• The angular acceleration required is:

$$\alpha = \frac{4\pi}{9} \approx 1.40\, \text{rad/s}^2$$

• The total torque required for the acceleration is:

$$\tau = \frac{212\pi}{9} + 16 \approx 90.00 \, \text{N} \cdot \text{m}$$

• The time to stop from 500 rpm is:

$$t_{\text{stop}} = \dfrac{50\pi}{3 \times 0.3018} \approx 173.44 \, \text{seconds}$$

• The number of revolutions from 500 rpm to 0 rpm is:

$$\text{Revolutions} = \frac{4542.01}{2\pi} \approx 722.88 \text{revolutions}$$

To solve this problem, we'll break it down into several steps:

Step 1: Convert RPM to Radians per Second

First, we need to convert the initial ( $\omega_i$ ) and final ( $\omega_f$ ) angular velocities from revolutions per minute (RPM) to radians per second:

$$\omega_i = 300 \, \text{rpm} = 300 \times \frac{2\pi}{60} \, \text{rad/s} = 10\pi \, \text{rad/s}$$

$$\omega_f = 500 \, \text{rpm} = 500 \times \frac{2\pi}{60} \, \text{rad/s} = \frac{50\pi}{3} \, \text{rad/s}$$

Step 2: Calculate Angular Acceleration

The angular acceleration ( $\alpha$ ) required to increase the speed from $\omega_i$ to $\omega_f$ in 15 seconds can be determined using the formula:

$$\alpha = \frac{\omega_f - \omega_i}{t}$$

Substituting the values:

$$\alpha = \frac{\frac{50\pi}{3} - 10\pi}{15} = \frac{\frac{50\pi - 30\pi}{3}}{15} = \frac{20\pi}{45} \, \text{rad/s}^2 = \frac{4\pi}{9} \, \text{rad/s}^2$$

Step 3: Calculate the Total Torque Required

The torque ( $\tau$ ) required to achieve the desired angular acceleration is given by:

$$\tau = I \cdot \alpha + \tau_{\text{friction}}$$

Where:

• $I = 53 \, \text{kg} \cdot \text{m}^2$ is the moment of inertia.
• $\tau_{\text{friction}} = 16 \, \text{N} \cdot \text{m}$ is the frictional resistance.

Substituting in the values:

$$\tau = 53 \cdot \frac{4\pi}{9} + 16$$

Calculate:

$$\tau = \frac{212\pi}{9} + 16 \approx 74.00 + 16 = 90.00 \, \text{N} \cdot \text{m}$$

Step 4: Calculate Time to Stop Spinning

If the friction is constant and the applied torque is removed, the deceleration will be due to the frictional torque alone. Thus, the angular deceleration ( $\alpha_f$ ) is:

$$\alpha_f = \frac{\tau_{\text{friction}}}{I} = \frac{16}{53} \, \text{rad/s}^2 \approx 0.3018 \, \text{rad/s}^2$$

To find the time ( $t_{\text{stop}}$ ) it takes to stop from 500 rpm, use the equation:

$$\omega_f = \omega_i - \alpha_f t_{\text{stop}}$$

$$t_{\text{stop}} = \dfrac{\omega_i - \omega_f}{\alpha_f}$$

Here, $\omega_f = 0$:

$$t_{\text{stop}} = \dfrac{\omega_i}{\alpha_f}$$

$$t_{\text{stop}} = \dfrac{50\pi}{3 \times 0.3018} \approx 173.44 \, \text{seconds}$$

Step 5: Calculate the Number of Revolutions While Slowing Down

During deceleration, the angular deceleration has been calculated as:

$$\alpha_f \approx 0.3018 \, \text{rad/s}^2$$

By integrating the acceleration, and considering the initial velocity, the angular velocity is given by:

$$\omega(t) = -\alpha_f t+w_f \approx -0.3018t + \frac{50\pi}{3}$$

By integrating the angular velocity, we can calculate the angular displacement:

$$\theta(t) = -\frac{\alpha_f}{2} t^2 + w_ft \approx -\frac{1}{2}0.3018t^2 + \frac{50\pi}{3}t$$

At the end of the motion ( $t =t_{stop}$ ) the angular displacement is:

$$\theta = \omega_f t_{\text{stop}} - \frac{\alpha_f}{2} t_{\text{stop}}^2$$

Substituting the values:

$$\theta = \frac{50\pi}{3} \cdot 173.44 - \frac{1}{2} \cdot 0.3018 \cdot (173.44)^2$$

Calculate:

$$\theta \approx 9081.30 - 4539.28 \approx 4542.01 \, \text{rad}$$

Convert radians to revolutions:

$$\text{Revolutions} = \frac{4542.01}{2\pi} \approx 722.88$$

Thus, the flywheel will make approximately 728.3 revolutions while slowing down.