Question

A student throws a stick of length L up in the air. At the moment the stick leaves his hand, the speed of the stick's end in his hand is zero. The stick completes N turns as it is caught by the student at the initial release point. The height to which the centre of mass of the stick rose is:;
a) πNL/4

b) πNL/2

c) 2πNL/3

d) None

Solution

Let the height to which the Centre of Mass of the stick rose be h
Let the angular velocity of the stick be ω
Let the translational velocity of the centre of mass be v

If the stick is thrown such that its one end is at rest, it means that its motion is analogous to the case of no-slip motion.
Therefore, v = R * ω

where, R is the radius of the circular motion, i.e.,
R = L/2

(as the motion is about the centre of mass of the stick)

The height translated by the stick is,
h = (v\^2)/(2g)

(using the equations of motion)

The expression for g can be derived by equating the time taken, t for translation and rotational motions,

t = (2v)/g = θ/ω

where, θ is the angular displacement, which is equal to the number of turns taken multiplied by 2π

Therefore,

(2v)/g = (2πN)/ω

∴ g = (2vω)/(2πN)

Substituting g in the expression for height, we get

h = (πNL)/4

(using v = Rω = L/(2ω))

Therefore the answer is a) (πNL)/4