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Question: A stone tied to a string of length L is whirled in a vertical circle with the other end of the strin...

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 change in its velocity as it reaches a position, where the string is horizontal is

A

u2−2gl\sqrt{u^{2} - 2gl}

B

2gl\sqrt{2gl}

C

u2−gl\sqrt{u^{2} - gl}

D

2(u2−gl)\sqrt{2\left( u^{2} - gl \right)}

Answer

2(u2−gl)\sqrt{2\left( u^{2} - gl \right)}

Explanation

Solution

Here v2 – u2 = -2gl (i)

v2 = u2 – 2gl

Since the velocities are mutually perpendicular, change in velocity

∆v = u2+v2\sqrt{u^{2} + v^{2}} …. (ii)

= u2+u2−2gl\sqrt{u^{2} + u^{2} - 2gl}

(substituting the value of v2 from (i))

or ∆v =2(u2−gl)\sqrt{2\left( u^{2} - gl \right)}