Question

A particle of mass m is placed over a horizontal circular table rotating with an angular velocity ω about a vertical axis passing through its center. The distance of the object from the axis is r. Based on the above, find the force of friction f between the particle and the table.

Answer 1

The particle is stationary in the rotating reference frame rigidly fixed to the rotating table. In the list of all the forces acting on the particle, include the centrifugal force (pseudo force) acting on the particle radially outwards. 

Let us solve this problem from both the frames. The one is a frame fixed on the ground while the other is a frame fixed on the table itself.

From frame of reference fixed on ground (inertial) 

Here, N will balance its weight and force of friction, whereas f will provide the necessary centripetal force. 

Thus f = mrω²

From frame of reference fixed on table itself (non-inertial) 

In the FBD of particle with respect to table, in addition to the above three forces (N, mg and f) a pseudo force of magnitude mrω² will have to be applied in a direction away from the center. But one significant point here is that in this frame the particle is in equilibrium, i.e., N will balance its weight in vertical direction, whereas f will balance the pseudo force in horizontal direction. f = mrω² 

Thus, we observe that ‘f’ equals 2 mrω² from both the frames. Now, let us work up few more examples of circular motion.