Question

the angular momentum of a particle performing uniform circular motion is l if the kinetic energy of a particle is doubled the frequency is half then angular momentum becomes

Answer 1

Hope it helps you,

Please tell us in following situations:

There is any doubt,
There is any mistake,
There is any suggestion,
Need for more explanation in depth,
etc.

Thank you!

Answer 2

Answer: The correct answer is "the angular momentum becomes 4 times.

Explanation:

The expression for the angular momentum is as follows;

$L=I\omega$

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

In the given problem, the angular momentum of a particle performing uniform circular motion is l.

Put L=I and I=I'.

The expression for the angular momentum will become as;

$I=I'\omega$                                                                   ...... (1)

The expression for the rotational kinetic energy is as follows;

$K=\frac{1}{2}I'\omega ^{2}$

Rearrange the expression for I'.

$I'=\frac{2K}{\omega ^{2}}$

Put the value of I' in the equation (1).

$I=I'\omega$  

$I=(\frac{2K}{\omega ^{2}})\omega$

$I=\frac{2K}{\omega}$

Now, express the expression as;

$\frac{I'}{I}=(\frac{K'}{K})(\frac{\omega }{\omega'})$

As it is given in the problem, if the kinetic energy of a particle is doubled and the frequency is half.

Put K'=K and $\omega '=\frac{\omega }{2}$.

$\frac{I'}{I}=(\frac{2K}{K})(\frac{\omega }{\frac{\omega }{2}})$

$I'=4I$

Therefore, the angular momentum becomes 4 times.