Suppose m =378 g (including the mass holder) is released from rest. The pulley h
ID: 1513661 • Letter: S
Question
Suppose m =378 g (including the mass holder) is released from rest. The pulley has diameter d = 3.66 cm of the smallest wheel (note: the string is wrapped around this smallest wheel in the 3-step pulley), and you measure the angular acceleration of the pulley = 0.202 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.
- g=9.80 m/s2
Calculate the moment of inertia of the system.
Isystem= ______ kg-m2
Explanation / Answer
Apply Newton's second law in veritcal direction.
( the mass has two forces acting on it, gravity downwards and tension upwards )
Fy = m ay
T - mg = m ay
Consider a sum of moments about the pulley's center, here the tension produces a torque:
= I
T (d/2) = I
From the definition of acceleration, ay = r = (d/2), therefore the above equations are
T - mg = m (d/2) ...... ( 1 )
T (d/2) = I ...... ( 2 )
Substitute numerical values in first equation, leads
T - (0.378)(9.8) = 0.8378 (0.202 ) (0.0366 / 2 )
Solve for T
T = 3.7057 N
Substitute T value in equation ( 2 )
T (d/2) = I
3.7057 N (0.0366 / 2 ) = I ( 0.202 )
Solve for I
I = 0.336 kg.m2