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In the figure shown, a small solid ball of mass ‘m’ can move without sliding in a fixed cylindrical track of radius R in vertical plane. It is released from the top. The resultant force on the ball when it is at the lowest point of the track is given by k (mg) , then k is

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answer is 1.42.

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Detailed Solution

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From conservation of energy, the kinetic energy of ball at lowest portion is ( $v_{c} =$ speed of centre of ball)
$\frac{1}{2} m v_{c}^{2} + \frac{1}{2} \times \frac{2}{5} m v_{c}^{2} = m g R O r \frac{7}{10} m v_{c}^{2} = m g R$

Since net tangential force on sphere at lowest point is zero, net force on sphere at lowest position is
$= \frac{m v_{c}^{2}}{R} = \frac{10}{7} m g u p w a r d s$

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