Physics Two Dimensional Motion Question 31
Question: A particle of mass $ m $ is moving in a circular path of constant radius $ r $ such that its centripetal acceleration $ a _{c} $ is varying with time t as, $ a _{c}=k^{2}rt^{2} $ , The power delivered to the particle by the forces acting on it is
[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
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Answer:
Correct Answer: B
Solution:
Here the tangential acceleration also exits 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 $
BETA
