Kerala Engineering Entrance 2007 Questions on Rotational Motion

The following questions appeared in KEAM (Engineering) 2007 question paper:

(1) A sphere of mass ‘m’ and radius ‘r’ rolls on a horizontal plane without slipping with speed ‘u’. Now if it rolls up vertically, the maximum height it would attain would be

(a) 3u²/4g (b) 5u²/2g (c) 7u²/10g (d) u²/2g (e) 11u²/9g

The initial kinetic energy (E) of the sphere while rolling on the horizontal surface is given by

E = ½ mu² + ½ Iω² where I is the moment of inertia of the sphere about its diameter[(which is equal to (2/5)mr²] and ‘ω’ is the angular velocity (which is u/r).

[Note that the first term is the translational kinetic energy and the second term is the rotational kinetic energy].

Therefore, E = ½ mu² + ½ ×(2/5)mr²u²/r² = 7mu²/10.

Since this energy is converted into gravitational potential energy (mgh) to attain the maximum height h, we have

7mu²/10 = mgh, from which h = 7u²/10g.

(2) If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of the full day would be

(a) 3 hours (b) 1.5 hours (c) 6 hours (d) 4 hours (e) 2 hours

If the radius of the earth becomes ‘n’ times he present value, without change in the mass, the duration of the day will become 24n² hours so that the answer to the above question is 24×(1/4)² = 1.5 hours.