Kinematics Question 491
Question: A stone tied to a string of length $ L $ it’s 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 it’s lowest position and has a speed $ u $ . The magnitude of the change in it’s velocity as it reaches a position where the string is horizontal it’s[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)} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{1}{2}mu^{2}-\frac{1}{2}mv^{2}=mgL $
therefore $ v=\sqrt{u^{2}-2gL} $
$ |\vec{v}-\vec{u}|=\sqrt{u^{2}+v^{2}}=\sqrt{u^{2}+u^{2}-2gL}=\sqrt{2(u^{2}-gL)} $
