In the arrangement shown in figure the mass of the ball is `eta` times as that of the rod. The length of the rod is `L` the masses of the pulleys and the threads as well as the friction, are negligible. The ball is set on the same level as the lower end of the rod and then released. How soon will the ball be opposite the upper and of the rod?

Explanation

Suppose, the ball goes up with acceleration `w_1` and the rod comes down with the acceleration `w_2`.
As the length of the thread is constant, `2w_1=w_2` (1)
From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives,
`T-mg=mw_1` (2)
and `Mg-T^'=Mw_2` (3)
but `T=2T` (because pulley is massless) (4)
From Eqs. (1), (2), (3) and (4)
`w=((2M-m)g)/(m+4M)=((2-eta)g)/(eta+4)` (in upward direction)
and `w_2=(2(2-eta)g)/((eta+4))` (downwards)
From kinematical equation in projection form, we get
`l=1/2(w_1+w_2)t^2`
as `w_1` and `w_2` are in the opposite direction.
Putting the values of `w_1` and `w_2`, the sought time becomes
`t=sqrt(2l(eta+4)//3(2-eta)g)=1*4s`