Question

A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the center. 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 (assume $g$ as acceleration due to gravity)

• A. $\sqrt{u^{2} - 2 g L}$
• B. $\sqrt{2 g L}$
• C. $\sqrt{u^{2} - g L}$
• D. $\sqrt{2 \left(u^{2} - g L\right)}$

Answer 1

Answer: (D)

From energy conservation
$v^{2}=u^{2}-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
$\left|\Delta \overset{ \rightarrow }{v \, }\right|=\sqrt{u^{2} + v^{2}}$

Substituting value of $ \, v^{2}$ from Eq.(1)
$\left|\Delta \overset{ \rightarrow }{v \, }\right|=\sqrt{u^{2} + u^{2} - 2 g L}$
$\left|\Delta \overset{ \rightarrow }{v \, }\right|=\sqrt{2 \left(u^{2} - g L\right)}$