Question

A rough disc of mass $m$ rotates freely with an angular velocity $\omega$. If another rough disc of mass $\frac{m}{2}$ and of same radius but spinning in opposite sense with angular speed $\omega$ is kept on the first disc. Then:

• A. The final angular speed of the disc is $\frac{\omega }{3}$.
• B. The net work done by friction is zero.
• C. The fiction does a positive work on the lighter disc.
• D. The net wok done by friction is $\frac{-m{r}^2{\omega }^2}{3}$.

Answer 1

Answer: (A), (D)

The net wok done by friction is $\frac{-m{r}^2{\omega }^2}{3}$.

Let the angular velocity of first disc be $\omega \circlearrowright$ and angular velocity of second disc be $\omega \circlearrowleft$.

Let the moment of inertia of first disc be $I_1$ and for second disc be $I_2$.

Since no external force is acting on the system, the angular momentum will be conserved. Let the final angular velocity of the system will be ${\omega }^{\prime }$

According to the law of conservation of angular momentum

$I_1\omega -I_2\omega =(I_1+I_2){\omega }^{{}^{\prime }}$

Negative sign is taken for the angular momentum of second disc because it is moving in the opposite direction to the first disc.

Substituting the values,

$\frac{1}{2}m{r}^2\omega -\frac{1}{2}\left(\frac{m}{2}\right)r^2\omega =(\frac{1}{2}m{r}^2+\frac{1}{4}m{r}^2){\omega }^{{}^{\prime }}$

$\Rightarrow \frac{m{r}^2\omega }{4}=\frac{3}{4}m{r}^2{\omega }^{\prime }$

$\therefore {\omega }^{\prime }=\frac{\omega }{3}$

Hence, option (a) is correct.

Work done by friction : from work energy theorem,

$\text{Work done by the friction}=\text{Loss in K.E}$

$\Rightarrow W_f=K{E}_f-K{E}_{in}$

$\Rightarrow W_f=\frac{1}{2}(I_1+I_2)({\omega }^{\prime }{)}^2-[\frac{1}{2}{I}_1{\omega }^2+\frac{1}{2}{I}_2{\omega }^2]$

$\Rightarrow W_f=\frac{1}{2}[\frac{1}{2}m{r}^2+\frac{1}{2}\times \frac{m}{2}{r}^2]\frac{{\omega }^2}{9}-[\frac{1}{2}\times \frac{1}{2}m{r}^2{\omega }^2+\frac{1}{2}\times \frac{1}{2}\times \frac{m}{2}{r}^2{\omega }^2]$

$\Rightarrow W_f=\frac{m{r}^2{\omega }^2}{24}-m{r}^2{\omega }^2[\frac{1}{4}+\frac{1}{8}]$

$\Rightarrow W_f=m{r}^2{\omega }^2[\frac{1}{24}-\frac{3}{8}]$

$\Rightarrow W_f=m{r}^2{\omega }^2(-\frac{8}{24})$

$\Rightarrow W_f=-\frac{m{r}^2{\omega }^2}{3}$

So, option (d) is also correct.

Hence, option (a) and (d) are the correct options.