Question

7) Derive and expression for the kinetic energy
a body of mass M rotating uniformly about a
given axis. Show that the rotational kinetic
energy =1/2M×(L/K) square​

Answer 1

Answer:

Given:

A body has been provided of mass M,

rotating in a given axis.

To Prove:

Rotational KE = ½ M (L/K)²

Proof:

Let's consider a rigid body rotating around a given axis with angular velocity of ω.

Now , we know that a body is formed by many particles.

As the rigid body rotates, all the particles follow Uniform Circular Body :

Proof:

Let velocity of 1 particle be v1

So v1 = r1 × ω

Similar expressions for all the particles.

So total KE for all particles :

$= \dfrac{1}{2}m_{1} {v_{1}}^{2} + \dfrac{1}{2}m_{2} {v_{2}}^{2}....(n \: times)$

$= \dfrac{1}{2} m_{1} {(r_{1} \omega)}^{2} + \dfrac{1}{2} m_{2} {(r_{2} \omega)}^{2} ...$

$= \dfrac{1}{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2} + ...) { \omega}^{2}$

$= \dfrac{1}{2} \Sigma(m_{i} {r_{i}}^{2} ) { \omega}^{2}$

$= \dfrac{1}{2} I { \omega}^{2}$

We know that : L = I × ω

=> ω = L/I

Now putting this value :

$= \dfrac{1}{2} I { (\dfrac{L}{I} )}^{2}$

$= \dfrac{1}{2} ( \dfrac{ {L}^{2} }{I} )$

Now considering K = radius of gyration.

$= \dfrac{1}{2} ( \dfrac{ {L}^{2} }{m {K}^{2} } )$

$= \dfrac{1}{2m} ( { \dfrac{L}{K} )}^{2}$

So proved.