Question

A particle performing uniform circular motion has angular momentum $L$. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is:
A.)$\dfrac{L}{4}$
B.)$2L$
C.)$4L$
D.)$\dfrac{L}{2}$

Answer 1

We know that angular momentum $L$ can be written as $I \omega$. A good starting point would be to derive an expression for the angular momentum in terms of only frequency and kinetic energy. In other words, obtain an expression for the frequency in terms of angular velocity $\omega$, and the kinetic energy in terms of the moment of inertia I and $\omega$ and substitute them in the first angular momentum equation.

Formula used:
Angular momentum $L = I \omega$, where $I$ is the moment of inertia and $\omega$ is angular velocity.
Angular velocity $\omega = 2 \pi f$ where f is the frequency of circular motion.
Kinetic energy $KE = \dfrac{1}{2} I \omega^2$

Complete step-by-step answer:
Let us first try and understand the situation.
We have a particle that is performing repetitive circular motion. So, the number of times the particle completes a circular path will be its frequency of motion (measured in $s^{-1}$), which in turn depends on how fast the particle is moving with directional changes as it moves in a circle (which entails $360 ^{\circ}$ or $2 \pi radians)$, and we call this the angular velocity.
Thus, we have angular velocity given by

$\omega$ = distance it travels x the number of times it travels the distance = $2 \pi f$

For linear motion, we know that the kinetic energy possessed by the moving body is given as:

$KE_{linear} = \dfrac{1}{2}mv^2$ where m is the mass (inertia in linear motion) and v is the linear velocity.

However, since a particle moving in circular motion is subjected to a rotational acceleration, we cannot just consider the mass but instead we consider the mass’s resistance to this rotational acceleration, which we call as the moment of inertia. Thus, out rotational kinetic energy becomes:

$KE_{rotational} = \dfrac{1}{2} I \omega^2 \Rightarrow I = \dfrac{2\;KE}{\omega^2}$
Now, we know that linear momentum P is given as:
$P = mv$
Thus, our angular momentum L is given as:
$L = I \omega$
Substituting the expressions for I and $\omega$ that we obtained above we get
$L = \dfrac{2\; KE_{rotational}}{\omega^2} \times \omega = \dfrac{2 KE_{rotational}}{\omega} = \dfrac{2 KE_{rotational}}{2 \pi f}$
$\implies L = \dfrac{ KE_{rotational}}{\pi f}$
Now if frequency is doubled ($f^{\prime} = 2f$) and kinetic energy halved $(KE^{\prime}_{rotational} = \dfrac{ KE_{rotational} }{2})$ then:
$L^{\prime} = \dfrac{= \dfrac{ KE_{rotational} }{2}}{\pi 2f} = \dfrac{ KE_{rotational}}{\pi f} \times \dfrac{1}{4}$

$\implies L^{\prime} = \dfrac{L}{4}$

So, the correct answer is “Option A”.

Note:
(I)It is important to understand the distinction between mass and moment of inertia. We know that in general, inertia is the object’s resistance against acceleration.
(II)In linear (translational) motion this is called mass [kg]. The larger the mass the tougher it is to push something to move it or slow it down.
(III)In rotational cases, this is called moment of inertia [kg $m^2$]. The larger the moment of inertia, the tougher it is to wing a wheel up to spin fast or slow it down. Quantitatively it is given as
$I = \Sigma mr^2$ where m is the mass of the rotating body with respect to the distance r of the mass from the axis of rotation.