Kinematics Question 546
Question: If the equation for the displacement of a particle moving on a circular path it’s given by $ (\theta )=2t^{3}+0.5 $ , where $ \theta $ it’s in radians and $ t $ in seconds, then the angular velocity of the particle after 2 sec froM is start it’s[AIIMS 1998]
Options:
A) 8 rad/sec
B)12 rad/sec
C) 24 rad/sec
D)36 rad/sec
Show Answer
Answer:
Correct Answer: C
Solution:
$ \omega =\frac{d\theta }{dt}=\frac{d}{dt}(2t^{3}+0.5)=6t^{2} $ at t =2 s, $ \omega =6\times {{(2)}^{2}}=24rad/s $
