Question

A solid cylinder down an inclined plane. Its mass is 2 kg and radius 0.1 m/ If the height inclined plane is 4 m, what is its rotational kinetic energy, when it reaches the foot of the plane?

Answer 1

We know that,

Potential energy = K.E of translation + K.E of rotation

$Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$

$Mgh=\frac{1}{2}M\left({\omega }^2{R}^2\right)+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2$

$Mgh=\frac{3}{4}M{\omega }^2{R}^2$

${\omega }^2=\frac{4gh}{3R^2}$

Now, K.E of rotation is

$K=\frac{1}{2}I{\omega }^2$

$K=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2$

$K=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\times \frac{4gh}{3R^2}$

$K=\frac{Mgh}{3}$

$K=\frac{2\times 9.8\times 4}{3}$

$K=26.13J$

Hence, this is the required solution