Question

The angular momentum of a particle performing uniform circular motion is L. If the kinetic energy of the particle is doubled and frequency is halved, then angular momentum becomes
A.\[\dfrac{L}{2}\]
B.\[2L\]
C.\[\dfrac{L}{4}\]
D.\[4L\]

Answer 1

Hint: At the point when you set an inflexible body in the pivot, it will, in general, keep turning at a similar speed, around a similar axis - except if followed up on by a torque. We state that pivoting bodies have angular momentum. Angular momentum is preserved, and the main path for an article to quit turning is if its angular momentum is moved to another item. (That is the thing that torque is: the exchange of angular momentum.)

Complete answer:
Angular momentum is defined by the following formula \[L=I\omega ...........(i)\]
Where I will be the moment of inertia of the body.
So, for the kinetic energy:
\[K=\dfrac{1}{2}I{{\omega }^{2}}\]
(from equation(i))
\[L=\dfrac{2K}{\omega }.......(ii)\]
Now we have the given angular frequency which has been has been halved
So, let \[\omega '=\dfrac{\omega }{2}\]
The kinetic energy is doubled
So \[K'=2K\]
Now we have the angular frequency has as L’ is the new angular momentum, then using (ii) we can write,
\[L'=\dfrac{2K'}{\omega '}\]
\[\Rightarrow L'=4\dfrac{2K}{\omega }\]
\[\Rightarrow L'=4L\]
So, angular momentum becomes four times its original value.
The angular momentum of a rigid object can be defined as the product of the moment of inertia and the angular velocity.
It is clearly analogous to the linear momentum and is subjected to the fundamental units of the conservation of angular momentum principle if there is no external torque on the object.

The correct answer is D.

Note:
Angular velocity and the angular momentum are vector quantities, and they have both magnitude and the direction. The direction of the angular velocity and angular momentum is clearly perpendicular to the plane of the rotation.