Question

A thin disc and a thin ring both have mass M and radius R both rotate about axes through COM and perpendicular to their surface at same angular velocity who have higher kinetic energy

Answer 1

Given:

A thin disc and a thin ring both have mass M and radius R both rotate about axes through COM and perpendicular to their surface at same angular velocity.

To find:

Who will have higher rotational kinetic energy.

Calculation:

General Expression for rotational kinetic energy

$\boxed{ \bf{KE = \dfrac{1}{2} \times I \times { \omega}^{2} }}$

I => Moment Of Inertia ,

$\omega$=>Angular Velocity

For ring:

$\rm{KE_{1} = \dfrac{1}{2} \times I \times { \omega}^{2} }$

$= > \rm{KE_{1} = \dfrac{1}{2} \times (M{R}^{2}) \times { \omega}^{2} }$

$= > \rm{KE_{1} = \dfrac{M{R}^{2} { \omega}^{2}}{2}}$

For disc:

$\rm{KE_{2} = \dfrac{1}{2} \times I \times { \omega}^{2} }$

$= > \rm{KE_{2} = \dfrac{1}{2} \times ( \dfrac{M{R}^{2}}{2}) \times { \omega}^{2} }$

$= > \rm{KE_{2} = \dfrac{M{R}^{2} { \omega}^{2}}{4}}$

So, KE2 < KE1

Hence , Ring will have higher rotational kinetic energy as compared to the thin disc.