A student throws a horizontal stick of length `L` up into the air. At the moment it leaves her hand the speed of stick's closes end is zero. The stick completes `N` turns as it is caought by the student at the initial release point. Find the height `h` to which the centre of mass of the rod rises.

Answer

`(pi NL)/(4)`

Explanation

`V_(cm)=omega(l)/(2)`
`t=(2V_(cm))/(g)=(omegal)/(g)` time of flight
`T=(2pi)/(omega)` time period of one revolution
`t=NT ` `(omega l)/(g)=N(2pi)/(omega)`
`omega^(2)l^(2)=2Npilg`
`H=(V_(cm)^(2))/(2g)=(omega^(2)l^(2))/(4xx2g)=(2Npilg)/(4xx2g)=(Npil)/(4)`