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 $t=2 \mathrm{~s}$ ?

<p><strong>Moment of Inertia and Angular Motion</strong></p> <p>For a solid circular disk, the moment of inertia is given by</p> <p>$I = \frac{1}{2}MR^2.$</p> <p>The angular position is defined as</p> <p>$\theta(t) = 5t^2 - 8t.$</p> <p><strong>Angular Velocity and Acceleration</strong></p> <p>Differentiate with respect to time to obtain the angular velocity:</p> <p>$\omega(t) = \frac{d\theta}{dt} = 10t - 8.$</p> <p>Differentiating again, the angular acceleration is:</p> <p>$\alpha(t) = \frac{d\omega}{dt} = 10.$</p> <p>At time $t = 2 \, \text{s}$:</p> <p><p>Angular velocity: </p> <p>$\omega(2) = 10(2) - 8 = 12 \, \text{rad/s}.$</p></p> <p><p>Angular acceleration: </p> <p>$\alpha(2) = 10 \, \text{rad/s}^2.$</p></p> <p><strong>Torque Calculation</strong></p> <p>The torque applied by the external force is related to the moment of inertia and angular acceleration:</p> <p>$\tau = I \alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2.$</p> <p><strong>Power Delivered</strong></p> <p>Power delivered by a torque is given by:</p> <p>$P = \tau \omega.$</p> <p>At $t = 2 \, \text{s}$, substitute the values:</p> <p>$P = 5MR^2 \times 12 = 60MR^2.$</p> <p>Thus, the power delivered by the applied torque at $t = 2 \, \text{s}$ is:</p> <p>$\boxed{60MR^2}.$</p>