JEE Main · 2011 · 10 Apr 2011 · Offline · Rotational Mechanics

A mass

m

hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass

m

and radius

R

. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass

m

, if the string does not slip on the pulley, is

(A) $g$
(B) $\frac{2}{3} g$ ✓ Correct
(C) $\frac{g}{3}$
(D) $\frac{3}{2} g$

Answer: (B)

Solution

For the falling mass,

m g - T = ma

. For the disc pulley, the torque

T R

equals

I α = \frac{1}{2} m R^{2} \cdot \frac{a}{R}

, so

T = \frac{1}{2} ma

. Substituting into the first equation,

m g - \frac{1}{2} ma = ma

, hence

m g = \frac{3}{2} ma

and the acceleration is

a = \frac{2 g}{3}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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