Physics Experiment Report 2001 Aim of the experiment

Physics Experiment Report 2001

Aim of the experiment ? Which one will reach the bottom first? Same mass & radius h h

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

Procedure n n n 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

Data and Results n n md = mr Rd = Rr Length of the disk and the ring that passes the sensor L = 8. 94*10 -2 m

Data and Results For Disk Time/t (s) Speed = L / t (ms-1) 0. 0946 0. 0955 0. 1022 0. 0974 Average speed = 0. 94548 0. 93657 0. 87689 0. 91838 0. 91931

Data and Results For Ring t 1 t 2 t 3 1. 3855 0. 0107 0. 1025 1. 6685 1. 5740 0. 0108 0. 1053 2. 0759 1. 9822 0. 0108 0. 1045 t 2+t 3 1. 4713 t=t 1 -

Data and Results For Ring Time/t (s) Speed = l / t (ms-1) 0. 1025 0. 1053 0. 1045 0. 87261 0. 84941 0. 85591 Average speed = 0. 85931

Data and Results Observation n n So it can be seen that the final speed of the disk is larger than that of the ring. i. e. Vd > Vr ……. ………(1)

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

Conclusion First investigation Formula to calculate the moment of inertia of the disk is I = ½ * ( m * R 2 ) Now md = 0. 2 kg , Rd = 0. 06 m, Id = ½ * 0. 2 * ( 0. 06 ) 2 = 3. 6 * 10 -4

Conclusion First investigation Formula to calculate the moment of inertia of the ring is I = ½ * m ( R o 2 + R i 2 ) Now mr = 0. 2 kg , Ro = 0. 06 m, Ri = 0. 052 m Ir = ½ * 0. 2 * ( 0. 062 + 0. 0522 ) = 6. 304 * 10 -4

Conclusion First investigation ∴ Id < Ir . . . ………(2) Vd > Vr . . . ………(1) From (1) and (2), we can see that: Smaller is the moment of inertia; larger is the final speed.

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

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

Conclusion Second investigation Now let’s discuss the disk first. md = 0. 2 kg Id = 3. 6 * 10 -4 n n h = 6. 5 cm = 0. 065 m Vd = 0. 91931 ms-1 R = 0. 06 m 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 n Energy is conserved ( for the disk ).

Conclusion Second investigation For the ring mr = 0. 2 kg Ir = 6. 304 * 10 -4 h = 6. 5 cm = 0. 065 m Vr = 0. 85931 ms-1 R = 0. 06 m LHS = mgh = 0. 2 * 10 * 0. 065 = 0. 13 J RHS = ½ mv 2 + ½ I (V/R)2 = 0. 13 J n Energy is conserved ( for the ring).

Conclusion First & Second investigation 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.

Error Resources / Experiment Precautions n n 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

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