A stone tied 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 the change in its velocity as it reaches a position where the string is horizontal is

Question

Dr. Abubacker · Accepted Answer

(d) In this case motion of stone is in vertical circle of radius $L$ and centre at $O$
The change in velocity is
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$\mathrm{\Delta}\overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{v}} - \overrightarrow{\mathbf{u}} = v\widehat{\mathbf{j}} - u\widehat{\mathbf{i}}$
$\left| \mathrm{\Delta}\overrightarrow{\mathbf{v}} ight| = \sqrt{(v)^{2} + ( - u)^{2}}$
$	herefore\ \  = \sqrt{v^{2} + u^{2}}$
According to work-energy theorem,
$W = \mathrm{\Delta}K$
or $W_{T} + W_{g} = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$ …(i)
$W_{T} =$ work done by the force of tension = 0
$W_{g} =$ work done by the fore of gravity
$= mgL$ (path independent)
From Eq. (i), $0 - mgL = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$
$\because\ \ \ v^{2} = u^{2} - 2gL$
$	herefore\ \ \left| \mathrm{\Delta}\overrightarrow{\mathbf{v}} ight| = \sqrt{v^{2} + u^{2}} = \sqrt{2(u^{2} - gL)}$

Explanation

Reviewed by Dr. Abubacker

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