A body of mass m was released from the position just above the surface of the Moon of mass M and radius r. What is its acceleration of free fall? A GmM r 2 C GMm r B Gm r D GM r 2 2. The gravitational field strength at a point outside a planet at 8.44 × 106 m from its centre is 1.0 N kg–1. If the planet is a sphere, and the acceleration due to gravity on its surface is 9.0 m s–2, what is the radius of the planet? A 1.04 × 105 m C 2.81 × 106 m B 9.38 × 105 m D 4.22 × 106 m 3. The Earth’s gravitational field on its surface is 10 N kg–1 and 5 N kg–1 at a distance x from its centre. Which of the following gives the value of x in terms of the radius R of the Earth? A x = 1 2 R C x = 2R B x = 2 R D x = 4R 4. The Earth is considered as a uniform sphere rotating about its axis. The weight of a body on the equator compared to its weight at the N-pole is A greater, because points on the equator move at a greater speed than the N-pole. B greater, because the weight at the equator is the sum of the gravitational pull and the centripetal force. C smaller, because the gravitational pull must provide the centripetal force and the weight. D smaller, because the gravitational pull at the N-pole is greater than at the equator. 5. A satellite is at a height of 144 km from the Earth’s surface. The radius of the Earth is 5 760 km. The gravitational force on the satellite compared to its weight W on the Earth’s surface is A 1.10 W C 0.95 W B 1.05 W D 0.90 W 6. The satellite Measat 1 is at a height h above the surface of the Earth. If the radius of the Earth is r, and the acceleration due to gravity on the surface of the Earth is g, the period of orbit of the satellite is A 2π r + h g C 2π (r + h)2 gr B 2π r 2 g(r + h) D 2π (r + h)3 gr 2 7. The gravitational pull on a satellite on the surface of the Earth of radius R is W. What is the gravitational pull on the satellite when it is at a height R 50 above the Earth? A 1.02 W C 0.96 W B 0.98 W D zero 8. Which of the following is true about the gravitational field strength outside the surface of the Earth? A Uniform for all points outside the Earth’s surface. B Directly proportional to the distance from the centre of the Earth. C Inversely proportional to the distance from the centre of the Earth. D Inversely proportional to the square of the distance from the centre of the Earth. 9. The period of a simple pendulum on the Earth is 2.0 s. What is the period of this pendulum on the Moon? (Mass of Moon = 7.35 × 1022 kg, mass of Earth = 6.0 × 1024 kg, radius of Moon = 1 720 km, radius of Earth = 6.4 × 106 m) 10. The period of a simple pendulum on the Equator is 1.00 s. What is its period at the south pole? (Radius of Earth = 6.4 × 106 m, g at the south pole is 9.81 m s–2) 11. (a) Show that the gravitational field strength at a height h above the surface of a planet of mass M and radius R is g = GM (R + h)2. (b) The radius of the Earth is 3.7 times the radius of the Moon, and the mass of 132 6

Physics Term 1 STPM Chapter 6 Gravitation the Earth is 81 times the mass of the Moon. Distance of the Moon from the Earth is 3.84 × 108 m. The gravitational field strength on the Earth’s surface is 9.81 N kg–1. (i) Calculate the gravitational field strength on the surface of the Moon. (ii) Along the line that joins the Earth to the Moon, there is a point where the resultant gravitational field strength is zero. What is the distance of this point from the centre of the Earth. 12. (a) The radius of the Moon’s orbit around the Earth is 3.82 × 108 m and the period is 2.36 × 106 s. (i) What is the acceleration of the Moon? (ii) Explain why the Moon does not fall and hit the Earth? (b) By comparing the acceleration of the Moon with the acceleration due to gravity on the Earth’s surface, show that the Moon’s acceleration is consistent with Newton’s law of gravitation. (Radius of Earth = 6.36 × 106 m) 6.3 Gravitational Potential Students should be able to: • define the potential at a point in a gravitational field • derive and use the formula V = – GM r • use the formula for potential energy U = – GMm r • show that ∆U = mg∆r = mgh is a special case of U = – GMm r for situations near to the surface of the Earth • use the relationship g = – dV dr • explain, with graphical illustrations, the variations of gravitational field strength and gravitational potential with distance from the surface of the Earth Learning Outcomes 1. The vector quantity associated with a point in a gravitational field is the gravitational field strength, E. 2. There is a scalar quantity associated with a point in a gravitational field. It is the gravitational potential, V. 3. The gravitational potential, V of a point in a gravitational field is the work done per unit mass by the pull of gravity to bring a body from infinity to that point. P Earth m F x M Figure 6.6 2008/P1/Q8, 2009/P1/Q9, 2010/P1/Q9, 2010/P2/Q2, 2011/P1/Q9, 2015/P1/Q7, 2017/P1/Q10 133 6

Physics Term 1 STPM Chapter 6 Gravitation 4. The gravitational pull of the Earth on a mass m at a distance x from the Earth, F = GMm x2 . The work done by the force F to bring the mass through a small distance dx closer to the Earth is dU = F dx = GMm x2 dx To bring the mass m from infinity to a point distance r from the Earth, The work done, U = ∫∞ r dU = ∫∞ r GMm x2 dx = – [GMm x ] r ∞ = – GMm r ............................... a Hence, the gravitational potential V at a distance r from the Earth, V = U m V = – GM r 5. The gravitational potential at infinity is zero. 6. Figure 6.7 shows the variation of gravitational potential V, and gravitational field strength g with distance r from the Earth for r ≥ R, radius of the Earth. r R V, g g = – GM– r 2 V = – – GM r 0 Figure 6.7 Gravitational Potential Energy, U 1. The gravitational potential energy, U of a mass m at a point in a gravitational field is the work done by the gravitational pull to bring the mass from infinity to the point. 2. Equation a above gives the gravitational potential energy U of a mass m at a distance r from the Earth of mass M. Gravitational potential energy, U = – GMm r For a mass m on the surface of the Earth, r = R, radius of Earth, gravitational potential energy, U = – GMm R Exam Tips Do not forget the negative sign for (a) the gravitational potential, V = – GM r , and (b) the gravitational potential energy, U = – GMm r 134 6

Physics Term 1 STPM Chapter 6 Gravitation 3. The relation between the gravitational potential energy and (i) gravitational potential V is U = mV (ii) gravitational force F is F = – dU dx or U = – ∫∞ r F dx Similarly, gravitational field strength g is related to gravitational potential V by g = – dV dx V = – ∫∞ r g dx Example 6 The figure on the right shows a section of the Earth’s gravitational field. The broken lines represent the equipotential surfaces which are perpendicular to the field lines. (a) (i) The gravitational potential at which of the points P or Q is higher? Explain your answer. (ii) Calculate the change in the potential energy of a spacecraft of mass 4 000 kg that travels from P to S. (b) The equipotential surfaces for a difference of 0.5 × 107 J kg–1 are not spaced uniformly. Calculate (i) the distance PQ, (ii) the distance QR, if mG = 4.0 × 1014 N m2 kg–1 where m = mass of Earth, G = gravitational constant. Then, explain why the equipotential surfaces are not spaced evenly. Solution: (a) (i) VP – VQ = (–5.0 × 107 ) – (–5.5 × 107 ) J kg–1 = 5.0 × 106 J kg–1 > 0 Hence, the gravitational potential VP > VQ. (ii) Change in the gravitational potential energy = m(VS – VP) = 4 000(–6.0 × 107 – (–5.0 × 107 )) J = –4.00 × 1010 J (b) If the distance of P from the centre of the Earth is rP , Using VP = – Gm rP ∴ rP = –4.0 × 1014 –5.0 × 107 = 8.00 × 106 m Similarly, rQ = –4.0 × 1014 –5.5 × 107 = 7.27 × 106 m and rR = –4.0 × 1014 –6.0 × 107 = 6.67 × 106 m P Earth –5.0 x 107 J kg–1 –5.5 x 107 J kg–1 –6.0 x 107 J kg–1 Q R S PQ = rP – rQ = (8.00 – 7.27) × 106 m = 7.30 × 105 m QR = (7.27 – 6.67) × 106 m = 6.00 × 105 m The equipotential surfaces are not spaced evenly because although (VP – VQ) = (VQ – VR), PQ > QR 135 6

Physics Term 1 STPM Chapter 6 Gravitation Example 7 A stationary object is released from a point P at distance 4R from the centre of the Earth which has radius R and mass M. What is the speed of the object when it hits the Earth’s surface? Solution: Using the principle of conservation of energy, Gain in kinetic energy = Loss in gravitational potential energy 1 2 mv 2 = – GMm 4R – (– GMm R ) = 3GMm 4R v = 3GM 2R Difference in Gravitational Potential Energy 1. When a body of mass m is lifted through a height h << R, radius of the Earth, its gravitational potential energy increases by mgh. 2. This is consistent with the concept of gravitational potential energy. 3. When the body is on the Earth’s surface, its gravitational potential energy U0 = – GmM R M = mass of Earth When the body is at height h << R, its gravitational potential energy U1 = – GmM (R + h) 4. Increase in gravitational potential energy U1 – U0 = – GmM (R + h) – (– GmM R ) = GmM R – GmM R(1 + h R ) = GmM R [1 – 1 1 + h R ] = GmM R [1 – (1 + h R ) –1 ] = GmM R [1 – (1 – h R + ...)] (neglecting terms in h2 R2 and higher because h << R). = GmM R2 h = mgh P 136 6

Physics Term 1 STPM Chapter 6 Gravitation 1. A meteor of mass 500 kg is attracted by the Earth’s gravitational field. When it moves from a point Q to another point P which is nearer the Earth, the work done by the Earth’s gravitational field is –6.0 × 109 J. The gravitational potential at the point P is –5.0 × 107 J kg–1. What is the value of the gravitational potential at the point Q? A –1.2 × 107 j kg–1 C –5.8 × 107 j kg–1 B –3.8 × 107 j kg–1 D –6.2 × 107 j kg–1 2. Which of the following graphs show correctly the variation of the gravitational force F on a point mass, and the gravitational potential energy U with the distance r from the Earth? A B C D 3. A satellite is shifted from a stable orbit to another orbit which is higher. Which of the following quantities increases? A Gravitational force B Gravitational potential energy C Linear speed D Centripetal acceleration 4. P and Q are two points at distance x (x > R, radius of the Earth) and 2x respectively from the centre of the Earth. The gravitational potential at P is –800 kJ kg–1. What is the work done to take a 1 kg mass from P to Q? A 400 kJ C 800 kJ B 600 kJ D 2 400 kJ 5. Which of the following graphs best represents the relation between ∆U, the change in the gravitational potential energy of a shell fired from a gun, and x, the height of the shell above the surface of the Earth? A C B D 6. The graph on the right shows the variation of the gravitational potential V with the distance x from the centre of a planet. What does the gradient at point P on the curve represent? A The gravitational potential energy B The gravitational field strength C The gravitational pull D The gravitational potential 7. Two points P and Q are at distance R and 2R from the centre of the Earth. The gravitational potential at P is –12 MJ kg–1. P Quick Check 4 137 6

Physics Term 1 STPM Chapter 6 Gravitation P Q What is the gain in the kinetic energy of a mass of 5 kg, which is released from rest at point Q, when it reaches point P? A 6 MJ C 30 MJ B 12 MJ D 60 MJ 8. A stationary object is released from the point P with distance 3R from the centre of the Moon of radius R and mass m. P Find an expression in terms of G, m and R for the speed of the object when it hits the Moon’s surface. 9. The distance between a body A of mass 8 000 kg and a body B of mass 6 000 kg is 0.25 m. (a) Calculate the gravitational field strength due to the two bodies at a point P with distance 0.20 m from A and 0.15 m from B. (b) What is the gravitational potential at the point P due to the two bodies? 10. The gravitational potential due to the Earth at various distances from the Earth’s surface is given in the table below. Distance from the Earth’s surface/m Gravitational potential/MJ kg–1 0 –62.72 390 000 –59.12 400 000 –59.03 410 000 –58.94 Infinity 0 (a) If a satellite of mass 700 kg falls from a height of 400 000 m to the Earth’s surface, what is the loss in gravitational potential energy? (b) Deduce a value for the Earth’s gravitational field strength at a height of 400 000 m. 11. The gravitational potential V at a point in the Earth’s field is given by the equation V = – GM r r > R, radius of Earth where M = mass of the Earth and r = distance of the point from the centre of the Earth. (a) Explain (i) the meaning of gravitational potential. (ii) why the potential has a negative value. (b) Calculate the gain in the potential energy of a satellite of mass 200 kg between launching and when it is in orbit at a height of 1.2 × 106 m above the Earth’s surface. (R = 6.40 × 106 m) 6.4 Satellite Motion in a Circular Orbit Students should be able to: • solve problems involving satellites moving in circular orbits in a gravitational field • explain the concept of weightlessness Learning Outcomes 1. The Earth has only one natural satellite. Now, there are many artificial satellites orbiting around the Earth. 2. The first artificial satellite, Sputnik 1 was launched in 1957. Malaysia launched her first satellite, Measat 1 in January 1996 and Measat 2 in November 1996. 3. To launch an artificial satellite, it is first brought to the required height using a multi-stage rocket. 2009/P1/Q8, 2014/P1/Q8, 2016/P1/Q5,6,19 Earth Satellite v M r Figure 6.8 Satellite system of the Earth 138 6

Physics Term 1 STPM Chapter 6 Gravitation 4. The satellite is then fired in a horizontal direction relative to the Earth’s surface with a velocity v given by: F = GMm r 2 = mv 2 r Velocity of satellite, v = GM r where M = mass of Earth r = radius of satellite orbit 5. For a satellite orbiting close to the Earth, its radius of orbit r is approximately equal to the radius of the Earth R. Hence v = GM R (r = R) = g0R2 R (GM = g0R2 ) = g0R = 9.81 × 6.4 × 106 R = 6.4 × 106 m = 8 × 103 m s–1 = 8 km s–1 6. The period of revolution, T = 2πr v = 2π × (6.4 × 106 ) 8 × 103 = 5 × 103 s ≈ 83 minutes Satellite Satellite r Earth E v E r √ r v Earth GM GM r √ (a) (b) Figure 6.9 7. Figure 6.9(a) shows what happens when the satellite is launched with a velocity greater than required, v > GM r . The orbit of the satellite is no longer a circle, but an ellipse with the Earth at the near focus of the ellipse. 8. If the satellite is launched with a velocity less than that required, v < GM r , and the satellite does not crash into the Earth, the orbit is also elliptical but with the Earth at the far focus of the ellipse. (Figure 6.9(b)). 139 6

Physics Term 1 STPM Chapter 6 Gravitation Geosynchronous Satellite Figure 6.10 Geosynchronous satellite 1. A geosynchronous satellite appears to remain stationary, relative to points on the Earth. 2. This type of satellite is used for telecommunication purposes. 3. By remaining in one fixed position relative to the Earth, transmitters on receivers on the ground can be pointed at them, without the hassle of having to be adjusted every now and then. 4. For a satellite to have a geosynchronous orbit, • the satellite must orbit the Earth in the same direction as the rotation of the Earth about its axis. • the period of the satellite orbit must equal the period of rotation of the Earth about its axis, that is 24 hours or 8.64 × 104 s. • the orbit must be above the equator. The plane of the satellite orbit contains the plane of the equator. • the radius of the orbit r is calculated as follows GMm r 2 = mrω2 = mr( 2π T ) 2 r3 = GM( T2 4π2 ) = g0R2 T2 4π2 GM = g0R2 = 9.81 × (6.4 × 106 )2 (8.64 × 104 )2 4π2 r = 4.24 × 107 m M = mass of Earth m = mass of satellite ω = 2π T INFO How Satellite TV Works? 140 6

Physics Term 1 STPM Chapter 6 Gravitation Relation between Period and Radius of Orbit 1. Figure 6.11 shows a satellite of mass m in circular orbit of radius r about the Earth of mass M. 2. The centripetal force required for the circular motion of the satellite is the gravitational attraction of the Earth on the satellite F = GMm r 2 = mrω2 ω = angular velocity of planet = mr( 2π T ) 2 = 2π T , T = period T2 r3 = 4π2 GM = constant That is T2 r3 = constant 3. The value of the constant depends on the mass at the centre of the circular orbit, that is the mass of the Earth. 4. The Moon is the only natural satellite of the Earth with an orbital radius of 3.8 × 108 m and period of 2.4 × 106 s. This information can be used to estimate the mass of the Earth using the expression T2 r3 = 4π2 GM . Mass of the Earth, M = 4π2 r3 GT2 = 4π2 (3.8 × 108 )3 (6.67 × 10–11)(2.4 × 106 )2 kg = 6 × 1024 kg 5. Figure 6.12 shows two satellites in circular orbits around the Earth. The relationship between the radii of orbits r1 and r2, and the periods T1 and T2 is T2 1 r3 1 = T2 2 r3 2 6. This relationship was stated by Kepler for planets orbiting the Sun, and is known as Kepler’s law. The law applies for any system of bodies in circular orbit about a central mass. Weightlessness 1. Weightlessness is the sensation of having no weight experienced by an astronaut in a spacecraft orbiting the Earth. 2. For an object in a spacecraft orbiting the Earth, the centripetal acceleration of the spacecraft and objects in it is equal to the acceleration due to gravity g′. Satellite Earth Figure 6.11 Exam Tips T 2 r 3 = constant or T1 2 r1 3 = T2 2 r2 3 is true for 2 bodies in circular orbits about the same body. Figure 6.12 Satellite 1 Satellite 2 Earth r 1 r 2 mg’ g’ Spacecraft m R Figure 6.13 141 6

Physics Term 1 STPM Chapter 6 Gravitation 3. Using F = ma for a mass in the spacecraft, mg′ – R = ma a = g′ = mg′ R = 0 4. The reaction R of the floor of the space craft on the object is zero. Hence, the astronaut experiences the sensation of no weight. Example 8 A satellite of mass 50 kg is at a point S, a distance of 1.50 × 109 m from the centre of the Earth as shown in the figure. The mass of the Sun = 2.00 × 1030 kg, mass of Earth = 6.00 × 1024 kg and distance of Earth from the Sun = 1.50 × 1011 m. (a) Calculate (i) the gravitational pull of the Earth on the satellite, (ii) the gravitational pull of the Sun on the satellite, (iii) the magnitude and direction of the resultant force on the satellite, (iv) the acceleration of the satellite. (b) If the satellite is in circular orbit around the Sun, (i) calculate the period of the satellite round the Sun to the nearest year. (ii) then, describe the motion of the satellite relative to the Earth. (iii) give an advantage of having a satellite in this orbit. Solution: (a) (i) Gravitational pull of the Earth on the satellite F1 = GM1m r1 2 = (6.67 × 10–11)(6 × 1024)(50) (1.50 × 109 )2 = 8.89 × 10–3 N (ii) Gravitational pull of the Sun on the satellite F2 = GM2m r2 2 = (6.67 × 10–11)(2 × 1030)(50) (1.50 × 1011– 1.50 × 109 )2 = 0.296 N (iii) Resultant gravitational force on the satellite F2 – F1 = (0.296 – 8.89 × 10–3) N towards the Sun = 0.287 N (iv) Acceleration of satellite towards the Sun a = 0.287 50 (a = F m ) = 5.74 × 10–3 m s–2 (b) (i) a = rω2 = r ( 2π T ) 2 T2 = 4π2 r a = 4π2 × (1.5 × 1011 – 1.5 × 109 ) 5.74 × 10–3 T = 3.20 × 107 s = 1.0 year Earth Sun S 1.50 1011 m 1.50 109 m (ii) Since the period of orbit of the satellite is the same as the period of the Earth’s rotation about the Sun, 1.0 year, the Earth, satellite and Sun is always colinear. (iii) Advantage : The distance of the satellite from the Earth is constant. 142 6

Physics Term 1 STPM Chapter 6 Gravitation Example 9 A system consists of two stars A and B of mass 8.0 × 1012 kg and 4.0 × 1012 kg separated by a distance of 7.5 × 108 m. The two stars move in circular orbits about the centre of mass of the system which remains stationary. (a) (i) Explain why the centre of mass of the system is stationary. (ii) Find the position of centre of mass for the system. (b) Show on a sketch, the circular orbits of the stars. Show the positions of the stars at two different instants. (c) Calculate (i) the speeds of the stars A and B. (ii) their period of orbits. (d) Describe the motion of the stars if the mass of A is very large compared to that of B. Solution: (a) (i) The centre of mass of the system remains stationary because there is no external force on the system. (ii) Using x – = Σ(mixi) (Σmi) Distance of centre of mass from the star A, x = (8 × 1012) × 0 + (4 × 1012) × (7.5 × 108 ) (8.0 + 4.0) × 1012 = 2.5 × 108 m (b) A B B A G A’s orbit B’s orbit x (r – x) (c) (i) The centripetal force for the circular motion of the stars is the gravitational pull between A and B, F = Gm1m2 r 2 For A, m1v1 2 x = Gm1m2 r 2 Speed of A, v1 = Gm2x r 2 = (6.67 × 10–11)(4.0 × 1012)(2.5 × 108 ) (7.5 × 108 )2 = 3.44 × 10–4 m s–1 143 6