<p> Assignment 7 : Motion in a Circle</p><p>Assume acceleration due to free fall = 9.81 m s-2</p><p>1. [N02/P3/2] (a) An object rotates in a vertical circle. Which of the following quantities are constant when the object has constant speed? </p><p>(i) Period, (ii) Frequency, (iii) Angular velocity, (iv) Acceleration, (v) Kinetic energy, (vi) Potential energy [4]</p><p>(i) Period,  constant (ii) Frequency,  constant (iii) Angular velocity,  constant (iv) Acceleration,  varying because the direction constantly changes (v) Kinetic energy,  constant because the speed is constant (vi) Potential energy  varying because the height constantly changes [4] All correct ans – 4 marks 1-2 wrong ans - 3 marks 3-4 wrong ans - 2 marks 5 wrong ans – 1 marks 6 wrong ans – 0 marks</p><p>(b) Fig. 2.1(a) below shows a ride in an amusement park.</p><p>The passengers are strapped firmly to a vertical panel. The floor of the panel stays horizontal while it is being moved by two arms rotating with angular velocity 3.1 rad s -1. Each passenger rotates in a vertical circle of radius 2.3 m. Calculate (i) the time taken for one revolution of the arms, (ii) the linear speed of each passenger, (iii) the centripetal acceleration of each passenger. [5]</p><p>1 2p (b) (i) w = T 2p 3. 1= T T = 2. 03 s [1]</p><p>(ii) v= rw =2. 3 3 . 1 [1] =7. 13 ms-1 [1]</p><p>2 (iii) ac = rw = 2. 3 3 . 12 [1] = 22. 1 ms-2 [1]</p><p>(c) Consider a person of mass 60 kg on the ride described in (b) at the instant when the panel is at the bottom of the path, as shown in Fig. 2.1 (b). (i) State the direction of the acceleration of this person. (ii) Calculate the resultant force necessary to cause this acceleration. (iii) Draw a force diagram showing the weight W of the person and the force P which the panel exerts on the person. (iv) Calculate P. [7]</p><p>(c) (i) For uniform circular motion, The acceleration is always toward the centre of the circle. In this case, the direction is upwards. [1] Note: It was not acceptable to write ‘towards the fixed axis’, since clearly there are two fixed axes. </p><p>(ii) Fc = mac Fc = 60 × 22.1 [1] = 1330 N [1]</p><p>(iii) P</p><p>[2] W (iv) Since the centripetal acceleration is pointing upwards, the centripetal force must also be pointing upwards, i.e. P must be larger than W for the resultant force to be pointing upwards.</p><p>Fc = P + (- W) 1330 = P + ( - 60 × 9.81) [1] P = 1910 N [1]</p><p>(d) Using energy considerations, suggest why it is difficult to drive a fairground ride such as described in (b) at a constant speed. [4] </p><p>2 Since the ride is to go at a constant angular speed, the ride must be moving at constant speed. The KE must remain constant [1] On the other hand, the PE rises on the way up and falls on the way down [1]. The motor would therefore have to supply varying power on the way up [1] and something would have to absorb energy on the way down (maybe friction, a brake or the motor acting as a dynamo) [1]. </p><p>From Examiners’ Report : Many students did not use the word ‘energy’ in their answers at all.</p><p>Mr Ng’s version : More work (energy) is needed to raise the people on the way up than on the way down since gravity acts downwards. On the way down, gravity helps in bringing the people downwards. Thus, the motor would have to work harder on the way up than on the way down. It is difficult for motor to produce varying power output.</p><p>3</p>