Q. A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripctal acceleration $a_{c}$ is varying with time $t$ as $a_{c} = k^{2}rt^{2}.$ the power 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 $a_{c} = \frac{v^{2}}{r} = k^{2}rt\ \ \ \because v = krt$
$\therefore\ v = krt$
The integral acceleration is $a_{t} = \frac{dv}{dt} = \frac{d(krt)}{dt} = kr$
The work done by centripetal force will be zero
So power is delivered to the particle by only tangential force which acts in the same direction of instantaneous velocity
$\therefore\ Power\ = F_{y}v = ma_{t}krt = m(kr)(krt) = mk^{2}r^{2}t$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.