Question

The rotational kinetic energy of a body is E. In the absence of external torque, if mass of the body is halved and radius of gyration is doubled, then its rotational kinetic energy will be:-

Answer 1

Whenever there is an absence of external torque, always remember angular momentum $L$ is conserved. Therefore for this question we can use conservation of angular momentum where moment of inertia can be written in the terms of radius of gyration $I = m{k^2}$ where $m$ is mass of the body and k is radius of gyration.
$L = I\omega$ , $I = m{k^2}$ , $K.E = \dfrac{1}{2}I{\omega ^2}$

Complete step by step answer:
The initial rotational kinetic energy of the body is given in the question as $K.{E_1} = E$ .then it was said that there was an absence of external torque which means the total angular momentum is constant which implies the angular momentum is conserved.
CONSERVATION OF ANGULAR MOMENTUM-
$\Rightarrow L = I\omega$ , where $I$ is moment of inertia and $\omega$ is angular velocity and $L$ is angular momentum
${L_1} = {I_1}{\omega _1}$ ………initial conditions (1)
${L_2} = {I_2}{\omega _2}$ ………in the absence of torque (2)
As angular momentum is conserved we write ${L_1} = {L_2}$
$\Rightarrow {I_1}{\omega _1} = {I_2}{\omega _2}$ …...... (3)
$I = m{k^2}$ , where $I$ is moment of inertia and $m$ is mass of body and $k$ is radius of gyration.
Now we can write
${I_1} = {m_1}k_1^2$ ……….. (4)
${I_2} = {m_2}k_2^2$ ……….. (5)
UNDER ABSENCE OF TORQUE,
Given mass of body is halved so we can write ${m_2} = \dfrac{{{m_1}}}{2}$
Given radius of gyration is double so we can write ${k_2} = 2{k_1}$
Now, substituting these in equation 4 and 5 and taking ratios we get,
$\Rightarrow$ $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{m_1}k_1^2}}{{{m_2}k_2^2}} = \dfrac{{{m_1}k_1^2}}{{\dfrac{{{m_1}}}{2}{{\left( {2{k_1}} \right)}^2}}} = 2 \times \left( {\dfrac{1}{4}} \right) = \dfrac{1}{2}$
From equation 3 we can write
$\Rightarrow$ $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{\omega _2}}}{{{\omega _1}}} = \dfrac{1}{2}$ $\Rightarrow$ $2{I_1} = {I_2}$ , ${\omega _1} = 2{\omega _2}$
Now initial rotational kinetic energy is $K.{E_1} = \dfrac{1}{2}{I_1}\omega _1^2$ $= E$ and final rotational kinetic energy can be written as $K.{E_2} = \dfrac{1}{2}{I_2}\omega _2^2$
$\dfrac{{K.{E_1}}}{{K.{E_2}}} = \dfrac{{\dfrac{1}{2}{I_1}\omega _1^2}}{{\dfrac{1}{2}{I_2}\omega _2^2}} = \dfrac{{{I_1}\omega _1^2}}{{{I_2}\omega _2^2}} = \dfrac{{{I_1}{{\left( {2{\omega _2}} \right)}^2}}}{{2{I_1}\left( {\omega _2^2} \right)}} = 2$
$\dfrac{{K.{E_1}}}{{K.{E_2}}} = \dfrac{E}{{K.{E_2}}} = 2$
$\Rightarrow K.{E_2} = \dfrac{E}{2}$
Therefore, in the absence of external torque, if mass of the body is halved and radius of gyration is doubled, then its rotational kinetic energy will be halved.

Note:
Torque can be defined as the rate of change of angular momentum, analogous to force. The net external torque on any system is always equal to the total torque on the system; in other words, the sum of all internal torques of any system is always 0 (this is the rotational analogue of Newton's Third Law). Here we need to understand that angular momentum is always conserved or total angular momentum is constant in the absence of external torque.