Problem 19 - University Physics with Modern Physics

Question

At \(t=0\) a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0 \(\mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{s}\) . From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

Step-by-Step Solution

Answer

(a) 540 rad; (b) 12.29 s; (c) -8.17 rad/s².

Step 1: Calculate the angle turned while accelerating

To find the angle turned while the wheel is accelerating, use the kinematic equation: \( \theta_1 = \omega_0 t + \frac{1}{2} \, \alpha \, t^2 \), where \( \omega_0 = 24 \, \text{rad/s} \), \( \alpha = 30 \, \text{rad/s}^2 \), and \( t = 2 \, \text{s} \). Substituting the values, we get \( \theta_1 = 24 \times 2 + \frac{1}{2} \times 30 \times 2^2 \). Calculating this gives \( \theta_1 = 48 + 60 = 108 \, \text{rad} \).

Step 2: Find total angle turned

The total angle the wheel turns from \( t = 0 \) until it stops is the sum of the angle turned while accelerating and the angle turned while coasting. The wheel coasts through 432 rad. Therefore, the total angle \( \theta_{\text{total}} = \theta_1 + 432 = 108 + 432 = 540 \, \text{rad} \).

Step 3: Calculate the angular velocity at the end of acceleration

To find the angular velocity when the circuit breaker trips, use the equation \( \omega = \omega_0 + \alpha t \). We have \( \omega_0 = 24 \, \text{rad/s} \), \( \alpha = 30 \, \text{rad/s}^2 \), and \( t = 2 \, \text{s} \). Substituting these, \( \omega = 24 + 30 \times 2 = 24 + 60 = 84 \, \text{rad/s} \).

Step 4: Find time to stop during coasting

Now, use the kinematic equation for angular motion: \( \omega^2 = \omega_1^2 + 2 \alpha_2 \theta_2 \). Since the wheel stops, \( \omega = 0 \), and \( \omega_1 = 84 \, \text{rad/s} \), \( \alpha_2 \) is the angular acceleration while coasting, and \( \theta_2 = 432 \, \text{rad} \). Solving for \( \alpha_2 \), we have \( 0 = 84^2 + 2 \times \alpha_2 \times 432 \), leading to \( \alpha_2 = -\frac{84^2}{2 \times 432} = -8.1667 \, \text{rad/s}^2 \).

Step 5: Calculate total stopping time

The time to stop, \( t_2 \), can be found using \( \omega = \omega_1 + \alpha_2 t_2 \), with \( \omega = 0 \), \( \omega_1 = 84 \, \text{rad/s} \), and \( \alpha_2 = -8.1667 \, \text{rad/s}^2 \). Solving gives \( 0 = 84 - 8.1667 \times t_2 \), so \( t_2 = \frac{84}{8.1667} \approx 10.29 \, \text{s} \).

Step 6: Determine total time elapsed

The total time from \( t = 0 \) to when the wheel stopped is \( t_{\text{total}} = t_1 + t_2 \), where \( t_1 = 2 \, \text{s} \) and \( t_2 \approx 10.29 \, \text{s} \). Therefore, \( t_{\text{total}} \approx 2 + 10.29 \approx 12.29 \, \text{s} \).

Key Concepts

Kinematics

Kinematics is the science of analyzing motion without considering the forces that cause it. In the context of angular motion, kinematics describes how rotational movements occur. For instance, consider the initial problem where a grinding wheel spins with an angular velocity of 24 rad/s. It also experiences a constant angular acceleration of 30 rad/s² for a period of time. These plans are similar to how we handle linear motion, but with specific changes to handle its circular path:

Here is the kinematic equation used in the problem:

• To find the angle turned (\( \theta_1 = \omega_0 t + \frac{1}{2} \alpha t^2 \) ), where:
  • \( \omega_0 \) is the initial angular velocity.
  • \( \alpha \) is the angular acceleration.
  • \( t \) is the time interval.

Using these parameters, you can calculate how far the wheel turns during the acceleration phase. The use of these equations helps in predicting movement at any fixed time along a circular path. This part of physics is vital for understanding rotational systems in devices and machinery.

Angular Acceleration

Angular acceleration refers to how quickly the angular velocity of an object changes over time. In other words, it's how fast an object speeds up or slows down its spinning motion. This rate of change is crucial for understanding rotational dynamics. In the example of a grinding wheel from the exercise, it first accelerates with an angular acceleration of 30 rad/s². Angular acceleration has a considerable impact on how the wheel speeds up initially.

However, once the circuit breaker trips, the wheel experiences a different kind of angular acceleration as it slows to a stop. It uses the following equation:

• \( \omega = \omega_1 + \alpha_2 t_2 \) , where:
  • \( \omega \) is the final angular velocity (zero when it stops).
  • \( \omega_1 \) is the angular velocity after acceleration.
  • \( \alpha_2 \) is the angular deceleration.
  • \( t_2 \) is the time taken to stop.

Recognizing these changes helps in determining how long and under what conditions an object comes to rest. Thus, being familiar with these concepts contributes greatly to solving real-world issues that involve changing speeds in rotational motion.

Rotational Dynamics

Rotational dynamics deals with the forces and torques that result in rotational motion. Understanding these concepts helps us explore how and why objects spin. In the exercise, the wheel experienced different phases of motion: an initial acceleration and then deceleration to a stop. These changes were influenced by the torque, or rotational force, acting upon the wheel.

During deceleration, it's important to calculate the total time until the wheel stops. For this, the problem employs the following formula relating to angular motion:

• \( 0 = \omega_1^2 + 2\alpha_2\theta_2 \) , where:
  • \( \omega_1 \)is the angular velocity before the deceleration phase.
  • \( \alpha_2 \) is the angular acceleration (negative, since it decelerates).
  • \( \theta_2 \) is the angle turned during the coasting phase.

Understanding these dynamics assists us in predicting the motion and energy transitions within systems that periodically undergo changes in rotation. This is especially applicable to engines, turbines, and other machinery whereby rotational forces dominate.