Rolling Without Slipping On An Inclined Plane MCQ - Practice Questions with Answers

When a body of mass m and radius R rolls down an inclined plane having angle of inclination and at height ‘h’

By conservation of mechanical energy

$$
m g h=\frac{1}{2} m V^2\left(1+\frac{K^2}{R^2}\right)
$$

Where $V=$ Velocity at the lowest point

$$
\text { And, } V=\sqrt{\frac{2 g h}{1+\frac{K^2}{R^2}}}
$$

Similarly using $\quad V^2=u^2+2 a s$
Acceleration $=a=\frac{g \sin \Theta}{1+\frac{K^2}{R^2}}$
And angular acceleration $=\alpha=R a$
And we know that $\tau=I \alpha$
And torque due to friction force $=\tau_f=f R=I \alpha=m K^2(R a)$

So, $f=\frac{m g \sin \Theta}{1+\frac{R^2}{K^2}}$
As $f=\mu N=\mu m g \cos \theta$
So Condition for pure rolling on inclined plane

$$
\mu_s \geq \frac{\tan \Theta}{1+\frac{R^2}{K^2}}
$$
 

Where $\mu_s=$ limiting coefficient of friction
And let $\mathrm{t}=$ time taken by the body to reach the lowest point
Sousing $V=u+a t$

We get,

$$
\mathbf{t}=\frac{1}{\sin \theta} * \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}} *\left[1+\frac{\mathbf{K}^2}{\mathbf{R}^2}\right]}
$$