Kinematics Question 766
Question: A cyclist goes round a circular path of circumference 34.3 m in $ \sqrt{22} $ sec. the angle made by him, with the vertical, will be [MH CET 2000]
Options:
A) $ 45^{o} $
B) $ 40^{o} $
C) $ 42^{o} $
D) $ 48^{o} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 2\pi r=34.3 $
therefore $ r=\frac{34.3}{2\pi } $ and $ v=\frac{2\pi r}{T}=\frac{2\pi r}{\sqrt{22}} $ Angle of binding $ \theta ={{\tan }^{-1}}( \frac{v^{2}}{rg} )=45{}^\circ $
