Question

In the arrangement shown in figure above the 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. The rod length equals $l$. The friction force between the ball and the thread if $t$ seconds after the beginning of motion the ball got opposite the upper end of the rod is $F_{fr}=\frac{xlMm}{(M-m)t^2}$. Find $x$.

Answer 1

As the thread is not tied with $m$, so if there were no friction between the thread and the ball $m$, the tension in the thread would be zero and as a result both bodies will have free fall motion. Obviously in the given problem it is the friction force exerted by the ball on the thread, which becomes the tension in the thread. From the condition or language of the problem $w_M>w_m$ and as both are directed downward so, relative acceleration of $M=w_M-w_m$ and is directed downward. Kinematical equation for the ball in the frame of rod in projection form along upward direction gives:
$l=\frac{1}{2}(w_M-w_m)t^2$ (1)
Newton's second law in projection form along vertically downward direction for both, rod and ball gives,
$Mg-fr=M{w}_M$ (2)
$mg-fr=m{w}_m$ (3)
Multiplying equation (2) by $m$ and equation (3) by $M$ and then subtracting equation (3) from (2) and after using equation (1) we get:
$fr=\frac{2lMm}{(M-m)t^2}$