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_{n e t}=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_{n e t}=K_T+K_R=\frac{1}{2} m V^2\left(1+\frac{K^2}{R^2}\right)
$

2. Net Velocity at a point-

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

Where, $\vec{V}_{r o t}=r w$

By conservation of mechanical energy

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

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

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

Similarly using $\quad V^2=u^2+2 a s$

$
\text { Acceleration }=a=\frac{g \sin \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 $=\tau_f=f R=I \alpha=m K^2(R a)$
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 an inclined plane

$
\mu_s \geq \frac{\tan \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 $V=u+a t$

We get,

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