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Question:

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which it has rotated, is given as kθ². If its moment of inertia is I then the angular acceleration of the disc is:

k2Iθ

kIθ

k4Iθ

2kIθ

Solution:

Correct option is D.2kIθKinetic energy=12Iω2=kθ2ω2=2kθ2I⇒ω2kIθ(1Differentiate (1) wrt timedωdt=α=2kI(dθdt)α=2kI2kIθα=2kIθ