A hollow rod has frictionless inner walls. Inside the rod are two small spheres that are kept on either side of the centre of rod. The rod is rotated about its central axis which is perpendicular to the plane of rotation. If the rod had been rotated by an impulsive torque which gave it an instantaneous angular velocity of 5rad/s, what will be the angular velocity after some finite but long time ‘t’? Assume no external forces act on rod after the impulse. Also state what will happen to the spheres in the centre of the rod. The rod has a mass =2kg & length =10cm. The two spheres have a mass =1kg each.

(a) 2.5 rad/s, spheres will stay where they are as there is no friction to move them

(b) 0 rad/s, spheres will move outwards and decrease velocity of rod to zero

(c) 1.25 rad/s, spheres will move to opposite ends of rod

(d) 2.5 rad/s, spheres will move to opposite ends of rod

This question was posed to me in an online interview.

This intriguing question comes from Angular Momentum in case of Rotations about a Fixed Axis topic in division System of Particles and Rotational Motion of Physics – Class 11

Select the correct answer from above options

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Correct option is (c) 1.25 rad/s, spheres will move to opposite ends of rod

To explain I would say: When the rod gets an impulse, it gets an initial angular velocity. Now, the two spheres will start going outwards till they reach the end of the rod on either side as the walls of rod’s ends will provide the required centripetal force. Also, to find final angular velocity we can conserve angular momentum as there is no external torque.

I1w1 = I2w2,

where I1, I2, w1, w2 are the initial & final moment of inertias & angular velocities resp.

I1 = MI^2/12 + 0 = 2*0.01/12 = 0.01/6 kgm^2.

I1 = MI^2/12 + 2*mI^2/4

= (0.01/6) + (0.5*1*0.01)

= (0.01/6) + (0.01/2)

= 0.04/6 = 0.02/3 kgm^2.

∴ I1w1 = I2w2 0.01/6 * 5 = 0.02/3 *w2

∴ w2 = 1.25 rad/s.