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: (a) k/4Iθ (b) k/Iθ (c) k/2Iθ (d) 2k/Iθ

Answer: (d). 1/2Iω² = kθ² ⇒ ω² = (2kθ²)/(I) α = ω(dω)/(dθ) = 1/2(d(ω²))/(dθ) = 1/2·(4kθ)/(I) α = 2k/Iθ, so the answer is (d).

A stationary horizontal disc is free to rotate about its axis. When a… | JEE Main 2019 (9 Apr, Shift 1)

Q58JEE Main 2019 · 9 APR · SHIFT 1Difficult

Rotational Kinematics & Dynamics

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 $\theta$ , where $\theta$ is the angle by which it has rotated, is given as k $\theta$ 2. If its moment of inertia is I then the angular acceleration of the disc is:

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Answer: (d)

$\tfrac{1}{2}I\omega^2 = k\theta^2 \Rightarrow \omega^2 = \dfrac{2k\theta^2}{I}$
$\alpha = \omega\dfrac{d\omega}{d\theta} = \dfrac{1}{2}\dfrac{d(\omega^2)}{d\theta} = \dfrac{1}{2}\cdot\dfrac{4k\theta}{I}$
$\alpha = \dfrac{2k}{I}\theta$ , so the answer is (d).

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