The mass of the rod $M$ exceeds the mass 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 negligible. At the initial moment the ball was located opposite to lower end of the rod. When set free both bodies began moving with constant acceleration. Find the frictional force between the ball and the thread if $t$ seconds after the beginning of the motion the ball got opposite the upper end of the rod. The rod length equals $\mathcal{l}$ .

Let $T$ is the force in the thread due to friction between ball and thread. Suppose $A$ and a are the accelerations of rod and the ball respectively.

By Newton’s second law, we have

for rod,         $Mg-T = MA$    …(i)

for ball         $mg-T = ma$     …(ii)

Multiplying equation (i) by $m$ and (ii) by $M$ and subtracting equation (ii) from (i), we get

                  $T = \left(\frac{Mm}{M-m}\right)(A-a)$     …(iii)

From second equation of motion, we have

                  $t = 0+\frac{1}{2}(A-a)t^2$

or              $(A-a) = \frac{2\mathcal{l}}{t^2} .$       …(iv)

Now from equation (iii) and (iv), we get

                 $T =\frac{ 2Mm\mathcal{l}}{(M-m)t^2}$ .