To solve this problem, we will use the following kinematic equations for rotational motion:

1. Angular acceleration ($\alpha$): $\omega_f^2 = \omega_i^2 + 2\alpha\theta$ where:

   • $\omega_f$ is the final angular velocity,
   • $\omega_i$ is the initial angular velocity,
   • $\alpha$ is the angular acceleration,
   • $\theta$ is the angular displacement in radians.

2. Time elapsed ($t$): $\omega_f = \omega_i + \alpha t$ where:

   • $t$ is the time elapsed.

Step 1: Convert revolutions per minute (rev/min) to radians per second (rad/s)

• $\omega_i = 643 \, \text{rev/min}$
• $\omega_f = 387 \, \text{rev/min}$
• $\theta = 53 \, \text{revolutions}$

Since $1 \, \text{revolution} = 2\pi \, \text{radians}$ and $1 \, \text{minute} = 60 \, \text{seconds}$:

$\omega_{\text{rad/s}} = \omega_{\text{rev/min}} \times \frac{2\pi}{60}$

Now, let's compute the angular acceleration $\alpha$ and the time elapsed $t$.(a) The angular acceleration of the flywheel is approximately $-4.34 \, \text{rad/s}^2$. The negative sign indicates that the flywheel is decelerating.

(b) The time elapsed during the 53 revolutions is approximately $6.17 \, \text{seconds}$.

Would you like further details, or do you have any other questions?

Here are some related questions to consider:

1. How would the angular acceleration change if the flywheel were initially rotating faster?
2. What is the significance of a negative angular acceleration?
3. How would the time change if the flywheel decelerated at a slower rate?
4. How many revolutions would occur if the flywheel continued to decelerate at the same rate?
5. What would happen if there were a sudden increase in friction during the rotation?

Tip: When dealing with rotational motion, always pay attention to the direction of acceleration and velocities, as they can significantly affect the results.