If time taken by bead A and B to slide down on chords PQ and RQ are $t_{1}$ and $t_{2}$ respectively, then find the ratio $\frac{t_{1}}{t_{2}}$ .(diameter of vertical ring is d)

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Solution

Time taken by bead A and B to slide down or chords PQ and RQ are $t_{1}$ and $t_{2}$ respectively.
Finding ratio $= \frac{t_{1}}{t_{2}}$
$t_{1}^{2} = \frac{2 L}{g sin θ} t_{2}^{2} = \frac{3 L}{g sin θ}$
When bead rolls forward
$a = g sin θ - \frac{f}{m} \Rightarrow v = a t ∴ t_{1} : t_{2} = \sqrt{2} : \sqrt{3}$

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