Question

A mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I, and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates. The magnitude of the torque on the pulley is..
Choose one of these answers:
A. greater than mgR
B. less than mgR
C. equal to mgR
(explain in words)

Answer 1

Answer:B

Explanation:

Given

radius of Pulley is r

Pulley has a moment of inertia is I

Suppose mass is falling with an acceleration a and T be the Tension in string

$mg-T=ma$

$T=m(g-a)$

And $I\times \alpha =T\times R$

$I\times \frac{a}{R}=T\times R$

$I\times \frac{a}{R}=m(g-a)\times R$

$I=\frac{mR^2(g-a)}{a}$

and Torque is $I\alpha$

$\tau =\frac{mR^2(g-a)}{a}\times \alpha$

$\tau =\frac{mR^2(g-a)}{a}\times \frac{a}{R}$

$\tau =mR(g-a)$

Which is less than $mgR$