A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u

Question

A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u . The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is x u 2 - g L . The value of x is

TestHub · Accepted Answer

Applying conservation of mechanical energy at the point $A$ and $B$ , we get $\frac{1}{2} m u^{2} + 0 = \frac{1}{2} m v^{2} + m g L$

$\Rightarrow v = \sqrt{u^{2} - 2 g L}$

Now in vector form. ${\vec{v}}_{i} = u \hat{i}$ and ${\vec{v}}_{f} = v \hat{j}$

Therefore, change in velocity will be, $∆ \vec{V} = v \hat{j} - u \hat{i}$ and $|Δ \vec{V}| = \sqrt{u^{2} + v^{2}}$

$\Rightarrow Δ V = \sqrt{u^{2} + (u^{2} - 2 g L)}$

$= \sqrt{2 (u^{2} - g L)}$

Hence, $x = 2$

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