Suppose m = 599 g (including the mass holder) is released from rest. The pulley

Question

Suppose m = 599 g (including the mass holder) is released from rest. The pulley has diameter d = 3.56 cm, and you measure the angular acceleration of the pulley = 0.576 rad/s2. Assume
- There is no kinetic friction anywhere.
- The string does not slip on the pulley.
- The string is always perpendicular to the diameter of the pulley.

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

b) Calculate the tension in the string.
T = ___ N

c) Calculate the torque exerted by the string on the pulley.
= ___ N-m

d) Calculate the moment of inertia of the system.
Isystem ___ kg-m2

Explanation / Answer

Given that alpha = 0.576 rad/s^2

radius = diameter/2 = 3.56/2= 1.78 cm = 0.0178 m

A) linear accelaration a = r*alpha = 0.0178*0.576 = 0.0102528 m/s^2

B) writing equations of motion

m*g-F = m*a

F = m(g-a) = 0.599*(9.81-0.0102528) = 5.87 N

C) Torque T = r*F = 0.0178*5.87 = 0.104486 N-m

D) But Torque T = Isystem*alpha

Isystem = T/alpha = 0.104486/0.576 = 0.181 kg-m^2

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