1. 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. Additionally, the center of mass of a spherical bob is precisely at its geometric center, which simplifies calculations and provides more accurate measurements of the pendulum's length and period. 2. A simple pendulum in simple harmonic motion: Diagram: ` O (pivot) | | L (length of string) | | | A <--- B (equilibrium) ---> C (max displacement) (max displacement) ` (A simple diagram showing a pivot, a string of length L, and a bob swinging between two extreme positions A and C, passing through an equilibrium position B.) a) The maximum velocity* of the bob occurs at the equilibrium position (B), which is the lowest point of its swing. b) The maximum acceleration* of the bob occurs at the extreme positions (A and C), where the displacement from equilibrium is greatest. 3. Step 1: Convert the final angular velocity from revolutions per minute to radians per second. $$ \omega_f = 60 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} $$ $$ \omega_f = 2\pi \frac{\text{rad}}{\text{s}} $$ Step 2: Use the kinematic equation for angular motion to find the angular acceleration. 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. $$ r = 40 \text{ cm} = 0.40 \text{ m} $$ Step 2: Calculate the total angular displacement in radians. The stone makes 10 complete revolutions. Each revolution is $2\pi$ radians. $$ \Delta \theta = 10 \text{ rev} \times 2\pi \frac{\text{rad}}{\text{rev}} $$ $$ \Delta \theta = 20\pi \text{ rad} $$ Step 3: Calculate the angular velocity. $$ \omega = \frac{\Delta \theta}{t} $$ $$ \omega = \frac{20\pi \text{ rad}}{3.8 \text{ s}} $$ $$ \omega \approx 16.53 \frac{\text{rad}}{\text{s}} $$ Step 4: Calculate the linear velocity. $$ v = r\omega $$ $$ v = (0.40 \text{ m}) \times \left( \frac{20\pi}{3.8} \frac{\text{rad}}{\text{s}} \right) $$ $$ v = \frac{8\pi}{3.8} \frac{\text{m}}{\text{s}} $$ $$ v \approx 6.59 \frac{\text{m}}{\text{s}} $$ 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 $\boxed{\text{}\frac{8\pi}{3.8} \frac{\text{m}}{\text{s}} \text{ or } 6.59 \frac{\text{m}}{\text{s}}\text{}}$