A particle of mass m is moving in a circular path of constant radius $r$ such that its centripetal acceleration (a) is varying with time t as $\mathrm{a}=\mathrm{k}^{2} \mathrm{rt}^{2}$. where k is a constant. The power delivered to the particle by the force acting on it is given as

(C) $\mathrm{mk}^{2} \mathrm{r}^{2} \mathrm{t}$

(D) $\mathrm{mk}^{2} \mathrm{rt}$

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Work, Power & Energy

A particle of mass m is moving in a circular path of constant radius

r

such that its centripetal acceleration (a) is varying with time t as

a = k^{2} rt^{2}

. where k is a constant. The power delivered to the particle by the force acting on it is given as

(A)zero
(B) $m k^{2} r^{2} t^{2}$
(C) $mk^{2} r^{2} t$ ✓ Correct
(D) $mk^{2} rt$

Answer: (C)

Solution

Given centripetal acceleration

a_{c} = k^{2} r t^{2} = \frac{v ^{2}}{r}

, we find the velocity:

v^{2} = k^{2} r^{2} t^{2}

, so

v = k r t

.

The tangential acceleration is

a_{t} = \frac{d v}{d t} = k r

.

The tangential force is

F_{t} = m a_{t} = mk r

.

Power is the dot product of force and velocity. Since only the tangential component of force does work (the centripetal force is perpendicular to velocity):

P = F_{t} \cdot v = (mk r) (k r t) = m k^{2} r^{2} t

Therefore, the answer is

m k^{2} r^{2} t

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Work, Power & Energy

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