Question

Derive an expression for the rotational kinetic energy of a rigid body rotating with angular velocity omaga and hence define moment of inertia.

Answer 1

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Answer 2

The expression for rigid body’s “rotational kinetic energy” is $\bold{\mathrm{K} . \mathrm{E}=\frac{1}{2} I \omega^{2}}$

SOLUTION:

A body undergoing a rotational dynamics with angular velocity ω will also poses a translational motion. Let the velocity for translation motion be ‘’v’’.

We know for translational motion the energy poses by body is kinetic and given as

$K . E=\frac{1}{2} m v^{2}$

Also we know that for a body obeying rotation  

v=rω

Substituting in translation

$=\frac{1}{2} m r^{2} \omega^{2}$

Also there is a relation for moment of inertia i.e $I=mr^2$

Substituting we get,

$\mathrm{K.E}=\frac{1}{2} I \omega^{2}$