Suppose that, in procedure 1, you measured the moment of inertia of the system a

Question

Suppose that, in procedure 1, you measured the moment of inertia of the system as I = 3.04 kg-m2. Assume the diameter of the pulley around which you wrap the string is d = 3.03 cm.

a) In trial 1:the system is initially spinning at angular speed o = 1.66 rad/s. After you drop a disk of mass 825 g on it, the system rotates at f = 1.58 rad/s. Find:
- the moment of inertia of the disk: _____kg-m2
- the radius of the disk: _____m

b) In trial 3: suppose the system is initially spinning at angular speed o = 2.46 rad/s. After you drop a cube of mass 969 g on it, the system rotates at f = 2.28 rad/s. Find:
- the moment of inertia of the cube: _____kg-m2
- the length of one side of the cube: _____cm

Explanation / Answer

(a) Let I1 be the moment of inertia of disk       Mass of disk is m = 825 g = 0.825 kg Since there is no external torque on the system, angular momentum conserved.        Total initial angular momentum = Total final angular momentum         I*o = (I + I1)f         I*o = If + I1f              I1 = I (o - f) / f              I1 = 0.154 kg.m2           But the moment of inetia of disk is I1 = (1/2)mr2                                     0.154 kg.m2 = (1/2)(0.825 kg)r2                                                     r = 0.61 m ---------------------------------------------------------------------- (b) Let I1 be the moment of inertia of disk       Mass of disk is m = 0.969 kg Since there is no external torque on the system, angular momentum conserved.        Total initial angular momentum = Total final angular momentum         I*o = (I + I1)f         I*o = If + I1f              I1 = I (o - f) / f              I1 = 0.24 kg.m2           But the moment of inetia of disk is I1 = (1/ 12)mL2                                       0.24 kg.m2 = (1/ 12)(0.969 kg)L2                                                      L = 1.72 m                                                                                             0.154 kg.m2 = (1/2)(0.825 kg)r2                                                     r = 0.61 m ---------------------------------------------------------------------- (b) Let I1 be the moment of inertia of disk       Mass of disk is m = 0.969 kg Since there is no external torque on the system, angular momentum conserved.        Total initial angular momentum = Total final angular momentum         I*o = (I + I1)f         I*o = If + I1f              I1 = I (o - f) / f              I1 = 0.24 kg.m2           But the moment of inetia of disk is I1 = (1/ 12)mL2                                       0.24 kg.m2 = (1/ 12)(0.969 kg)L2                                                      L = 1.72 m       Mass of disk is m = 0.969 kg Since there is no external torque on the system, angular momentum conserved.        Total initial angular momentum = Total final angular momentum         I*o = (I + I1)f         I*o = If + I1f              I1 = I (o - f) / f              I1 = 0.24 kg.m2           But the moment of inetia of disk is I1 = (1/ 12)mL2                                       0.24 kg.m2 = (1/ 12)(0.969 kg)L2                                                      L = 1.72 m                                       0.24 kg.m2 = (1/ 12)(0.969 kg)L2                                                      L = 1.72 m                                                               

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