Kinematics Question 550

Question: A pendulum bob on a 2 m string is d it’splaced 60o from the vertical and then released.What is the speed of the bob as it passes through the lowest point in it’s path [JIPMER 1999]

Options:

A) $ \sqrt{2}m/s $

B) $ \sqrt{9.8}m/s $

C) 4.43 m/s

D)$ 1/\sqrt{2}m/s $

Show Answer

Answer:

Correct Answer: C

Solution:

$ v=\sqrt{2gl(1-\cos \theta )}=\sqrt{2\times 9.8\times 2(1-\cos 60{}^\circ )} $

$ =4.43m/s $



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