In the arrangement shown in fig., mass of the rod M exceeds the mass m of the ball. The ball has an opening permitting it to slide along the thread with some friction. The mass of the pulley and the friction in its axle are negligible. At the initial moment, the ball was located opposite the lower end of the rod. When set free, both bodies began moving with constant accelerations. Find the friction force between the ball and the thread if t seconds after the beginning of motion, the ball got opposite to the upper end of the rod. The rod length equals l.

Explanation

The friction force between ball and thread is `f_(r)` Acceleration of both the bodies is downward.

`mg-T=ma_(1)`
`Mg-T=Ma_(2)`
But `f_(r)=T`
`mg -f_(r)=ma_(1)`
`a_(2)-a_(1)=((Mg-f_(r))/(M))-((mg-f_(r))/(M))`
According to problem
`l=1/2 (a_(2)-a_(1))t^(2)` or `(a_(2)-a_(1))=(2l)/(t^(2))`
`(m(Mg-f_(r))-M(mg-f_(r)))/(Mm)=(2l)/(r^2)`
`(M-m)f_(r)=(2lmM)/(t^2) implies f_(r)=(2lmM)/((M-m)t^(2))`.