Question

A uniform disc of M and radius R has a string wound around its periphery and tied to a ceiling as shown in figure.The acceleration of the center of mass after it is released

• A. g
• B. $\frac{g}{3}$
• C. $\frac{2g}{3}$
• D. $\frac{g}{2}$

Answer 1

Answer: (C)

$\frac{2g}{3}$

Equation of motion of the disc :
$Ma=Mg-T$ ....(1)
Also, torque about O, ${\tau }_o=TR=I\alpha$
$\therefore$ $TR=\left(\frac{1}{2}M{R}^2\right)\alpha$
$⟹T=\frac{M}{2}\alpha R$
For no slipping, $a=\alpha R$
We get $T=\frac{Ma}{2}$ ...(2)
From (1) and (2), $Ma=Mg-\frac{Ma}{2}$
$⟹$ $a=\frac{2g}{3}$