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A semicircular track of radius $R = 62.5 cm$ is cut in a block. Mass of block, having track, is $M = 1 kg$ and rests over a smooth horizontal floor. A cylinder of radius $r = 10 cm$ and mass $m = 0.5 kg$ is hanging by thread such that axes of cylinder and track are in same level and surface of cylinder is in contact with the track as shown in figure. When the thread is burnt, cylinder starts to move down the track. Sufficient friction exists between surface of cylinder and track, so that cylinder does not slip. Find force (in Newton) applied by block on the floor at that moment. $(g = 10 m / s^{2})$

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$10.57 N$

$13.57 N$

$15.57 N$

$21.57 N$

answer is D.

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Detailed Solution

From conservational of linear momentum

${mv}_{1} = {Mv}_{2}$

$∴$ Velocity of cylinder axis relative to block

$v_{r} = v_{1} + v_{2}$

Applying conservation of mechanical energy,

$mg (R - r) = \frac{1}{2} {mv}_{1}^{2} + \frac{1}{2} {Iω}^{2} + \frac{1}{2} {Mv}_{2}^{2}$

Here, $I = \frac{1}{2} {mr}^{2}$ and $ω = \frac{v_{r}}{r}$

Solving the above equations with given data. we get

$v_{1} = 2.0 m / s v_{2} = 1.0 m / s Further, N - mg = \frac{{mv}^{2}}{R - r} N = mg + \frac{{mv}_{r}^{2}}{R - r} = (0.5) (10) + \frac{(0.5) (3.0)^{2}}{0.525} = 13.57 N$

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