JEE Main20199 Apr 2019Morning ShiftPhysicsRotational MotionActual
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 k θ 2 . If its moment of inertia is I then the angular acceleration of the disc is:
Options
- A2 k I θ
- Bk 2 I θ
- Ck 4 I θ
- Dk I θ
Correct answer
A. 2 k I θ
Step-by-step solution
Kinetic Energy = k θ 2 1 2 I ω 2 = k θ 2 ω 2 = 2 k θ 2 I Differentiate both side w.r.t. θ . 2 ω d ω d θ = 4 k θ I ω d ω d θ = 2 k θ I α = 2 k θ I