Question

A uniform disc of mass $M$ and radius $R$ is attached to a block of mass $m$ by means of a light string and a light pulley fixed at the top of an inclined plane of inclination $\theta$. The string is wrapped around the disc. The disc rolls down the incline. If $M=6\text{m}$ and $\theta ={30}^{\circ }$, the acceleration of the centre of mass of the disc is: (Assume no slipping at any contact point)

• A. $\frac{g}{6}$
• B. $\frac{g}{13}$
• C. $\frac{g}{16}$
• D. $g$

Answer 1

The correct option is B $\frac{g}{13}$

Figure (a) and (b) shows the free body diagrams of the disc and the block.


Disc is rolling without sliding. Hence, $a_{cm}=\alpha R$
Here, $T=$ tension in the string
and $f=$ frictional force between the disc and the incline

From $\sum F_x=Ma$ (along the inclined plane),
For the disc: $Mg\sin \theta -T-f=M{a}_{cm}...\left(1\right)$
For block: $T-mg=ma...\left(2\right)$

Acceleration of the point on the periphery of the disc where string is attached $=a_{cm}+\alpha R=2a_{cm}$
[$\because$ for pure rolling $a_{cm}=\alpha R$]
From the string constraint, $a=2a_{cm}$. Therefore
$T-mg=2m{a}_{cm}...\left(3\right)$

Net torque on the disc is
$fR-TR=I\alpha$
$\Rightarrow fR-TR=\frac{M{R}^2}{2}\times \frac{a_{cm}}{R}$
$\Rightarrow f=T+\frac{1}{2}M{a}_{cm}...\left(4\right)$

Putting (4) in (1) we get
$Mg\sin \theta -2T=\frac{3}{2}M{a}_{cm}$ .....$\left(5\right)$
Using (3) and (5) we get
$Mg\sin \theta -2(mg+2m{a}_{cm})=\frac{3}{2}M{a}_{cm}$
$\Rightarrow a_{cm}=\left[\frac{(M\sin \theta -2m)}{(\frac{3}{2}M+4m)}\right]g...\left(6\right)$
Putting $M=6m$ and $\theta ={30}^{\circ }$ in (6),
we get $a_{cm}=\frac{g}{13}$
So the correct choice is (b).