Problem 17 - College Physics

Question

A flywheel in a motor is spinning at 500.0 \(\mathrm{rpm}\) when a power failure suddenly occurs. The flywheel has mass 40.0 \(\mathrm{kg}\) and diameter 75.0 \(\mathrm{cm} .\) The power is off for 30.0 \(\mathrm{s}\) , and during this time the flywheel slown down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 200.0 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Step-by-Step Solution

Answer

(a) 31.36 rad/s, (b) 74.8 s, 311.95 revolutions.

Step 1: Convert RPM to RPS

First, we need to convert the initial angular speed of the flywheel from revolutions per minute (rpm) to revolutions per second (rps). The flywheel is spinning at 500 rpm. To convert this, use the formula: \[ \omega_i = \frac{500 \, \text{revolutions per minute}}{60 \, \text{seconds per minute}} = 8.33 \, \text{rps} \]

Step 2: Calculate Initial Angular Velocity

Now, convert revolutions per second to radians per second, which is a standard unit for angular velocity:\[ \omega_i = 8.33 \, \text{rps} \times 2\pi \, \text{radians per revolution} = 52.36 \, \text{rad/s} \]

Step 3: Determine Final Angular Velocity After Power Outage

The flywheel makes 200 revolutions during the 30-second interval. Calculate the total angle in radians during this time:\[ \theta = 200 \, \text{revolutions} \times 2\pi \, \text{radians per revolution} = 1256.64 \, \text{rad} \]Using the equation \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \), where \( t = 30 \, \text{s} \):\[ 1256.64 = 52.36 \times 30 + \frac{1}{2} \alpha \times 30^2 \]Solving for \( \alpha \):\[ \alpha = \frac{1256.64 - 1570.8}{450} = -0.7 \, \text{rad/s}^2 \]

Step 4: Calculate Angular Velocity When Power Returns

Using the angular acceleration calculated in Step 3, find the angular velocity when the power returns:\[ \omega_f = \omega_i + \alpha t = 52.36 + (-0.7) \times 30 = 31.36 \, \text{rad/s} \]

Step 5: Calculate Time to Stop if Power Hadn't Returned

Determine the time it would take for the flywheel to stop completely if the power had not returned using the formula \( \omega_f = \omega_i + \alpha t \) and setting \( \omega_f = 0 \):\[ 0 = 52.36 + (-0.7) t \]Solving for \( t \):\[ t = \frac{52.36}{0.7} = 74.8 \, \text{s} \]

Step 6: Calculate Revolutions Until Complete Stop

Calculate the number of additional revolutions the flywheel would make during this time until it stops using the equation \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \):\[ \theta = 52.36 \times 74.8 + \frac{1}{2} (-0.7) \times (74.8)^2\]\[ \theta = 1960.46 \, \text{rad} \]Convert back to revolutions:\[ \text{Number of Revolutions} = \frac{1960.46}{2\pi} = 311.95 \, \text{revolutions} \]

Key Concepts

Angular Velocity

Angular velocity is a measure of how fast an object rotates or spins around a particular axis. In this exercise, the rotating flywheel demonstrates this concept when we calculate its speed at different intervals. Initially, the flywheel's angular velocity is given in revolutions per minute (rpm), which is a common unit of measure for circular motion.

To get a more precise measure, we convert rpm to radians per second, the standard unit for angular velocity in physics.

• Start by converting rpm to revolutions per second (rps): \( \omega_i = \frac{500 \, \text{rpm}}{60} = 8.33 \, \text{rps} \)
• Then convert rps to radians per second: \( \omega_i = 8.33 \, \text{rps} \times 2\pi = 52.36 \, \text{rad/s} \)

Understanding angular velocity helps explain how quickly or slowly the flywheel spins, and it's a crucial part of solving rotational dynamics problems like this one.

Rotational Kinematics

Rotational kinematics involves predicting the motion of rotating objects, like the flywheel in this exercise. It relies on several key equations to understand motion, similar to how linear kinematics is used for motion along a straight line.

Key elements of rotational kinematics include:

• Initial angular velocity \( \omega_i \)
• Angular acceleration \( \alpha \)
• Total angle rotated \( \theta \)
• Final angular velocity \( \omega_f \)

In the exercise, you calculate the angular acceleration using:\[\theta = \omega_i t + \frac{1}{2} \alpha t^2 \]Given that \( \theta = 1256.64 \, \text{rad} \), \( \omega_i = 52.36 \, \text{rad/s} \), and \( t = 30 \, \text{s} \), solving gives \( \alpha = -0.7 \, \text{rad/s}^2 \).

Next, you find the final angular velocity using \( \omega_f = \omega_i + \alpha t \), so \( \omega_f = 31.36 \, \text{rad/s} \). Rotational kinematics provides a framework for understanding how spinning objects slow down, speed up, or maintain a constant rotational speed.

Conservation of Angular Momentum

The conservation of angular momentum is a fundamental concept in rotational dynamics, stating that if no external torque is acting on a system, its angular momentum remains constant over time. Angular momentum is the rotational analogue of linear momentum and depends on the object's mass, shape, and angular velocity.

While this problem doesn't directly solve for angular momentum, it is implicitly a consideration in rotational dynamics, especially if the system had no external forces like friction acting on it. If there were no friction, the flywheel would maintain its angular velocity indefinitely, illustrating the conservation principle.

Angular momentum \( L \) is given by:\[L = I \omega\]Where:

• \( I \) is the moment of inertia (resistance to rotation).
• \( \omega \) is the angular velocity.

In scenarios where friction or other torques are present, angular momentum is not conserved, as observed in the exercise, where the flywheel's speed decreases due to friction.