A pulley slows down with an angular acceleration of -10pi rad/s2. If it stops after 20 s with a final angular speed of 20 rev/s, what was its initial angular speed and total angular displacement?

The pulley slows down, so the angular acceleration is negative.

• Final angular speed, $\omega_f = 20 \text{ rev/s}$
• Angular acceleration, $\alpha = -10\pi \text{ rad/s}^2$ (negative because it slows down)
• Time, $t = 20 \text{ s}$

First, convert the final angular speed from revolutions per second to radians per second:

$$\omega_f = 20 \frac{\text{rev}}{\text{s}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 40\pi \text{ rad/s}$$

Use the rotational kinematic equation:

$$\omega_f = \omega_i + \alpha t$$

Substitute the known values:

$$40\pi \text{ rad/s} = \omega_i + (-10\pi \text{ rad/s}^2)(20 \text{ s})$$ $$40\pi \text{ rad/s} = \omega_i - 200\pi \text{ rad/s}$$

Solve for $\omega_i$:

$$\omega_i = 40\pi \text{ rad/s} + 200\pi \text{ rad/s}$$ $$\omega_i = 240\pi \text{ rad/s}$$

The initial angular speed is $\boxed{\text{240\pi rad/s}}$.

Use the rotational kinematic equation:

$$\theta = \omega_i t + \frac{1}{2} \alpha t^2$$

Substitute the known values:

$$\theta = (240\pi \text{ rad/s})(20 \text{ s}) + \frac{1}{2}(-10\pi \text{ rad/s}^2)(20 \text{ s})^2$$ $$\theta = 4800\pi \text{ rad} + \frac{1}{2}(-10\pi \text{ rad/s}^2)(400 \text{ s}^2)$$ $$\theta = 4800\pi \text{ rad} - 2000\pi \text{ rad}$$ $$\theta = 2800\pi \text{ rad}$$

Since $1 \text{ revolution} = 2\pi \text{ radians}$:

$$\text{Number of revolutions} = \frac{\theta}{2\pi \text{ rad/rev}}$$ $$\text{Number of revolutions} = \frac{2800\pi \text{ rad}}{2\pi \text{ rad/rev}}$$ $$\text{Number of revolutions} = 1400 \text{ revolutions}$$

The number of revolutions made is $\boxed{\text{1400 revolutions}}$.