Question

A horizontal disc rotates freely with angular velocity omega about a vertical axis through its centre. A ring having the same mass and radius as the disc, is now gently placed coaxially on the disc. After some time, the two rotate with a common angular velocity. Then,
A. Some frictional force 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 $\dfrac{2}{3}$ of the initial angular velocity of the disc.
D. 2/3rd of the initial kinetic energy changes to heat.

Answer 1

The way in which we approach this question relies in trying to prove the 4 options and identifying the provable ones. We can start with the proving of frictional force, then continue with angular momentum using the law of conservation of momentum, then arrive at equating the third option followed by the fourth option.

Complete answer:
(A) The frictional force between the disc and the ring is determined by the difference in velocities between the two. It is given that the ring is placed gently on top of an already moving disc. Say, the disc already rotates with an angular velocity v, then the ring moving along with it possessing the same velocity must also possess a relative velocity of zero in comparison with the disc. In relative motion, frictional force always plays a role and hence this concludes that there must be frictional force acting between the two. Hence option A is true.

(B) Angular velocity of a body possessing any form of inertia is given by:
$L = i\omega$
For a circular disc rotating about its vertical central axis, inertia is given by the equation:
${I_d} = \dfrac{{M{R^2}}}{2}$
And for a circular ring rotating about its vertical central axis is given by the equation:
${I_r} = M{R^2}$
Keeping these equations in mind, we can proceed to prove that the total change in momentum is zero, which conserves angular momentum.Since both the species spin on a coinciding axis with the same angular velocity with no external force. The torque of the system (moment of force) is 0, and torque is defined as the rate of change of angular momentum per time or dL/dt. Since torque is 0, dL should also be 0. (dt cannot be zero). Hence option B is also true.

(C) Using the equations mentioned above, we can compare the values of initial momentum (before the placing of the ring) and final momentum using mathematical equations as shown below:
${L_i}$ is the initial momentum and the momentum of the disc alone.
${I_d}\omega {\text{ }} = {\text{ }}\dfrac{{M{R^2}\omega }}{2}$
${L_f}$ is the final momentum and the sum of momenta of the disc and the ring ${I_d}\omega + {I_r}\omega = \left( {\dfrac{{M{R^2}}}{2} + M{R^2}} \right)\omega '$
Equating ${L_i}$ and ${L_f}$, we get:
$\dfrac{{M{R^2}}}{2} = \left( {\dfrac{{M{R^2}}}{2} + M{R^2}} \right)\omega '$ Or $\omega ' = \dfrac{\omega }{3}$
Hence, the final common angular velocity of the system is only 1/3rd the initial disc’s angular velocity. Therefore, option C is false.

(D) Kinetic energy using angular velocity and inertia is given by the equation:
$KE = \dfrac{1}{2}I{\omega ^2}$
In this system, we have a combination of a ring and a disc, therefore their inertias must be added for a combined inertia.
$KE = \dfrac{1}{{2{{\omega '}^2}}}\left( {\dfrac{{M{R^2}}}{2} + M{R^2}} \right)$
Which is the final kinetic energy.
The initial kinetic energy is given by the disc only and that has to be equated with the final energy to find the heat loss in the process.
$K{E_i} = \dfrac{1}{{2{\omega ^2}}}\left( {M{R^2}} \right)$
Equating these two equations after substituting $\omega '$ by $\dfrac{\omega }{3}$ will result in $K{E_f} = \dfrac{{K{E_i}}}{3}$ which proves the fourth option to be true as only 1/3rd remains as energy and the rest is dissipated as heat.

Note: Moment of inertia is a measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element's distance from the axis.
$I = \sum {{m_i}{r_i}^2}$