A semi-cylinder is cut out of a block and the block is kept on the horizontal surface. A small body of mass m is released at the top of the cylinder shaped hole. (Friction is negligible.) If the block

Question

A semi-cylinder is cut out of a block and the block is kept on the horizontal surface. A small body of mass m is released at the top of the cylinder shaped hole. (Friction is negligible.) If the block is able to move on the horizontal surface without friction, then the force exerted on the block by the small body at the lowest point is 7 m g /2 . Find the ratio of M / m ?

TestHub · Accepted Answer

There are no external horizontal forces, so momentum is conserved along the $x$ -axis. If $v_{1}$ is the speed of the block and $v_{2}$ the speed of the body at the lowest point,

$0 = m v_{2} - M v_{1} \dots (i)$

and the work--energy theorem gives

$m g R = \frac{1}{2} m v_{2}^{2} + \frac{1}{2} M v_{1}^{2} \dots (ii)$

From (i) and (ii),

$\Rightarrow 2 m g R = m v_{2}^{2} + \frac{m ^{2}}{M} v_{2}^{2} \Rightarrow v_{2}^{2} = \frac{2 g RM}{M + m}, v_{1} = \frac{m}{M} \frac{2 g RM}{M + m}$

Making the FBD of the particle with respect to the wedge at the lowest point (relative speed $v_{1} + v_{2}$ , circular path of radius $R$ ),

$\frac{7 m g}{2} - m g = \frac{m ( v _{1} + v _{2} ) ^{2}}{R}$

Substituting the values,

$(v_{1} + v_{2})^{2} = \frac{2 g RM}{M + m} (1 + \frac{m}{M})^{2} = \frac{2 g R ( M + m )}{M} \Rightarrow \frac{5 m g}{2} = \frac{2 m g ( M + m )}{M} \Rightarrow \frac{M}{m} = 4$

Physics - Center of Mass Question with Solution | TestHub

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