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:;
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!!
!!(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!!