Question

Four identical rods of mass $M=6\text{kg}$ each are welded at their ends to form a square and then welded to a massive ring having mass $M=4\text{kg}$ and radius $R=1\text{m}$. If the system is allowed to roll down the incline of inclination $\theta ={30}^{\circ }$.


The acceleration of the system will be

• A. $\frac{g}{2}$
• B. $\frac{g}{4}$
• C. $\frac{7g}{24}$
• D. $\frac{g}{8}$

Answer 1

The correct option is C $\frac{7g}{24}$

Taking the free body digram.
And $f$ be the friction force at point of contact of sphere;

$a_{cm}=R\alpha$ ....(1)
using free body digram,
$(4M+m)gsin\theta -f=(4M+m)a_{cm}$ ....(2)
where, $f=$ frictional force.
Now, using the torque at point $f$
$\tau =I\alpha =R\times f$
here,$\alpha =\frac{a_{cm}}{R}$
$f=\frac{I{a}_{cm}}{R^2}$ ......(3)

Using eq. (2) and (3)
$(4M+m)gsin\theta -\frac{I{a}_{cm}}{R^2}=(4M+m)a_{cm}$

Given, $M=6kg,m=4kg,\theta ={30}^o$

so, $\left(28\right)g\times \frac{1}{2}-\frac{20a_{cm}}{1}=28a_{cm}=a_{cm}=\frac{7g}{24}$