First slide

Vertical circular motion

Question

A stone is tied to a string of length 'l' and is whirled in a vertical circle with the other end of the string as 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 velocity as it reaches a position where the string is horizontal ( g being acceleration due to gravity) is

Moderate

A

$\sqrt{2 (u^{2} â gl)}$
B

$\sqrt{u^{2} â gl}$
C

$u â \sqrt{u^{2} â 2 gl}$
D

$\sqrt{2 gl}$

Solution

Applying the law of conservation of energy at points B and A, we have
$\frac{1}{2} {mu}^{2} = \frac{1}{2} {mv}^{2} + mgl$
or $V = \sqrt{(u^{2} â 2 gl)}$ â¦(1)
This magnitude of change of velocity figure is given by
$\begin{array}{l} | Îv | = | v â u | \\ = \sqrt{(v^{2} + u^{2} â 2 uvcos â¡ 90^{â})} \\ = \sqrt{(v^{2} + u^{2})} . . . . (2) \end{array}$
From eqs, (1) and (2), we get
$\begin{array}{l} | Îv | = \sqrt{(u^{2} â 2 gl + u^{2})} \\ = \sqrt{[2 (u^{2} â gl)]} \end{array}$

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