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In the arrangement shown in figure the m...

In the arrangement shown in figure the mass of the ball is `eta` times as that of the rod. The length of the rod is `L` the masses of the pulleys and the threats as well as the friction, are negligible. The ball is set on the same level as the lower and of the rod and then released. How soon will the ball be opposite the upper and of the rod?

Text Solution

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From constraint relation we can see that the acceleration of the rod is double then that of the acceleration 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 ((2l)/(3a))` ...(i)
Let mass of ball be `m` and that of rod is `M` the dynamic of these are
For rod `Mg -T = M(2a)` ...(i)
For rod `2T - mg = ma` ...(ii)
Substituting `m = eta M` and solving Eqs (ii) and (iii), we get
`a = ((2 - eta)/(eta + 4)) g`
From Eq (i), we have
`t = sqrt((2l(eta + 4))/(3 g(2 - eta)))`
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