Question

In the arrangement shown in figure above the mass of ball 1 is $\eta =1.8$ times as great as that of rod 2. The length of the latter is $l=100cm$. The masses of the pulleys and the threads, as well as the friction, are negligible. The ball is set on the same level as the lower end of the rod and then released. If t be the time taken by the ball to reach the opposite end of bar, find 5t

Answer 1

Suppose, the ball goes up with acceleration $w_1$ and the rod comes down with the acceleration $w_2$.
As the length of the thread is constant, $2w_1=w_2$ (1)
From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives,
$T-mg=m{w}_1$ (2)
and $Mg-T^{\prime }=M{w}_2$ (3)
but $T=2T$ (because pulley is massless) (4)
From equations (1), (2), (3) and (4)
$w_1=\frac{(2M-m)g}{m+4M}=\frac{(2-\eta )g}{\eta +4}$ (in upward direction)
and $w_2=\frac{2(2-\eta )g}{\eta +4}$ (downwards)
From kinematical equation in projection form, we get
$l=\frac{1}{2}(w_1+w_2)t^2$
as, $w_1$ and $w_2$ are in the opposite direction.
Putting the values of $w_1$ and $w_2$, the sought time becomes
$t=\sqrt{\frac{2l(\eta +4)}{3(2-\eta )g}}=1.4s$

$⟹5t=7$