Q. 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
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
Explanation
(b) Here the tangential acceleration also exists which requires power
Given that $a_{C} = k^{2}rt^{2}$ and $a_{C} = \frac{v^{2}}{r}\therefore\frac{v^{2}}{r} = k^{2}rt^{2}$
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 $P = F \times v = mkr \times krt = mk^{2}r^{2}t$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.