Thursday, May 09, 2013

NEET 2013 Questions (MCQ) on Rotational Motion

God used beautiful mathematics in creating the world.
– Dirac, Paul  Adrien Maurice

The following two questions on rotational motion, which were included in the NEET 2013 question paper, are worth noting:

(1) 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

(1) solid sphere

(2) hollow sphere

(3) disc

(4) ring

Initially the object has rotational and translational kinetic energy but zero gravitational potential energy. At the maximum height of 3v2/4g it has zero kinetic energy since the entire kinetic energy is converted into gravitational potential energy. Thus we have

            ½ I ω2 + ½ mv2 + 0 = mg(3v2/4g) where ‘I’ is the moment of inertia of the object about the axis of rotation, ‘ω’ is the angular velocity, ‘m’ is the mass and ‘v’ is the linear velocity of the object.

Simplifying, ½ I ω2 = ¼  mv2

Since ω =v/R the above equation becomes

            Iv2/R2 = ½ mv2

Therefore I = mR2/2

This means that the object is a disc [Option (3)].

(2) A rod PQ of M  and length L is hinged at end P. The rod is held horizontally by a massless string tied to point Q as shown in the figure. When the string is cut, the initial acceleration of the rod is

(1) g/L

(2) 2g/L

(3) 2g/3L

(4) 3g/2L

When the string is cut, the rod rotates about the end P and the torque responsible for the rotation is MgL/2.

[The weight Mg of the rod acts through the centre of gravity located at the middle of the rod. The distance of the line of action of the weight from the axis of rotation (the lever arm for the torque) is L/2]

Since torque is equal to Iα where I is the moment of inertia and α is the angular acceleration we have

            Iα = MgL/2 ……………..(i)

The moment of inertia of the rod about the axis of rotation through its end is ML2/3 as given by the parallel axes theorem.

[Moment of inertia of a uniform rod about a central axis perpendicular to its length is ML2/12. Moment of inertia about a parallel axis through the middle is ML2/12 + M(L/2)2 = ML2/3].

Substituting for I in equation (i),  we have

            (ML2/3)α = MgL/2

Therefore α = 3g/2L

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