First slide

Vertical circular motion

Question

If $v = \sqrt{(5 gl)}$ , what will be the velocity of the bob at the top of vertical circle

Easy

A

$\sqrt{(gl / 2)}$
B

$\sqrt{(gl)}$
C

$\sqrt{(2 gl)}$
D

$\sqrt{4 gl}$

Solution

$v^{2} = u^{2} + 2 gh$
Here $u = \sqrt{(5 gl)}, g = - g and h = 2 l$
$∴ V^{2} = 5 gl - 4 gl = gl or V = \sqrt{(gl)}$

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