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 $\theta(t)=5 t^{2}-8 t$, 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 $\mathrm{t}=2 \mathrm{~s}$ ?

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Rotational Mechanics

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) = 5 t^{2} - 8 t

, 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) $60 MR^{2}$ ✓ Correct
(B) $72 MR^{2}$
(C) $108 MR^{2}$
(D) $8 MR^{2}$

Answer: (A)

Solution

Given

θ (t) = 5 t^{2} - 8 t

, the angular velocity and acceleration are

ω = \frac{d θ}{d t} = 10 t - 8, α = \frac{d ω}{d t} = 10 rad/s^{2}

The moment of inertia of the disk about the given axis is

I = \frac{M R ^{2}}{2}

, so the applied torque is

τ = I α = \frac{M R ^{2}}{2} (10) = 5 M R^{2}

Power delivered is

P = τ ω

; at

t = 2

ω = 10 (2) - 8 = 12 rad/s

, so

P = 5 M R^{2} \times 12 = 60 M R^{2}

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Rotational Mechanics

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