A circular disk of radius R meter and mass M kg is rotating around...

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}

Options

A $72 \mathrm{MR}^2$
B $8 \mathrm{MR}^2$
C $108 \mathrm{MR}^2$
D $60 \mathrm{MR}^2$

Correct answer

D. $60 \mathrm{MR}^2$

Step-by-step 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

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