Kinematics Question 486
Question: A particle of mass $ m $ is moving in a circular path of constant radius $ r $ such that it’s centripetal acceleration $ a _{c} $ it’s varying with time t as, $ a _{c}=k^{2}rt^{2} $ , The power delivered to the particle by the forces acting on it’s[IIT 1994]
Options:
A) $ 2\pi mk^{2}r^{2}t $
B) $ mk^{2}r^{2}t $
C) $ \frac{mk^{4}r^{2}t^{5}}{3} $
D) Zero
Show Answer
Answer:
Correct Answer: B
Solution:
Here the tangential acceleration also ex it’s which requires power.
Given that $ a _{C}=k^{2}rt^{2} $ and $ a _{C}=\frac{v^{2}}{r} $ $ \frac{v^{2}}{r}=k^{2}rt^{2} $
or $ v^{2}=k^{2}r^{2}t^{2} $ or $ v=krt $
Tangential acceleration $ a=\frac{dv}{dt}=kr $
Now force $ F=m\times a=mkr $
So power $ P=F\times v=mkr\times krt=mk^{2}r^{2}t $
