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
