A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that θ(t) = 5t² - 8t, where θ(t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t = 2 s? (a) 60MR² (b) 72MR² (c) 8MR² (d) 108MR²

Answer: (a). ω = (dθ)/(dt) = 10t - 8 = 12 rad/s at t = 2 s; α = 10 rad/s². τ = Iα = 1/2MR² × 10 = 5MR² P = τω = 5MR² × 12 = 60MR² So the answer is (a) 60MR².

A circular disk of radius R meter and mass M kg is rotating around… | JEE Main 2025 (23 Jan, Shift 2)

Q17JEE Main 2025 · 23 JAN · SHIFT 2Medium

Rotational Kinematics & Dynamics

A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that $\theta(t) = 5t^2 - 8t$ , where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$ . How much power is delivered by the applied torque, when $t = 2\text{ s}$ ?

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Answer: (a)

$\omega = \frac{d\theta}{dt} = 10t - 8 = 12\text{ rad/s}$ at $t = 2$ s; $\alpha = 10\text{ rad/s}^2$ .
$\tau = I\alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2$
$P = \tau\omega = 5MR^2 \times 12 = 60MR^2$
So the answer is (a) $60MR^2$ .

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