Question

A solid cylinder of mass $20kg$ rotates about its axis with angular speed $100rad{s}^{-1}$. The radius of the cylinder is $0.25m$. What is the kinetic energy associated with the rotation of cylinder? What is the magnitude of angular momentum of cylinder about its axis?

• A. $62.5J-s$
• B. $70.4J-s$
• C. $79.6J-s$
• D. $60.5J-s$

Answer 1

The correct option is A $62.5J-s$

If $I$ be the moment of inertia of the cylinder about its axis, then

$I=\frac{1}{2}M{R}^2=\frac{1}{2}\times 20\times {\left(0.25\right)}^2=0.625kg-m^2$

$\therefore$ KE associated with the rotating cylinder is given by,

KE $=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.625\times {\left(100\right)}^2=3125J$

Also using the relation,

$L=\sqrt{2I\left(KE\right)}$

We have $L=\sqrt{2\times 0.625\times 3125}$

$L=62.5J-s$