A tangential for of 8.0 N acts on a wheel which com plates its first revolution from rest in 20 s s the force acts 0.30 m from the center of the wheel. Calculate a. The angular acceleration b. The moment of inertia of the wheel

**Step 1. Determine the Angular Acceleration** The wheel starts from rest and completes one full revolution in \( t = 20~\text{s} \). One full revolution corresponds to an angular displacement of \[ \theta = 2\pi~\text{radians}. \] For constant angular acceleration, the angular displacement is given by \[ \theta = \frac{1}{2}\alpha t^2. \] Solving for the angular acceleration \( \alpha \), \[ \alpha = \frac{2\theta}{t^2}. \] Substitute the known values, \[ \alpha = \frac{2(2\pi)}{(20)^2} = \frac{4\pi}{400} = \frac{\pi}{100}~\text{rad/s}^2. \] Thus, the angular acceleration is \[ \boxed{\alpha \approx 0.0314~\text{rad/s}^2}. \] --- **Step 2. Calculate the Moment of Inertia** A tangential force \( F = 8.0~\text{N} \) acts at a distance \( r = 0.30~\text{m} \) from the center. The torque \( \tau \) produced by this force is given by \[ \tau = F \cdot r. \] Substitute the values, \[ \tau = 8.0~\text{N} \times 0.30~\text{m} = 2.4~\text{N}\cdot\text{m}. \] The relation between torque and angular acceleration is \[ \tau = I \alpha, \] where \( I \) is the moment of inertia. Solving for \( I \), \[ I = \frac{\tau}{\alpha}. \] Now substitute the known values, \[ I = \frac{2.4~\text{N}\cdot\text{m}}{\frac{\pi}{100}~\text{rad/s}^2} = \frac{2.4 \times 100}{\pi} = \frac{240}{\pi}~\text{kg}\cdot\text{m}^2. \] Thus, the moment of inertia is approximately \[ \boxed{I \approx 76.4~\text{kg}\cdot\text{m}^2}. \]