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 $\mathrm{\theta }$(t) = 5t² – 8t, where $\mathrm{\theta }$(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 = 2s?

We are given the angular position function of a rotating disk:

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

We need to find the power delivered by the applied torque at $t = 2$ seconds.

Step 1: Find Angular Velocity

Angular velocity $\omega(t)$ is the first derivative of $\theta(t)$:

$$\omega(t) = \frac{d\theta}{dt} = \frac{d}{dt} (5t^2 - 8t)$$

 $$\omega(t) = 10t - 8$$

At $t = 2$:

$$\omega(2) = 10(2) - 8 = 20 - 8 = 12 \text{ rad/s}$$

Step 2: Find Angular Acceleration

Angular acceleration $\alpha(t)$ is the first derivative of $\omega(t)$:

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

 $$\alpha(t) = 10 \text{ rad/s}^2$$

Since $\alpha$ is constant, the torque applied is also constant.

Step 3: Find Torque

The torque $\tau$ is given by:

$$\tau = I \alpha$$

where $I$ is the moment of inertia of a circular disk about its perpendicular axis:

$$I = \frac{1}{2} M R^2$$

So,

$$\tau = \left(\frac{1}{2} M R^2 \right) (10)$$

 $$\tau = 5 M R^2$$ 

Step 4: Find Power

The power delivered by the applied torque is:

$$P = \tau \omega$$

Substituting the values:

$$P = (5 M R^2) (12)$$

 $$P = 60 M R^2 \text{ watts}$$ 

Final Answer:

The power delivered by the applied torque at $t = 2s$ is $60 M R^2$ watts.