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, \(a_c = k^2rt^2\). The power delivered to the particle by the forces acting on it is (\(K\) = constant)

• A. \(m^2k^2r^2t^2\)
• B. \(mk^2r^2t\)
• C. \(2πmk^2r^2t\)
• D. \(\frac{mk^4r^2t^5}{3}\)

Hint:

Find $v$ from $a_c=v^2/r$, then use $P=F_tv$ with $F_t=m\,dv/dt$.

Answer 1

The Correct Option is B

Step 1: Find the speed
\(a_c=\frac{v^2}{r}=k^2rt^2\), so \(v^2=k^2r^2t^2\) and \(v=krt\).

Step 2: Tangential acceleration
\(a_t=\frac{dv}{dt}=kr\). The tangential force is \(F_t=mkr\). The centripetal force does no work.

Step 3: Power
\(P=F_tv=mkr\times krt=mk^2r^2t\). Option (B).

Step 4: Why not the others
(D) would come from integrating the wrong expression, and (A) and (C) have wrong powers of \(m\) or extra factors of \(\pi\).

Final Answer:
The power is \(mk^2r^2t\), option (B). \[ \boxed{\text{(B)}} \]