Give me a crazy complex question on angular acceleration

The formula for the moment of inertia of a solid disk is $I = \frac{1}{2}MR^2$.

$$I = \frac{1}{2}(5.0 \, \text{kg})(0.20 \, \text{m})^2$$ $$I = \frac{1}{2}(5.0 \, \text{kg})(0.040 \, \text{m}^2)$$ $$I = 0.10 \, \text{kg} \cdot \text{m}^2$$

The formula is $\tau = I\alpha$.

$$10.0 \, \text{N} \cdot \text{m} = (0.10 \, \text{kg} \cdot \text{m}^2)\alpha$$ $$\alpha = \frac{10.0 \, \text{N} \cdot \text{m}}{0.10 \, \text{kg} \cdot \text{m}^2}$$ $$\alpha = 100 \, \text{rad/s}^2$$

The formula for final angular velocity is $\omega_f = \omega_i + \alpha t$. Since the disk starts from rest, $\omega_i = 0$.

$$\omega_f = 0 + (100 \, \text{rad/s}^2)(5.0 \, \text{s})$$ $$\omega_f = 500 \, \text{rad/s}$$

The formula for angular displacement is $\theta = \omega_i t + \frac{1}{2}\alpha t^2$.

$$\theta = (0)(5.0 \, \text{s}) + \frac{1}{2}(100 \, \text{rad/s}^2)(5.0 \, \text{s})^2$$ $$\theta = 0 + \frac{1}{2}(100 \, \text{rad/s}^2)(25.0 \, \text{s}^2)$$ $$\theta = 1250 \, \text{rad}$$

There are $2\pi$ radians in one revolution.

$$\text{Revolutions} = \frac{1250 \, \text{rad}}{2\pi \, \text{rad/rev}}$$ $$\text{Revolutions} \approx 198.9 \, \text{rev}$$

a) The angular acceleration is $\boxed{100 \, \text{rad/s}^2}$.

b) The angular velocity after $5.0 \, \text{s}$ is $\boxed{500 \, \text{rad/s}}$.

c) The number of revolutions in $5.0 \, \text{s}$ is approximately $\boxed{198.9 \, \text{rev}}$.