The drive pulley of an electric motor takes 20 seconds to accelerate uniformly from rest to \( 2400 \mathrm{r} / \mathrm{min} \). Calculate: (a) The angular acceleration of the pulley. \( \left(12,57 \mathrm{rad} / \mathrm{s}^{2}\right) \) (b) The number of revolutions made by the pulley during the acceleration. (c) The angular displacement of the pulley during the acceleration. \( (2513 \mathrm{rad}) \)

To find the required values, we'll start from the known quantities and use the equations of motion for rotational dynamics. **(a)** To calculate the angular acceleration \( \alpha \), we first need to convert the final angular speed \( \omega_f = 2400 \mathrm{r/min} \) into radians per second: \[ \omega_f = 2400 \times \frac{2\pi \, \text{rad}}{1 \text{ min}} \times \frac{1 \text{ min}}{60 \text{ s}} = 2400 \times \frac{2\pi}{60} \, \text{rad/s} \approx 251.33 \, \text{rad/s}. \] The initial angular velocity \( \omega_i = 0 \) (starts from rest), and the time \( t = 20 \, \text{s} \). Using the formula for angular acceleration: \[ \alpha = \frac{\omega_f - \omega_i}{t} = \frac{251.33 - 0}{20} \approx 12.57 \, \text{rad/s}^2. \] **(b)** To find the number of revolutions made by the pulley during the acceleration, we can use the following relationship: \[ \theta = \omega_i t + \frac{1}{2} \alpha t^2. \] Substituting \( \omega_i = 0 \): \[ \theta = \frac{1}{2} \alpha t^2 = \frac{1}{2} \times 12.57 \times 20^2 = \frac{1}{2} \times 12.57 \times 400 = 2514 \, \text{rad}. \] To convert this to revolutions: \[ \text{Number of revolutions} = \frac{2514 \text{ rad}}{2\pi} \approx 400.0 \text{ revolutions}. \] **(c)** The angular displacement \( \theta \) has already been calculated as \( 2514 \, \text{rad} \). In summary: (a) \( \alpha \approx 12.57 \, \text{rad/s}^2 \) (b) \( \text{Number of revolutions} \approx 400.0 \) (c) \( \theta \approx 2514 \, \text{rad} \)