Rolling Without Slipping On An Inclined Plane - Practice Questions & MCQ

When a body of mass m and radius R rolls down an inclined plane having an angle of inclination $(\theta )$ and at height ' h '

Kinetic energy-

The total kinetic energy of the body is the sum of both translational and rotational kinetic energy.

$K_{net}=K_T+K_R=\frac{1}{2}m{V}^2+\frac{1}{2}I{\omega }^2$

Using $V=\omega R$ and $I=m{K}^2$

$K_{net}=K_T+K_R=\frac{1}{2}m{V}^2(1+\frac{K^2}{R^2})$

2. Net Velocity at a point-

${\overrightarrow{V}}_{\text{net }}={\overrightarrow{V}}_{\text{translation }}+{\overrightarrow{V}}_{\text{rotation }}$

Where, ${\overrightarrow{V}}_{rot}=rw$

By conservation of mechanical energy

$mgh=\frac{1}{2}m{V}^2(1+\frac{K^2}{R^2})$

Where $\mathrm{V}=$ Velocity at the lowest point

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

Similarly using $V^2=u^2+2as$

$\text{ Acceleration }=a=\frac{g\sin \mathrm{\Theta }}{1+\frac{K^2}{R^2}}$

And angular acceleration $=a=R\alpha$
And we know that $\tau =I\alpha$
And torque due to friction force $={\tau }_f=fR=I\alpha =m{K}^2(Ra)$
So, $f=\frac{mg\sin \mathrm{\Theta }}{1+\frac{R^2}{K^2}}$
As $f=\mu N=\mu mg\cos \theta$
So Condition for pure rolling on an inclined plane

${\mu }_s\ge \frac{\tan \mathrm{\Theta }}{1+\frac{R^2}{K^2}}$
 

Where ${\mu }_s=$ limiting coefficient of friction
And let $t=$ time taken by the body to reach the lowest point
So using $V=u+at$

We get,

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