Kinematics Question 768
Question: A cyclist riding the bicycle at a speed of $ 14\sqrt{3} $ ms?1 takes a turn around a circular road of radius $ 20\sqrt{3} $ m without skidding.Given g = 9.8 ms^2, What is his inclination to the vertical[Kerala (Engg.) 2001]
Options:
A) $ 30^{o} $
B) $ 90^{o} $
C) 45 o
D) $ 60^{o} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \theta ={{\tan }^{-1}}( \frac{v^{2}}{rg} )={{\tan }^{-1}}[ \frac{{{(14\sqrt{3})}^{2}}}{20\sqrt{3}\times 9.8} ]={{\tan }^{-1}}[\sqrt{3}] $
$ =60{}^\circ $
