Question

The arrangement shown in figure mass of the rod is $M$ ad mass of bead is $m$ and $M>m$. The bead slides with some friction. The mass of the pulley and friction in its axle are negligible. At the initial moment the bead was located opposite the lower end. If set free both the bodies move with constant accelerations. The frictional force between the bead and the thread so that $t$ second after release ball reaches the upper end of the rod is given as $F_f=\frac{xmMl}{(M-m)t^2}$. Find $x$, Length of the rod is $l$.

Answer 1

Free body diagram is shown in fig.

From fig. 1(a);

$Mg-T=M{a}_1$ .................(1)

$T-f=0$

$T=f$ .................(2)

$f-mg=m{a}_2$ .................(3)

By using (2), (1) becomes;

$Mg-f=M{a}_1$

$g-\frac{f}{M}=a_1$ .................(4)

and eq. (3) becomes

$\frac{f}{m}-g=a_2$ .................(5)

From eq. (4) and (5);

$f(\frac{1}{m}-\frac{1}{M})=a_1+a_2$ .................(6)

Now, relative motion of bead w.r.t tower is $(a_1+a_2)$

there force, we have

$l=\frac{1}{2}(a_1+a_2)t^2$ $=\frac{1}{2}f(\frac{1}{m}-\frac{1}{M})t^2$

$\Rightarrow F=\frac{2mMl}{(M-m)t}$

Comparing given equations;

x =2