2. Physics Experiment Report 2001

3. Aim of the experiment ? W hich one will reach the bottom first? hh Same mass & radius

4. Aim of the experiment relationship of the moment of inertia and the translation velocity of a ring and disk conservation of energy Investigating

5. Procedure Set up the runway as shown on the table Attach the photogate at the lower end of the runway Suppose the height of the runway be h. Release the ring from the upper end measure the velocity when the ring reach the floor

6. Data and Results m d = m r R d = R r Length of the disk and the ring that passes the sensor L = 8.94*10 -2 m

7. Data and Results For Disk Time/ t (s)Speed = L / t (ms -1 ) 0.0946 0.94548 0.0955 0.93657 0.1022 0.87689 0.0974 0.91838 Average speed = 0.91931

8. t1 t2 t3 t= t1-t2 + t3 1.47131.38550.0107 0.1025 1.66851.57400.0108 0.1053 2.07591.98220.0108 0.1045 Data and Results For Ring

9. T ime/ t (s)Speed = l / t (ms -1 ) 0.10250.87261 0.10530.84941 0.10450.85591 Average speed = 0.85931 Data and Results For Ring

10. Data and Results Observation So it can be seen that the final speed of the disk is larger than that of the ring. i.e. V d > V r …….………(1)

11. Conclusion Do you know which one with higher value of moment of inertia ? The disk or the ring? First investigation

12. Conclusion Formula to calculate the moment of inertia of the disk is I = ½ * ( m * R 2 ) Now m d = 0.2 kg, R d = 0.06 m, I d = ½ * 0.2 * ( 0.06 ) 2 = 3.6 * 10 -4 First investigation

13. Formula to calculate the moment of inertia of the ring is I = ½ * m ( R o 2 + R i 2 ) Now m r = 0.2 kg, R o = 0.06 m, R i = 0.052 m I r = ½ * 0.2 * ( 0.06 2 + 0.052 2 ) = 6.304 * 10 -4 Conclusion First investigation

14. ∴ I d < I r...…….………(2) V d > V r...…….………(1) From (1) and (2), we can see that: Conclusion Smaller is the moment of inertia; larger is the final speed. First investigation

15. Total kinetic energy of the disk or the ring K.E.= ½ mv 2 + ½ I  2 = ½ mv 2 + ½ I (V/r) 2 Loss in potential energy P.E.= mgh Conclusion Second investigation

16. So we should prove the following equation with the measured data. mgh = ½ mv 2 + ½ I Conclusion Second investigation

17. Now let’s discuss the disk first. m d = 0.2 kgh = 6.5 cm = 0.065 mR = 0.06 m I d = 3.6 * 10 -4 V d = 0.91931 ms -1 LHS = mgh = 0.2 * 10 * 0.065 = 0.13 J RHS= ½ mv 2 + ½ I (V/R) 2 = ½ * 0.2 * ( 0.91931 ) 2 + ½ * 3.6 * 10 -4 * ( 0.91931/0.06 ) 2 = 0.13 J Conclusion Second investigation Energy is conserved ( for the disk ).

18. For the ring m r = 0.2 kgh = 6.5 cm = 0.065 m R = 0.06 m I r = 6.304 * 10 -4 V r = 0.85931ms -1 LHS = mgh = 0.2 * 10 * 0.065 = 0.13 J RHS = ½ mv 2 + ½ I (V/R) 2 = 0.13 J Conclusion Second investigation Energy is conserved ( for the ring).

19. Conclusion mgh = ½ mv 2 + ½ I (V/R) 2 From this equation, we can see that smaller moment of inertia ( I ) gives larger value of V. First & Second investigation

20. Error Resources / Experiment Precautions Air resistance acting on the disk/ring causes energy loss The disk/ring is not rolling in a straight line Human measuring error The disk/ring may be rolling down with slipping which will cause energy loss by the friction

21. The End That’s the end for our presentation How can we adjust the ring or disk so the they reach the bottom at the same time? Bye!