/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A high-speed flywheel in a motor... [FREE SOLUTION] | 91Ó°ÊÓ

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A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm. The power is off for 30.0 s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 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?

Short Answer

Expert verified
(a) 31.45 rad/s. (b) 75.13 s, 312.4 revolutions.

Step by step solution

01

Calculating Initial Angular Velocity

The initial angular velocity \( \omega_i \) is given by converting the initial speed from RPM to rad/s. Since 1 revolution is \(2\pi\) radians, \(1\text{ minute} = 60\text{ seconds}\), we have:\[\omega_i = \frac{500 \text{ rev/min} \times 2\pi \text{ rad/rev}}{60\text{ s/min}} \approx 52.36 \text{ rad/s}\]
02

Determining the Angular Displacement

The angular displacement \( \theta \) when the power is off for 30 seconds and the flywheel makes 200 revolutions is given by:\[ \theta = 200 \times 2\pi\text{ rad/rev} = 400\pi \text{ rad} \]
03

Using the Kinematic Equation to Find Final Angular Velocity

We use the kinematic equation \(\theta = \omega_i t + \frac{1}{2} \alpha t^2\), where \(\alpha\) is the angular deceleration. Given \(\theta = 400\pi \), \(\omega_i = 52.36 \), and \(t = 30\) seconds, we need to solve for \( \omega_f \):First calculate \( \alpha \) using:\[ \alpha = \frac{2(\theta - \omega_i t)}{t^2} = \frac{2(400\pi - 52.36 \times 30)}{30^2} \approx -0.697 \text{ rad/s}^2 \]Now, the final angular velocity \( \omega_f \) is:\[\omega_f = \omega_i + \alpha t = 52.36 + (-0.697 \times 30) \approx 31.45 \text{ rad/s} \]
04

Calculating Time to Stop

Using \(\omega_f = 0\) to indicate the wheel stops, we use the formula \(\omega_f = \omega_i + \alpha t\):\[ 0 = 52.36 + (-0.697) t \Rightarrow t = \frac{52.36}{0.697} \approx 75.13 \text{ seconds} \]
05

Calculating Additional Revolutions Before Stopping

Using the time to stop \(t = 75.13\) seconds, we find the additional angular displacement \(\theta_c\) which equals the total angular displacementminus the displacement during the 30 second power failure:\[\theta_c = \omega_i t_c + \frac{1}{2} \alpha t_c^2 = 52.36 \times 75.13 + \frac{1}{2} \times (-0.697) \times (75.13)^2 \]\[\theta_c \approx 1963.5 \text{ rad} \]Thus, the additional revolutions \(n_c\) before stopping is:\[n_c = \frac{1963.5}{2\pi} \approx 312.4 \text{ revolutions}\]
06

Final Results

(a) The rate at which the flywheel is spinning when the power comes back on is approximately 31.45 rad/s. (b) The flywheel would have taken approximately 75.13 seconds to stop and made about 312.4 more revolutions during this time if the power had not come back on.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Flywheel Dynamics
Flywheels are fascinating devices commonly used in machinery to store energy as rotational motion. They are often heavy, with significant mass, allowing them to maintain rotational inertia, which is the resistance to changes in rotational speed. This quality makes them ideal for applications requiring consistent rotational speed.
  • A flywheel in a motor acts as an energy reservoir, smoothing out the energy pulses that come from the power source.
  • By storing energy when available and releasing it when needed, it helps in maintaining a steady spin rate even if power input fluctuates.
In the exercise, the flywheel's initial motion is stored as kinetic energy due to its mass and diameter. When a power failure occurs, the flywheel's stored energy allows it to continue spinning, albeit gradually slowing due to friction at the axle bearings. Understanding these dynamics helps in analyzing the changes in rotation, particularly in terms of angular velocity and deceleration.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins. It is the rotational equivalent of linear velocity. Measured in radians per second (rad/s), angular velocity links the concept of rotational motion to the time it takes for an object to complete its rotations.
  • Initial angular velocity is determined when the motion begins. In the case of the flywheel, this is derived from its spinning at 500 revolutions per minute (rpm).
  • To convert from rpm to rad/s, we apply: \[ \omega_i = \frac{500 \times 2\pi}{60} \approx 52.36 \text{ rad/s} \]
Analyzing the flywheel's angular velocity before and after a power outage gives insights into how quickly it decelerates due to how much energy remains in the system.
Angular Deceleration
Angular deceleration is the rate of decrease in angular velocity over time, reflecting how quickly a spinning object is slowing down. In rotational systems, deceleration can occur due to factors such as friction.
  • The angular deceleration can be calculated by rearranging the kinematic equation given knowledge of initial velocity, time, and angular displacement. \[ \alpha = \frac{2(400\pi - \omega_i \times 30)}{30^2} \approx -0.697 \text{ rad/s}^2 \]
  • A negative sign in angular deceleration indicates a reduction in speed over time.
In the example, we see how to calculate the angular deceleration by knowing the final and initial velocities and the period during which the flywheel slows down. This understanding helps in predicting performance after power failures.
Revolution Calculations
Revolution calculations help determine how many complete turns an object makes around a central axis. This is significant in mechanics for understanding motion over time, especially when power supplies are interrupted.
  • To find the number of revolutions during a period, compute the angular displacement and convert radians into revolutions using: \[ n = \frac{\text{Angular Displacement}}{2\pi} \]
  • In the problem, during the 30-second outage, the flywheel makes 200 revolutions, equivalent to an angular displacement of \(400\pi\) radians.
  • Further calculations allow us to estimate future revolutions if the spin continues uninterrupted. Using the stopping time calculated, we find \[ n_c \approx 312.4 \text{ revolutions (while slowing to a halt)} \]
Revolution calculations grant valuable insights into the performance characteristics of flywheels, enabling better design and maintenance of machinery relying on them.

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Most popular questions from this chapter

9.75\(\cdots .\) It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 \(\mathrm{kg} / \mathrm{m}^{3}\) ) in the shape of a 10.0 -cm-thick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm about an axis perpendicular to the disk at its center? (b) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

A uniform disk of radius \(R\) is cut in half so that the remaining half has mass \(M\) (Fig. E9.34a).(a) What is the moment of inertia of this half about an axis perpendicular to its plane through point \(A\) ? (b) Why did your answer in part (a) come out the same as if this were a complete disk of mass \(M ?\) (c) What would be the moment of inertia of a quarter disk of mass \(M\) and radius \(R\) about an axis perpendicular to its plane passing through point \(B(\) Fig. \(E 9.34 b) ?\)

You are to design a rotating cylindrical axle to lift \(800-\mathrm{N}\) buckets of cement from the ground to a rooftop 78.0 \(\mathrm{m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 \(\mathrm{cm} / \mathrm{s}\) when it is turning at 7.5 \(\mathrm{rpm}\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

CP CALC The angular velocity of a flywheel obeys the equation \(\omega_{z}(t)=A+B t^{2},\) where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75\((\) for \(A)\) and 1.50 (for \(B )\) . (a) What are the units of \(A\) and \(B\) if \(\omega_{z}\) is in rad \(s ?\) (b) What is the angular acceleration of the wheel at (i) \(t=0.00\) and (ii) \(t=5.00\) s? (c) Through what angle does the flywheel turn during the first 2.00 s?

A safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)

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