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

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Solution

Consider a rigid body rotating with a constant angular velocity $\to ω$ about an axis passing through the point $O$ .
As the body rotates, all the particles perform uniform circular motion.
The linear speed of the particle with mass $m_{1}$ is $V_{1} = r_{1} ω$ . Therefore, its kinetic energy is
$E_{1} = \frac{1}{2} m_{1} V_{1}^{2} = \frac{1}{2} m_{1} r_{1}^{2} ω^{2}$
Similarly, the kinetic energy of the particle with mass $m_{2}$ is $E_{2} = \frac{1}{2} m_{2} V_{2}^{2} = \frac{1}{2} m_{2} r_{2}^{2} ω_{2}$ and so on. The rotational kinetic energy of the body is
$E_{r o t} = E_{1} + E_{2} + . . . . + E_{N}$
$= \frac{1}{2} m_{1} r_{2}^{2} ω^{2} + \frac{1}{2} m_{2} r_{2}^{2} ω^{2} + . . . + \frac{1}{2} m_{N} r_{N}^{2} ω^{2}$
$= \frac{1}{2} [m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + . . . . + m_{N} r_{N}^{2}] ω^{2}$
$= \frac{1}{2} (N \sum i = 1 m_{i} r_{i}^{2}) ω^{2}$
$∴ E_{r o t} = \frac{1}{2} I ω^{2}$
$K E = \frac{1}{2} I ω^{2}$
$L = I ω, ω = \frac{L}{I}$
$K E = \frac{1}{2} I \times {(\frac{L}{I})}^{2}$
$= \frac{1}{2} I \frac{L^{2}}{I^{2}} (∵ I = M K^{2})$
$= \frac{1}{2} \frac{L^{2}}{I}$
$= \frac{1}{2} \frac{L^{2}}{M K^{2}}$
$= \frac{1}{2 M} {[\frac{L}{K}]}^{2}$

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