Question

A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the same mass and radius as the disc, is now gently placed on the disc. After sometime, the two rotate with a common angular velocity, then

• A. some friction exists between the disc and the ring
• B. the angular momentum of the 'disc plus ring' is conserved
• C. the final common angular velocity is $\left(2/3\right)$rd of the initial angular velocity of the disc
• D. $\left(2/3\right)$rd of the initial kinetic energy changes to heat

Answer 1

Answer: (A), (B), (D)

some friction exists between the disc and the ring
the angular momentum of the 'disc plus ring' is conserved
$\left(2/3\right)$rd of the initial kinetic energy changes to heat

Initially the disc slides on the ring, thus a friction force will act between the disc and the ring but it is an internal force.

Let initial angular velocity of the disc be $w$ and common angular velocity of the system be $w^{\prime }.$

As ${\tau }_{ext}=0$ , thus $L$ is conserved.

$L_i=L_f$

$I_{D}w+0=(I_D+I_R)w^{\prime }$

$\frac{1}{2}m{r}^{2}w+0=(\frac{1}{2}m{r}^2+m{r}^2)w^{\prime }$

$⟹w^{\prime }=\frac{w}{3}$

Initial K.E, $E_i=\frac{1}{2}{I}_D{w}^2=\frac{1}{2}\times \frac{1}{2}m{r}^2{w}^2=\frac{1}{4}m{r}^2{w}^2$

Final K.E, $E_f=\frac{1}{2}[I_D+I_R](w^{\prime }{)}^2=\frac{1}{2}\times [\frac{1}{2}m{r}^2+m{r}^2]\frac{w^2}{9}=\frac{1}{12}m{r}^2{w}^2$

Thus $\frac{\Delta E}{E_i}=\frac{[\frac{1}{4}-\frac{1}{12}]m{r}^2{w}^2}{\frac{1}{4}m{r}^2{w}^2}=\frac{2}{3}$

Thus $\left(2/3\right)rd$ of the initial kinetic energy changes to heat.