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 3v

^{2}/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 3v

^{2}/4g it has zero kinetic energy since the entire kinetic energy is converted into gravitational potential energy. Thus we have
½
I ω

^{2}+ ½ mv^{2}+ 0 = mg(3v^{2}/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 ω

^{2}= ¼ mv^{2}
Since ω =v/R the above equation becomes

Iv

^{2}/R^{2}= ½ mv^{2}
Therefore I = mR

^{2}/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) 2

*g/L*
(3) 2

*g/*3*L*
(4) 3

*g/*2*L*
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

*I*α where*I*is the moment of inertia and α is the angular acceleration we have*I*α =

*MgL*/2 ……………..(i)

The moment of inertia of the rod about the axis
of rotation through its end is

*ML*^{2}/3 as given by the parallel axes theorem.
[Moment of inertia of a uniform rod about a
central axis perpendicular to its length is

*ML*^{2}/12. Moment of inertia about a parallel axis through the middle is*ML*^{2}/12 +*M*(*L/*2)^{2}=*ML*^{2}/3].
Substituting for

*I*in equation (i), we have
(

*ML*^{2}/3)α =*MgL*/2
Therefore α = 3

*g/*2*L*
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