Question

A round uniform body of radius $R,$ mass $M$ and moment of inertia $I$, rolls down (without slipping) an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is

• A. $\frac{gsin\theta }{1+\frac{I}{M{R}^2}}$
• B. $\frac{gsin\theta }{1+\frac{M{R}^2}{I}}$
• C. $\frac{gsin\theta }{1-\frac{I}{M{R}^2}}$
• D. $\frac{gsin\theta }{1-\frac{M{R}^2}{I}}$

Answer 1

The correct option is A $\frac{gsin\theta }{1+\frac{I}{M{R}^2}}$

Assume that no energy is used up against friction, the loss in potential energy is equal to the total gain in the kinetic energy.
ie, $Mgh=\frac{1}{2}I\left(\frac{v^2}{R^2}\right)+\frac{1}{2}M{v}^2$
or $\frac{1}{2}{v}^2(M+\frac{I}{R^2})=Mgh$
or $v^2=\frac{2Mgh}{M+\frac{I}{R^2}}=\frac{2gh}{1+\frac{I}{M{R}^2}}$
If s be the distance covered along the plane, then
$h=\text{s sin}\theta$
$v^2=\frac{2gssin\theta }{1+\frac{I}{M{R}^2}}$
Now, $V^2=2as$
$\therefore 2as=\frac{2gssin\theta }{1+\frac{I}{M{R}^2}}$
or $a=\frac{gsin\theta }{1+\frac{I}{M{R}^2}}$