JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
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 = k²rt², where k is a constant. The power delivered to the particle by the force acting on it is given as
- (a) zero
- (b) mk²r²t²
- (c) mk²r²t
- (d) mk²rt
Answer and solution
Answer (c) mk²r²t
Solution
Physics sir · 4 steps
Get the speed from the centripetal acceleration, then only the tangential force does work.
- Find the speed. Centripetal acceleration is v²/r, so v²/r = k² r t². This gives v² = k² r² t², so v = krt.
- Which force works. The centripetal force is perpendicular to the velocity, so it does no work. Only the tangential force, which changes the speed, delivers power.
- Tangential force. Tangential acceleration is aₜ = dv/dt = kr, so the tangential force is Fₜ = mkr. It acts along the velocity.
P = Fₜ v = (mkr)(krt) = mk²r²t
- So. You get the same result from the kinetic energy K = 1/2mk²r²t², because dK/dt = mk²r²t. The answer is (c).
The idea to keep. In circular motion, only the tangential force does work. Find v(t) from v²/r, then power is d/dt(1/2mv²).
Answer from the NTA final answer key. Solution by PrepWiser.