Question

A uniform disc of mass $m$ and radius $R$ is rotating with angular velocity $\omega$ on a smooth horizontal surface. Another identical disc is moving translationally with velocity $v$ as shown. When they touch each other, they stick together. The angular velocity of centre of mass of the system after contact will be

• A. Zero
• B. $\frac{2v-R\omega }{6R}$
• C. $\frac{v}{R}$
• D. $\frac{\omega }{2^2}$

Answer 1

The correct option is D $\frac{\omega }{2^2}$

Given: A uniform disc of mass m and radius R is rotating with angular velocity ω on a smooth horizontal surface. Another identical disc is moving translationally with velocity v as shown. When they touch each other, they stick together.

To find the angular velocity of centre of mass of the system after contact

Solution:

Let $I,\omega$ be the moment of inertia and angular velocity before collision.

And let $I^{\prime },{\omega }^{\prime }$ be the moment of inertia and angular velocity when they stick

Then

Applying the conservation of energy, we get

$\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2=\frac{1}{2}{I}^{\prime }{\omega }^{\prime 2}⟹\frac{m{r}^2}{2}{\omega }^2+m{\omega }^2{r}^2=\frac{3}{2}m{r}^2\times 2{\omega }^{\prime 2}⟹\frac{{\omega }^2}{2}+{\omega }^2=3{\omega }^{\prime 2}⟹{\omega }^{\prime 2}=\frac{1}{2}{\omega }^2⟹{\omega }^{\prime }=\frac{\omega }{\sqrt{2}}$

is the angular velocity of centre of mass of the system after contact