Question

A uniform disc of mass $m$ and radius $R$ rotates about a fixed vertical smooth axis passing through its centre with angular velocity $\omega$ . A particle of same mass $m$ and having velocity $2$$\omega$$R$ towards centre of the disc collides with the disc moving horizontally and sticks to its rim. Then,

• A. The angular velocity of the disc will become $\frac{\omega }{3}.$
• B. The angular velocity of the disc will become $\frac{5\omega }{3}.$
• C. The impulse on the particle due to disc is $\frac{\sqrt{37}}{3}m\omega R$
• D. The impulse on the disc due to hinge is $\frac{\sqrt{37}}{3}m\omega R$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The angular velocity of the disc will become $\frac{\omega }{3}.$
C The impulse on the particle due to disc is $\frac{\sqrt{37}}{3}m\omega R$
D The impulse on the disc due to hinge is $\frac{\sqrt{37}}{3}m\omega R$

Since the particle of mass hits the rim radially and sticks to it, there is no torque acting on the particle about the center O.
Hence, angular momentum of the system will be conserved.
$L=I\omega$ is conserved.
$L_i=L_f$
$\Rightarrow \frac{m{R}^2}{2}\omega =(I_{disc}+I_{particle}){\omega }_f=(\frac{m{R}^2}{2}+m{R}^2){\omega }_f=\frac{3}{2}m{R}^2{\omega }_f$
$\Rightarrow {\omega }_f=\frac{\omega }{3}$
Impulse on the particle due to disc:
Refer to the diagram.
$\overrightarrow{p_i}$ and $\overrightarrow{p_f}$ are the momentum of the particle before and after collision with the disc.
$\overrightarrow{p_i}=2m\omega R\hat{i}$
$\overrightarrow{p_f}=-m\left(\frac{\omega }{3}\right)R\hat{j}$
$\Delta p=\overrightarrow{p_f}-\overrightarrow{p_i}=-2m\omega R\hat{i}-m\left(\frac{\omega }{3}\right)R\hat{j}=-m\omega R(2\hat{j}+\frac{1}{3}\hat{i})$
Impulse $|\overrightarrow{p_f}-\overrightarrow{p_i}|=\left|m\omega R\right(2\hat{j}+\frac{1}{3}\hat{i}\left)\right|=m\omega R\sqrt{2^2+(\frac{1}{3}{)}^2}=m\omega R\frac{\sqrt{37}}{3}=\frac{\sqrt{37}}{3}m\omega R$
Impulse on the disc due to hinge will be balancing the impulse on the disc due to particle which is $\frac{\sqrt{37}}{3}m\omega R$.