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A block of mass M is situated on a smooth horizontal frictionless table. A thread tied to the block passes through a hole in the table and carries a mass m at its other end. If the length of the thread above the table is l,what should be the value of m so that it may remain suspended at a constant height and the block M moves in a circular path with an angular velocity $ω$ on the table?

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$\frac{{Mlω}^{2}}{3 g}$

$\frac{{Mlω}^{2}}{5 g}$

$\frac{{Mlω}^{2}}{g}$

$\frac{2 {Mlω}^{2}}{g}$

answer is A.

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Detailed Solution

The situation is shown in figure. For the mass m to be stationary, the tension in the string should provide the necessary centripetal force on the rotating mass M.
Now

$T = mg and T = {Mlω}^{2} or mg = M {lω}^{2}$

or $m = \frac{{Mlω}^{2}}{g}$

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