Sunday, September 16, 2007

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.

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