Question

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

• A. $78.4J$
• B. $39.2J$
• C. $\frac{78.4}{3}J$
• D. $19.6J$

Answer 1

The correct option is C $\frac{78.4}{3}J$

At the top of the inclined plane the cylinder has only potential energy(PE) since it starts from rest.
At the bottom of the inclined plane the cylinder has zero potential energy and has both translational and rotational kinetic energy.
Potential energy $PE=Mgh$
Kinetic energy $KE=K_{tr}+K_{rot}=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$
Substituting $v=\omega R$, we get
$KE=\frac{1}{2}M{v}^2+\frac{1}{2}I(\frac{v}{R}{)}^2=\frac{1}{2}M{v}^2+\frac{1}{2}\frac{M{R}^2}{2}(\frac{v}{R}{)}^2=\frac{3}{4}M{v}^2$
Equating $PE$ with $KE$
$Mgh=\frac{3}{4}M{v}^2$
$\Rightarrow v^2=\frac{4gh}{3}$
$K_{rot}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{M{R}^2}{2}\frac{v^2}{R^2}=\frac{1}{2}\frac{M{R}^2}{2}\frac{4gh}{3R^2}=\frac{Mgh}{3}=\frac{2\times 9.8\times 4}{3}=\frac{78.4}{3}J$