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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 the 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(u^{2} - gL)}

Answer

2(u2−gL)\sqrt{2(u^{2} - gL)}

Explanation

Solution

From energy conservation

v2 = u2 – 2gL … (1)

Now since the two velocity vectors shown in figure are mutually perpendicular, hence the magnitude of change of velocity will be given by

| Dv→\overset{\rightarrow}{v}| = u2+v2\sqrt{u^{2} + v^{2}}

Substituting value of v2 from Eq. (1)

| Dv→\overset{\rightarrow}{v}| = u2+u2−2gL\sqrt{u^{2} + u^{2} - 2gL}

| Dv→\overset{\rightarrow}{v}| = 2(u2−gL)\sqrt{2(u^{2} - gL)}