Question

In the arrangement shown in the figure, 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. Find 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. The rod length equals $l$ .

Accepted Answer

Let the acceleration of the rod is $w_{M} m s^{- 2}$ , and the acceleration of the ball is $w_{m} m s^{- 2}$ and both are in downward direction.
Now, the relative acceleration of rod wrt ball is $w = w_{M} - w_{m}$ , directed downwards.

It is given that, in time $t s$ , the ball travels $l$ distance wrt rod, as the length of the rod is $l$ , and the initial velocity is zero. So we can write using equation of displacement $s = u t + \frac{1}{2} a t^{2}$ , where $s$ is the displacement, $u$ is the initial velocity, $a$ is the acceleration and $t$ is time.
$l = \frac{1}{2} (w_{M} - w_{m}) t^{2}$
$\Rightarrow w_{M} - w_{m} = \frac{2 l}{t^{2}} . . . . (1)$

The tension is a pulling force between rod and string as well as between ball and string. The contact force between ball and string is the frictional force which will be equal to the tension in the string.

Using Newton's second law in vertical direction on rod as shown in figure,
$M g - T = M w_{M} \Rightarrow M g - f_{r} = M w_{M}$
$\Rightarrow w_{M} = \frac{M g - f_{r}}{M} . . . . (2)$
By applying on ball,
$m g - f_{r} = m w_{m}$
$\Rightarrow w_{m} = \frac{m g - f_{r}}{m} . . . . . . . (3)$

Using the equation $(2)$ and $(3)$ ,
$w_{M} - w_{m} = \frac{m (M g - f_{r}) - M (m g - f_{r})}{M m} = \frac{f_{r} (M - m)}{M m} . . . . (4)$

By comparing the equations $(1)$ and $(4)$ ,

$\frac{f_{r} (M - m)}{M m} = \frac{2 l}{t^{2}}$

$\Rightarrow f_{r} = \frac{2 l M m}{(M - m) t^{2}}$ .

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