Question

Derive an expression for the kinetic energy of a body of mass M rotating uniformly about a given axis. Hence show that the rotational kinetic energy is
$= \dfrac{1}{{2M}} \times {\left( {\dfrac{L}{K}} \right)^2}$

Answer 1

Hint: A body consists of a finite number of particles. The change in any physical quantity of the body is basically the change experienced by all these particles together as one system. Additionally, kinetic energy is defined as half times the product of mass and square of the velocity of a particle. Thus the total kinetic energy of the body will be the sum of the kinetic energy of individual particles.

Formula Used:
\eqalign{ & {\text{Kinetic Energy, }}K = \dfrac{1}{2}M{V^2} \cr & {\text{Linear Velocity, }}V = \omega R \cr & {\text{where }}M{\text{ is the mass of the body,}} \cr & V{\text{ is the velocity of the body,}} \cr & \omega {\text{ is its angular velocity,}} \cr & {\text{and }}R{\text{ is the radius of circle in which body rotates}} \cr}

Complete step-by-step answer:
Consider a rigid body of mass M rotating uniformly about a given axis through a point O, with angular velocity $\vec \omega$. Because the entire body is rotating, all its particles that constitute it are also performing the same circular motion.
We know that the kinetic energy of one such particle 1 in the body will be given by:
${K_1} = \dfrac{1}{2}{m_1}{v_1}^2$
Furthermore, the relation between its linear velocity and angular velocity is given by:
${v_1} = \omega {r_1}$
Substituting value of linear velocity in the first equation we get:
${K_1} = \dfrac{1}{2}{m_1}{\left( {\omega {r_1}} \right)^2}$
Now, there are a total of n such particles throughout the body. So the kinetic energy of these entire particles will together be the kinetic energy of the body. In other words, the sum of all the kinetic energies of the particles will give the total kinetic energy of the body.
\eqalign{ & {K_{rotational}} = {K_1} + {K_2} + {K_3} + \ldots \ldots + {K_n} \cr & {K_{rotational}} = \dfrac{1}{2}{m_1}{\left( {\omega {r_1}} \right)^2} + \dfrac{1}{2}{m_2}{\left( {\omega {r_2}} \right)^2} + \dfrac{1}{2}{m_3}{\left( {\omega {r_3}} \right)^2} \ldots \ldots + \dfrac{1}{2}{m_n}{\left( {\omega {r_n}} \right)^2} \cr & {K_{rotational}} = \dfrac{1}{2}\left[ {{m_1}r_1^2 + {m_2}r_2^2 + {m_3}r_3^2 + \ldots \ldots + {m_n}r_n^2} \right]{\omega ^2} \cr & {K_{rotational}} = \dfrac{1}{2}\left( {\sum\limits_{i = 1}^N {{m_i}{r_i}} } \right){\omega ^2} \cr}
But we know that$I = \left( {\sum\limits_{i = 1}^N {{m_i}{r_i}^2} } \right)$
$\therefore {K_{rotational}} = \dfrac{1}{2}I{\omega ^2}$
Angular momentum, $L = I\omega$
$\Rightarrow \omega = \dfrac{L}{I}$
Substituting value of angular velocity in equation of Rotational kinetic energy, we get:
\eqalign{ & {K_{rotational}} = \dfrac{1}{2}I{\left( {\dfrac{L}{I}} \right)^2} \cr & = \dfrac{1}{2}I\dfrac{{{L^2}}}{{{I^2}}} \cr & = \dfrac{1}{2}\dfrac{{{L^2}}}{I} \cr & = \dfrac{1}{2}\dfrac{{{L^2}}}{{M{K^2}}}{\text{ }}\left[ {\because I = M{K^2}} \right] \cr & \therefore {K_{rotational}} = \dfrac{1}{{2M}}{\left[ {\dfrac{L}{K}} \right]^2} \cr}
Hence proved that the rotational Kinetic energy of a body of mass M rotating uniformly about a given axis is $\dfrac{1}{{2M}} \times {\left( {\dfrac{L}{K}} \right)^2}$

Note: In a system that is performing rotational motion, the moment of inertia of the body takes place of its mass and angular velocity takes place of its linear velocity in equations. And its inertia is basically the product of the mass of the body and radius of gyration.