2. 1© Manhattan Press (H.K.) Ltd. Centripetal acceleration Centripetal force Centripetal force 5.2 Centripetal acceleration and force Centripetal force experiment Centripetal force experiment

3. 2 © Manhattan Press (H.K.) Ltd. Centripetal acceleration 5.2 Centripetal acceleration and force (SB p. 170) Change in velocity (  v) = v B – v A (represented by QR) Particle experiences change of velocity (acceleration) along direction AO (pointing to centre of circle) centripetal acceleration Go to Common Error Go to Common Error

4. 3 © Manhattan Press (H.K.) Ltd. Centripetal acceleration 5.2 Centripetal acceleration and force (SB p. 170) An object in uniform circular motion experiences a centripetal acceleration which is the acceleration directed towards the centre of the circle.

5. 4 © Manhattan Press (H.K.) Ltd. Centripetal acceleration 5.2 Centripetal acceleration and force (SB p. 171) pointing to centre of circle Go to Common Error

6. 5 © Manhattan Press (H.K.) Ltd. Centripetal force 5.2 Centripetal acceleration and force (SB p. 171) Steady speed v but changing direction of motion - force acting on it (Newton’s 1 st Law) Go to Common Error No force in direction of motion (constant speed) Force towards centre of circular path centripetal force (F c )

7. 6 © Manhattan Press (H.K.) Ltd. Centripetal force 5.2 Centripetal acceleration and force (SB p. 172) Note: For a uniform circular motion: 1. Centripetal force is always perpendicular to the motion of the body. It does no work on the body and the kinetic energy of the body remains unchanged.

8. 7 © Manhattan Press (H.K.) Ltd. Centripetal force 5.2 Centripetal acceleration and force (SB p. 172) Note: 2. The centripetal force (F c ) is the force required to keep the body moving in a circle. It is provided by the external resultant force towards the centre. It is a functional name rather than a real force. The origins of the centripetal forces may be tension, friction or reaction forces.

9. 8 © Manhattan Press (H.K.) Ltd. Centripetal force experiment 5.2 Centripetal acceleration and force (SB p. 172) 1. Procedure - rubber bung whirled around in horizontal circle - measure time taken for 50 revolutions of bung - calculate angular velocity (  )

10. 9 © Manhattan Press (H.K.) Ltd. Centripetal force experiment 5.2 Centripetal acceleration and force (SB p. 173) 2. Analysis T cosθ = mg.......................... (1) T sinθ = mr  2.......................... (2) T sinθ = m ( sinθ)  2 T = m  2 T is provided by Mg F c = mr  

11. 10 © Manhattan Press (H.K.) Ltd. Centripetal force experiment 5.2 Centripetal acceleration and force (SB p. 173) 3. Error (i) there is a friction acting at the opening of the glass tube, (ii) the string is not inextensible, (iii) the rubber bung is not whirled with constant speed, and (iv) the rubber bung is not whirled in a horizontal circle. Go to Example 3 Example 3 Go to Example 4 Example 4

12. 11 © Manhattan Press (H.K.) Ltd. End

13. 12 © Manhattan Press (H.K.) Ltd. When a particle moves in uniform circular motion, it has constant speed but not constant velocity because its direction changes from time to time. Return to Text 5.2 Centripetal acceleration and force (SB p. 170)

14. 13 © Manhattan Press (H.K.) Ltd. Uniform circular motion is not a kind of uniformly accelerated motion since the acceleration is fixed only in magnitude, but not in direction. Return to Text 5.2 Centripetal acceleration and force (SB p. 170)

15. 14 © Manhattan Press (H.K.) Ltd. As centripetal acceleration (a) may be expressed as It is wrong to think that Since velocity is not a constant and v  r. Therefore, Centripetal acceleration is actually increased with radius r. Return to Text 5.2 Centripetal acceleration and force (SB p. 171)

16. 15 © Manhattan Press (H.K.) Ltd. The circular motion does not produce a centripetal force. The fact is that the centripetal force that causes the circular motion is actually a resultant of other forces. Return to Text 5.2 Centripetal acceleration and force (SB p. 171)

17. 16 © Manhattan Press (H.K.) Ltd. Q: Q:(a) In the centripetal force experiment mentioned above, what will happen when the string breaks while the bung is whirling? (b) What is the relationship between the vertical angle θ of the string and the speed of the bung? (c) Explain with the aid of a diagram, why a mass at the end of a light inelastic string cannot be whirled in circle in air with the string horizontal. Solution 5.2 Centripetal acceleration and force (SB p. 173)

18. 17 © Manhattan Press (H.K.) Ltd. Solution: (a)If the string breaks, the centripetal force disappears. The bung can no longer keep in the circular motion. It will fly away tangentially. ∴ The angle θ increases as the bung is whirled at a higher speed. 5.2 Centripetal acceleration and force (SB p. 174) Go to More to Know 1 More to Know 1

19. 18 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Return to Text (c) If the string is horizontal (Fig. (a)), there is no vertical force to balance the weight of the mass mg. Therefore, the string must make an angle θ with the vertical (Fig. (b)) so that the vertical component of T counteracts the weight mg. T cosθ = mg 5.2 Centripetal acceleration and force (SB p. 174) Fig. (a) Fig. (b)

20. 19 © Manhattan Press (H.K.) Ltd. 1. The equation tanθ = is very useful in answering questions about circular motion. 2. The vertical angle θ is independent of the mass m. 3. A specific vertical angle θ is ideal for one speed only. Return to Text 5.2 Centripetal acceleration and force (SB p. 174)

21. 20 © Manhattan Press (H.K.) Ltd. Q: Q:(a) A pendulum bob moves in a horizontal circle with constant angular velocity  as shown in the figure. Find, in terms of m, , and g, (i) the centripetal force acting on the bob, (ii) the tension T in the string, and (iii) the angle θ. 5.2 Centripetal acceleration and force (SB p. 174)

22. 21 © Manhattan Press (H.K.) Ltd. Q: Q:(b) A student suggests that the value of g can be determined by measuring the period t of the revolution of the bob for various values of θ. (i) Find an expression for t in terms of, g and θ. (ii) Suggest a graph which could be used to obtain the value of g. (iii) Discuss critically whether this is a good method for the determination of g. Solution 5.2 Centripetal acceleration and force (SB p. 175)

23. 22 © Manhattan Press (H.K.) Ltd. Solution : (a) (i) Centripetal force (F c ) = mr  2 = m  sinθ  2 (ii) Horizontal component of T, T sinθ = F c = m  sinθ  2 ∴ T = m   2 (iii) Vertical component of T, 5.2 Centripetal acceleration and force (SB p. 175)

24. 23 © Manhattan Press (H.K.) Ltd. Solution (cont’d) : Return to Text 5.2 Centripetal acceleration and force (SB p. 175) (iii) It is not a good method of determining g because it is difficult to maintain the angle θ at a fixed value, and it is difficult to measure the angle θ.