## Friday, April 20, 2007

### IIT-JEE 2007 Questions on Rotational Motion

A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v2/4g with respect to the initial position. The object is

(a) ring (b) solid sphere (c) hollow sphere (d) disc

The body has translational and rotational kinetic energies and these are completely converted in to gravitational potential energy at the maximum height so that we can write

½ Mv2 + ½ Iω2 = Mgh Where M is the mass, I is the moment of inertia and ω is the angular velocity of the body and h is the maximum height reached. Since ω = v/R where R is the radius of the rolling body, the above equation can be rewritten as

½ Mv2 + ½ I(v2/R2) = Mg×(3v2/4g)

From this, I = MR2/2, which is the value for a disc [Option (d)].

The following three questions which appeared in IIT-JEE 2007 question paper are Linked Comprehension Type multiple choice questions:

Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential enargy of a spring compressed by a distance x1. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x2. Both the discs rotate in the clockwise direction.

Question(1):

The ratio x1/x2 is

(a) 2 (b) ½ (c) √2 (d) 1/√2

Equating the potential energies of the springs to the kinetic energies of the discs, we have

½ k x12 = ½ I×4ω2 and

½ k x22 = ½ ×2I×ω2 for the two cases. Here ‘k’ is the spring constant.

From these equations, x1/x2 = √2

Question 2:

When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is

(a) 2Iω/3t (b) 9Iω/2t (c) 9Iω/4t (d) 3Iω/2t

Since the angular momentum is conserved, we have

I×2ω + 2I×ω = (I+2I)×ω’ where ω’ is the common angular velocity of the discs. [We have added the angular momenta since they are in the same direction].

From this, ω’ = (4/3)ω

Disc A will have an angular retardation of magnitude ‘α1’ during the time ‘t’ where as disc B will have an angular acceleration of different magnitude ‘α2’ during the time ‘t’.

Considering disc A, we have ω’ = 2ω α1t from which α1 = (ω’)/t = 2ω/3t since ω’ = (4/3)ω.

The average frictional torque exerted on A by B = Iα1 = 2Iω/3t

[An equal and opposite torque will be exerted on B by A. Check by finding α2 and hence 2Iα2].

Question 3:

The loss of kinetic energy during the above process is

(a) Iω2/2 (b) Iω2/3 (c) Iω2/4 (d) Iω2/6

Loss of kinetic energy = Initial kinetic energy – Final kinetic energy

= ½ I×(2ω)2 + ½ ×2I×ω2 – ½ ×3I×(4ω/3)2 = (Iω2)/3