Look at procedure 1. Suppose m = 887 g (including the mass holder) is released f

Question

Look at procedure 1. Suppose m = 887 g (including the mass holder) is released from rest. The pulley has diameter d = 4.23 cm, and you measure the angular acceleration of the pulley a = 0.128 rad/s2. Assume
- The system is frictionless.
- The string does not slip on the pulley.
- The string is always perpendicular to the diameter of the pulley.

a) Calculate a, the linear acceleration of m as it falls.
1 m/s2

b) Calculate T, the tension in the string.
2 N

c) Calculate t, the torque exerted by the string on the pulley.
3 N-m

d) Calculate I, the moment of inertia of the system.
4 kg-m2

Explanation / Answer

a) a = angular acceleration * radius of pulley = 0.128 * 0.02115 = 0.0027072 m/s^2 (negative) b)T = -0.5 M a = -0.5 * 0.887 * -0.0027072 = 0.0012006432 N c) Torque = T * R = 0.0012006432 * 0.02115 = 0.00002539 N-m d) I = torque / angular acceleration = 0.00002539 / 0.128 = 0.00019838 kg-m^2

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