Question

two bodies have their moment of inertia I and 2I respectively about their axis of rotational if their kinetic energies of rotation are equal. find the ratio of their angular velocity.

Answer 1

Given

1/2(i1)(omega1)^2= 1/2(i2)(omega2)^2

Putting i1= I and i2 = 2I

We get

omega1/omega2= ( root over)2/1

Answer 2

Answer: The ratio of the angular velocities is $\frac{\omega _{1}}{\omega _{2}}=\sqrt{\frac{2}{1}}$.

Explanation:

The expression for the rotational kinetic energy is as follows;

$KE=\frac{1}{2}I\omega ^{2}$

Here, I is the moment of inertia and $\omega$ is the angular velocity.

In the given problem, two bodies have their moment of inertia I and 2I respectively about their axis of rotational and their kinetic energies of rotation are equal.

$KE_{1}=KE_{2}$

$\frac{1}{2}I_{1}\omega _{1}^{2}=\frac{1}{2}I_{2}\omega _{2}^{2}$

Put $I_{1}=I$  and $I_{2}=2I$.

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

$\frac{\omega _{1}}{\omega _{2}}=\sqrt{\frac{2}{1}}$

Therefore, The ratio of the angular velocities is $\frac{\omega _{1}}{\omega _{2}}=\sqrt{\frac{2}{1}}$.