Kinematics Question 491
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[IIT 1998; CBSE PMT 2004]
Options:
A) $ \sqrt{u^{2}-2gL} $
B) $ \sqrt{2gL} $
C) $ \sqrt{u^{2}-gl} $
D) $ \sqrt{2(u^{2}-gL)} $
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Answer:
Correct Answer: D
Solution:
$ \frac{1}{2}mu^{2}-\frac{1}{2}mv^{2}=mgh$
therefore $ v=\sqrt{u^{2}+2gL} $
$ |\vec{v}-\vec{u}|=\sqrt{u^{2}+v^{2}}=\sqrt{u^{2}+v^{2}-2gL}=\sqrt{2(u^{2}-gL)} $
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