$\begin{aligned}
& \theta=5 \mathrm{t}^2-8 \mathrm{t} \
& \omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=10 \mathrm{t}-8 \
& \alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=10 \
& \therefore \mathrm{p}=\tau \omega \
& =(\mathrm{I} \alpha) \omega \
& =\left(\frac{\mathrm{mR}^2}{2}\right) \alpha \omega \
& =\left(\frac{\mathrm{mR}^2}{2}\right)(10)(10
\end{aligned}Put\mathrm{t}=2\mathrm{p}=60 \mathrm{mR}^2$
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…
JEE Main 2025 — Physics 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)=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 ?
Held on 23 Jan 2025 · Verified 6 Jul 2026.
Options
- A
72MR2
- B
8MR2
- C
108MR2
- D
60MR2
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