Moment of Inertia and Angular Motion

For a solid circular disk, the moment of inertia is given by

$$ I = \frac{1}{2}MR^2. $$

The angular position is defined as

$$ \theta(t) = 5t^2 - 8t. $$

Angular Velocity and Acceleration

Differentiate with respect to time to obtain the angular velocity:

$$ \omega(t) = \frac{d\theta}{dt} = 10t - 8. $$

Differentiating again, the angular acceleration is:

$$ \alpha(t) = \frac{d\omega}{dt} = 10. $$

At time $$ t = 2 \, \text{s} $$:

Angular velocity:

$$ \omega(2) = 10(2) - 8 = 12 \, \text{rad/s}. $$

Angular acceleration:

$$ \alpha(2) = 10 \, \text{rad/s}^2. $$

Torque Calculation

The torque applied by the external force is related to the moment of inertia and angular acceleration:

$$ \tau = I \alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2. $$

Power Delivered

Power delivered by a torque is given by:

$$ P = \tau \omega. $$

At $$ t = 2 \, \text{s} $$, substitute the values:

$$ P = 5MR^2 \times 12 = 60MR^2. $$

Thus, the power delivered by the applied torque at $$ t = 2 \, \text{s} $$ is:

$$ \boxed{60MR^2}. $$