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Figure shows a smooth cylindrical pulley of radius R with centre at origin of co-ordinates. An ideal thread is thrown over it on the two ends of ideal thread two identical masses are tied initially at rest with coordinates (R, 0) and (–R, –R) respectively. If mass at x-axis is given a slight upward jerk, it leaves contact with pulley at $(R \cos θ, R \sin θ)$ . Then $\frac{θ}{\sin θ}$ is equal to

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answer is 2.

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Detailed Solution

Conservation of energy $m g (R θ) - m g R \sin θ = 2 (\frac{1}{2} m v^{2}) .... (i)$ $F o r l o \sin g c o n t a c t$

$\frac{{mv}^{2}}{R} =msinθ v^{2} =gRsinθ .... (ii) From (i) a n d (i i) \Rightarrow m g R (θ - \sin θ) = m g R \sin θ θ = 2 \sin θ \frac{θ}{\sin θ} = 2$

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