Rotation 1 Question 4

4. 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

(2019 Main, 9 April I)

(a) k2Iθ

(b) kIθ

(c) k4Iθ

(d) 2kIθ

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Answer:

Correct Answer: 4. (d)

Solution:

  1. Given, kinetic energy =kθ2

We know that, kinetic energy of a rotating body about its axis =12Iω2

where, I is moment of inertia and ω is angular velocity.

 So, 12Iω2=kθ2 or ω2=2kθ2Iω=2kIθ

Differentiating the above equation w.r.t. time on both sides, we get

dωdt=2kIdθdt=2kIωω=dθdt

Angular acceleration,

α=dωdt=2kIω=2kI2kIθ [using Eq. (i)]  or α=2kIθ

Alternate Solution

 As, ω2=2kθ2I2ωdωdt=2kI2θdθdt or ωdω=2kIθdθωdωdθ(=α)=2kIθ or α=2kIθ



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