Spherical bobs are preferred in simple pendulum experiments because their symmetrical shape minimizes air resistance, ensuring that air drag has a negligible effect on the pendulum's motion. Addi

Given: $\omega_0 = 0 \text{ rad/s}$ (from rest), $\omega_f = 2\pi \text{ rad/s}$, $t = 4 \text{ s}$.

$$\omega_f = \omega_0 + \alpha t$$ $$2\pi \frac{\text{rad}}{\text{s}} = 0 \frac{\text{rad}}{\text{s}} + \alpha (4 \text{ s})$$ $$\alpha = \frac{2\pi \frac{\text{rad}}{\text{s}}}{4 \text{ s}}$$ $$\alpha = \frac{\pi}{2} \frac{\text{rad}}{\text{s}^2}$$ $$\alpha \approx 1.57 \frac{\text{rad}}{\text{s}^2}$$

The angular acceleration is $\boxed{\text{}\frac{\pi}{2} \frac{\text{rad}}{\text{s}^2} \text{ or } 1.57 \frac{\text{rad}}{\text{s}^2}\text{}}$

4. Step 1: Convert the radius to meters.

$$r = 40 \text{ cm} = 0.40 \text{ m}$$

The stone makes 10 complete revolutions. Each revolution is $2\pi$ radians.

$$\Delta \theta = 10 \text{ rev} \times 2\pi \frac{\text{rad}}{\text{rev}}$$ $$\Delta \theta = 20\pi \text{ rad}$$

$$\omega = \frac{\Delta \theta}{t}$$ $$\omega = \frac{20\pi \text{ rad}}{3.8 \text{ s}}$$ $$\omega \approx 16.53 \frac{\text{rad}}{\text{s}}$$

$$v = r\omega$$ $$v = (0.40 \text{ m}) \times \left( \frac{20\pi}{3.8} \frac{\text{rad}}{\text{s}} \right)$$ $$v = \frac{8\pi}{3.8} \frac{\text{m}}{\text{s}}$$ $$v \approx 6.59 \frac{\text{m}}{\text{s}}$$

The angular velocity is $\boxed{\text{}\frac{20\pi}{3.8} \frac{\text{rad}}{\text{s}} \text{ or } 16.53 \frac{\text{rad}}{\text{s}}\text{}}$ and the linear velocity is $\boxed{\text{}\frac{8\pi}{3.8} \frac{\text{m}}{\text{s}} \text{ or } 6.59 \frac{\text{m}}{\text{s}}\text{}}$