Question

A bob of mass m attached to an inextensible string of length ‘l’ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed, $\omega rad \ s^{-1}$ about the vertical. About the point of suspension:
A. Angular momentum changes in direction but not in magnitude.
B. Angular momentum changes both in direction and magnitude.
C. Angular momentum is conserved.
D. Angular momentum changes in magnitude but not in direction.

Answer 1

In Newtonian mechanics, the product of moment of inertia and angular speed about an axis is called angular momentum of the body. The rate of change of linear momentum is called torque. If there is no change in angular momentum, or angular momentum remains constant with time, then we can say that the net torque on the system is zero. In other words we can say that if the net torque on the system is zero, then angular momentum will remain the same.
Formula used:
$I\omega = L , \ \tau = \dfrac{dL}{dt}$

Complete answer:
As the bob is rotating in a horizontal circle, the only force acting on the bob is tension and centrifugal force. Due to centrifugal force, the string will remain taut and hence the tension will always pass through the point of suspension. Hence net torque on the bob will be zero.
As net torque is zero i.e.$\tau = 0$, thus $\dfrac{dL}{dt} = 0$
This means, ‘L’ must be constant. Hence ‘L’ is constant and is equal to $L = I \omega$
And $I= ml^2, \ so \ L = ml^2 \omega$
Hence the angular momentum of the bob will be constant.

So, the correct answer is “Option C”.

Note:
We have proved that if net torque on the system is zero then angular momentum of the system can’t change. This change is including both magnitude and direction. Students must not be confused about the direction of angular momentum. It is a special kind of vector quantity called axial vectors. Angular momentum has a direction perpendicular to both the velocity of the particle and its radius vector i.e. is velocity of the particle is denoted by $\vec v$ and radius vector is denoted by $\vec r$ than the vector product will give the direction of the angular momentum i.e. $\vec v \times \vec r$. Hence for the bob, direction of angular momentum is always perpendicular to the string hence always constant (in vertical direction).