Question

In the arrangement shown in the mass of the ball is $\eta$ times as great as that of the rod. The length of the rod is $l$, 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. How soon will the ball be opposite the upper end of the rod?

• A. $t=\sqrt{\frac{2l(\eta +4)}{3g(2-\eta )}}$
• B. $t=\sqrt{\frac{l(\eta +4)}{3g(2-\eta )}}$
• C. $t=\sqrt{\frac{2l(\eta +4)}{g(2-\eta )}}$
• D. $t=\sqrt{\frac{3l(\eta +4)}{2g(2-\eta )}}$

Answer 1

Answer: (A)

$t=\sqrt{\frac{2l(\eta +4)}{3g(2-\eta )}}$

From constraint relation we can see that the acceleration of the rod is double than that of the ball.

If ball is going up with an acceleration $a$, rod will be coming down with the acceleration $2a$.

Thus, the relative acceleration of the ball with respect to rod is $3a$ in upward direction.
If it takes time $t$ seconds to reach the upper end of the rod, we have $t=\sqrt{\frac{2l}{3a}}$ ...(i)
Let mass of ball be m and that of rod is M, the dynamic equation of these are For rod $Mg-T=M\left(2a\right)$ ...(ii)
For ball $2T-mg=ma$ ...(iii)
Substituting $m=\eta M$ and solving Eqs. (ii) and (iii), we get $a=\left(\frac{2-\eta }{\eta +4}\right)g$
From Eq. (i), we have $t=\sqrt{\frac{2l(\eta +4)}{3g(2-\eta )}}$