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 some time, the two rotate with a common angular velocity. Which of the following are true ?

• A. Same 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 $\frac{2}{3}\text{rd}$ of the initial angular velocity of the disc
• D. $\frac{2}{3}\text{rd}$ of the initial kinetic energy changes to heat

Answer 1

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

$\frac{2}{3}\text{rd}$ of the initial kinetic energy changes to heat

$\left(a\right)$ As the ring is placed on the rotating disc, they rub against each other exerting kinetic friction as shown. This kinetic friction porduces torque which opposes the motion of the disc and slows it down to angular velocity $\omega$. The kinetic friction on the ring accelerates it from rest to final angular velocity $\omega$. When $\omega$ for both becomes same, they stop rubbing.

$\left(b\right)$ As the torques of frictional forces are internal and no external force acting on the system so, the angular momentum of system remains conserved.

$\left(c\right)$ From the angular momentum conservation,

$I_{\text{disc}}{\omega }_0+I_{ring}{\omega }_{ring}=I_{\text{disc}}\omega +I_{\text{ring}}\omega$

where,
Moment of inertia of the disc$I_{disc}=\frac{M{R}^2}{2}$

Moment of inertia of the ring$I_{ring}=M{R}^2$

Initial angular velocity of disc $={\omega }_0$

Initial angular velocity of the ring, ${\omega }_{ring}=0$

Final angular velocity of disc and ring, ${\omega }_{ring}=0$

Substituting the values in the above equation we get,

$\omega =\frac{\left(\frac{M{R}^2}{2}\right){\omega }_0}{\frac{M{R}^2}{2}+M{R}^2}=\frac{{\omega }_0}{3}$

$\Rightarrow \omega \ne \frac{2{\omega }_0}{3}$

$\left(d\right)$ Loss in kinetic energy,

$\Delta K.E=K_i-K_f$

$\Rightarrow \Delta K.E=\frac{M{R}^2}{2}{\omega }_0^2-(\frac{M{R}^2}{2}+M{R}^2)\frac{{\omega }_0^2}{9}$

$\Rightarrow \Delta K.E=\frac{1}{3}\left(M{R}^2{\omega }_0^2\right)$

$\Rightarrow \Delta K.E=\frac{2}{3}\left(\frac{1}{2}M{R}^2{\omega }_0^2\right)=\frac{2}{3}{K}_i$

Hence, option (a), (b) and (d) are correct.