Question

A body of mass $m$ and radius $R$ rolls down an inclined plane of inclination $\theta$. If the linear acceleration of the centre of mass of the body is $\frac{2}{3}g\sin \theta$, then the body can be :

[Assume, there is no slipping between the body and incline]

• A. Disc
• B. Ring
• C. Solid cylinder
• D. Hollow cylinder

Answer 1

Answer: (A), (C)

Solid cylinder


Let $I$ be the moment of inertia of the given body. As it rolls down, it is given that acceleration of its COM
$a=\frac{2}{3}g\sin \theta$

From FBD, we can write that,

$mg\sin \theta -f=m\times \frac{2}{3}g\sin \theta$

$\Rightarrow f=\frac{1}{3}mg\sin \theta$ ($f$ is the frictional force)

Now for rotational motion, the torque about centre ${}^{\prime }{O}^{\prime }$ will be

${\tau }_0=f\times R$

$\Rightarrow I\alpha =\frac{1}{3}mg\sin \theta \times R........\left(1\right)$

For pure rolling, $a=\alpha R$

$\Rightarrow \alpha =\frac{a}{R}=\frac{2g\sin \theta }{3R}$

Substituting the value of $\alpha$ in $\left(1\right)$, we get

$I\times \frac{2g\sin \theta }{3R}=\frac{R}{3}mg\sin \theta$

$\Rightarrow I=\frac{1}{2}m{R}^2$

This the MOI of a disc or a solid cylinder.

Hence, options (a) and (c) are the correct answers.