Question

Derive an expression for kinetic energy of a rotating body with uniform angular velocity.

Answer 1

we know, translation kinetic energy , $K.E=\frac{1}{2}mv^2$,

when a body of mass ‘ m ’ moves with uniform angular velocity, $\omega$ and position of body from the axis of rotation r .
then, linear velocity of body $v=\omega r$

so, rotational kinetic energy of body is given by
$K.E=\frac{1}{2}m(\omega r)^2$

or, $K.E=\frac{1}{2}m\omega^2r^2$

or, $K.E=\frac{1}{2}(mr^2)\omega^2$

we know, moment of inertia is the product of mass and square of separation between position of particle and axis of rotation.
e.g.,. $I=mr^2$

so, $K.E=\frac{1}{2}I\omega^2$ it is the expression for kinetic energy of a rotating body with uniform angular velocity

Answer 2

Answer:

Hope this will help u guys