A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An…

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 $t=2 \mathrm{~s}$ ?
  1. $72 \mathrm{MR}^2$
  2. $8 \mathrm{MR}^2$
  3. $108 \mathrm{MR}^2$
  4. $60 \mathrm{MR}^2$

Solution

$\begin{aligned}
& \theta=5 \mathrm{t}^2-8 \mathrm{t} \\ & \omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=10 \mathrm{t}-8 \\ & \alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=10 \\ & \therefore \mathrm{p}=\tau \omega \\ & =(\mathrm{I} \alpha) \omega \\ & =\left(\frac{\mathrm{MR}^2}{2}\right) \alpha \omega \\ & =\left(\frac{\mathrm{MR}^2}{2}\right)(10)(10
\end{aligned}$
Put $\mathrm{t}=2$
$\mathrm{p}=60 \mathrm{MR}^2$

Asked in: JEE Main 2025 (23 Jan Shift 2)

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