Question

A circular disk of moment of inertia ${{I}_{t}}$ ​ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ${{\omega }_{i}}$ ​. Another disk of moment of inertia ${{I}_{b}}$ ​ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ${{\omega }_{f}}$ ​. The energy lost by the initially rotating disc to friction is
A: $\dfrac{1}{2}\dfrac{{{I}_{b}}^{2}}{({{I}_{t}}+{{I}_{b}})}{{\omega }_{1}}^{2}$
B: $\dfrac{1}{2}\dfrac{{{I}_{t}}^{2}}{({{I}_{t}}+{{I}_{b}})}{{\omega }_{1}}^{2}$
C: $\dfrac{{{I}_{b}}-{{I}_{t}}^{{}}}{({{I}_{t}}+{{I}_{b}})}{{\omega }_{1}}^{2}$
D: $\dfrac{1}{2}\dfrac{{{I}_{b}}{{I}_{t}}^{{}}}{({{I}_{t}}+{{I}_{b}})}{{\omega }_{1}}^{2}$

Answer 1

Hint : In this case, the rotation of the disc takes place along a fixed axis. Since rotation occurs here, we can find the application of angular momentum in this case. The energy that is said to be lost is the loss of kinetic energy. Hence we can find it with the help of the law of conservation of angular momentum.

Complete Step By Step Answer:
In the question we are given that, a circular disk of moment of inertia ${{I}_{t}}$ ​ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ${{\omega }_{i}}$ ​. Another disk of moment of inertia ${{I}_{b}}$ ​ is dropped coaxially on to the rotating disk. Initially the second disk has zero angular speed and eventually both the disks rotate with a constant angular speed ${{\omega }_{f}}$ ​.
The initial angular momentum would be ${{I}_{t}}{{\omega }_{1}}$
And the final angular momentum would be $({{I}_{t}}+{{I}_{b}}){{\omega }_{2}}$
Applying the law of conservation of angular momentum, we find that
{{I}_{t}}{{\omega }_{1}}=({{I}_{t}}+{{I}_{b}}){{\omega }_{2}} \\\ \Rightarrow {{\omega }_{2}}=\dfrac{{{I}_{t}}}{({{I}_{t}}+{{I}_{b}})}{{\omega }_{1}} \\\
The loss in kinetic energy is found by
\Delta K=\dfrac{1}{2}{{I}_{t}}{{\omega }_{1}}^{2}-\dfrac{1}{2}({{I}_{t}}+{{I}_{b}}){{(\dfrac{{{I}_{t}}}{({{I}_{t}}+{{I}_{b}})}{{\omega }_{1}}^{{}})}^{2}} \\\ =\dfrac{1}{2}\dfrac{{{I}_{b}}{{I}_{t}}^{{}}}{({{I}_{t}}+{{I}_{b}})}{{\omega }_{1}}^{2} \\\
Hence, option D is the correct answer among the given options.

Note :
We are applying the law of conservation of angular momentum here because the other elements like the plane in which it rotates are considered to be massless. If there are conditions regarding additional mass I the system then the conservation might not take place efficiently.
Also, we have to take the moment of inertia of the body given, we should keep in mind different bodies have different moments of inertia. Also, the moment of inertia depends upon the axis of rotation around which it is being calculated