Question

A horizontal circular plate is rotating about a vertical axis passing through its center with an angular velocity ${\omega _0}$. A man sitting at the center having two blocks in his hands stretches out his hands so that the movement of inertia of the system doubles. If the kinetic energy of the system is $K$ initially, its final kinetic energy will be:
A. $2K$
B. $\dfrac{K}{2}$
C. $K$
D. $\dfrac{K}{4}$

Answer 1

Hint: Find the relation between initial and final angular velocity. By using the fact that the final moment of inertia is twice the initial. Then use that relation to find the relation between initial and final kinetic energy.

Formula used:
$K.E. = \dfrac{1}{2}I{\omega ^2}$
Where
$K.E.$ is the kinetic energy.
$I$ is the inertia of the body
$\omega $ is the angular velocity.

Complete step by step answer:

The outer circle in the diagram is the rotation of the horizontal plate. When the man stretches his hands holding two blocks then because of that there will be a circular motion that is shown by the inner circle.
It is given in the question that ${\omega _0}$ is the initial angular velocity of the disk.
$K$ is the initial kinetic energy of the system.
Let $I$ be the inertial of the system
Let $\omega $ be the new angular velocity of the system when the man stretches his hands.
It is given in the question that the moment of inertia doubles.
$ \Rightarrow 2I\omega = I{\omega _0}$
$ \Rightarrow \omega = \dfrac{{{\omega _0}}}{2}$ . . . (1)
Now we know that the kinetic energy of a system in terms of moment of inertia is given by.
$K.E = \dfrac{1}{2}I{\omega ^2}$
Where,
$K.E.$ is the kinetic energy
$I$ is the inertia
$\omega $ is the angular velocity.
Substituting the given values for the initial kinetic energy of the system, in the above equation we get.
$K = \dfrac{1}{2}I{\omega _0}^2$
Let the final kinetic energy be ${K_f}$.
Then, using the formula of kinetic energy, we can write.
${K_f} = \dfrac{1}{2}I{\omega ^2}$
Substituting the value of $\omega $ from equation (1) into the above equation, we get
${K_f} = \dfrac{1}{2}(2I){\left( {\dfrac{{{\omega _0}}}{2}} \right)^2}$
$ = \dfrac{1}{2} \times \dfrac{{2I\omega _0^2}}{4}$
By rearranging it, we get
${K_f} = \dfrac{1}{2} \times \dfrac{1}{2}I{\omega _0}^2$
\[ \Rightarrow {K_f} = \dfrac{1}{2}K\] \[\left( {\because K = \dfrac{1}{2}I{\omega _0}^2} \right)\]
Thus, the final kinetic energy will be $\dfrac{K}{2}$.

Note:You need to understand that the angular momentum of the system will be affected because of the man, when he would stretch his hands. Hence the initial and final velocities will be different.