Question

A sphere is released on a smooth inclined plane from the top. When it moves down, what happens to its angular momentum?
A. Conserved about every point
B. Conserved about the point of contact only
C. Conserved about the centre of the sphere only
D. Conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball.

Answer 1

Hint: Before we proceed with the problem, it is important to know about torque and angular momentum. Torque is defined as the measure of force that can cause an object to rotate about an axis. Angular momentum is the measure of the rotational momentum of a rotating body or system.

Complete step by step solution:
Consider a sphere that is moving on a smooth inclined plane as shown in the diagram. Since it is a smooth inclined plane, there is no friction that comes into play and the sphere can never roll, rather it will just slip down.

Image: A sphere rolling on an inclined plane

The condition for the angular momentum should be conserved if the net torque should be zero i.e., \[{\tau _{net}} = 0\] means that the angular momentum is constant. Now we will see at which point it is conserved.

All the forces acting on the body are passing through the centre of mass. So, if we find the torque about the centre of mass of the sphere it would be zero i.e., here the \[{\rm{mgsin\theta }}\] is passing through the centre of the sphere, the \[\tau \] due to \[{\rm{mgsin\theta }}\] is zero and if you take any point parallel to the inclined plane which is passing through the centre of mass, here also the torque will be zero and the angular momentum will be conserved. Therefore, the angular momentum remains conserved at about any point on a line parallel to the inclined plane and passes through the centre of the ball

Hence, option D is the correct answer

Note:Option C is also the right answer if the term ‘only’ is excluded because the angular momentum is also conserved about any point on a line parallel to the inclined plane.