/*! 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) 100 rpm; (b) Stops in 37.5 s after total ~78.2 revolutions.

Step by step solution

01

Convert Initial Angular Speed

Initially, the flywheel spins at 500 revolutions per minute (rpm). To convert this speed to radians per second, we use the conversion factor: \( 1 \, \text{revolution} = 2\pi \, \text{radians} \) and \( 1 \, \text{minute} = 60 \, \text{seconds} \). So the initial angular velocity \( \omega_i \) is:\[ \omega_i = 500 \, \frac{\text{revolutions}}{\text{minute}} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = \frac{500 \times 2\pi}{60} \, \text{radians/second} \] This simplifies to \( \omega_i = \frac{500 \pi}{30} \approx 52.36 \, \text{radians/second} \).
02

Calculate Angular Deceleration

Given that the flywheel makes 200 revolutions during the power failure, the angular displacement \( \theta \) in radians is:\[ \theta = 200 \, \, \text{revolutions} \times 2\pi \, \text{radians} \times \frac{1 \, \text{revolution}}{1} = 400\pi \, \text{radians} \].Using the kinematic equation \[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]we can solve for angular acceleration \( \alpha \). Substituting known values (\( t = 30 \, \text{seconds} \) and \( \theta = 400\pi \)), we get: \[ 400\pi = (52.36)(30) + \frac{1}{2} \alpha (30)^2 \].This simplifies and rearranges to:\[ \alpha = \frac{2(400\pi - 1570.8)}{900} \text{ radians/second}^2 \approx -1.396 \text{ radians/second}^2 \].
03

Determine Final Angular Speed After Power Failure

Using another kinematic equation \[ \omega_f = \omega_i + \alpha t \] we substitute \( \omega_i = 52.36 \, \text{radians/second} \), \( \alpha = -1.396 \, \text{radians/second}^2 \), and \( t = 30 \, \text{seconds} \) to find \( \omega_f \). \[ \omega_f = 52.36 + (-1.396)(30) \].This calculates to:\[ \omega_f \approx 10.48 \, \text{radians/second} \].Convert back to rpm: \[ \omega_f = 10.48 \, \text{radians/second} \times \frac{60}{2\pi} \, \text{revolutions/minute} \approx 100 \, \text{rpm} \].
04

Calculate Time to Stop If Power Had Not Returned

If the power had not returned, the flywheel would continue to decelerate at \( \alpha = -1.396 \, \text{radians/second}^2 \) until it stops (\( \omega_f = 0 \)). Using:\[ 0 = \omega_i + \alpha t \],we solve for \( t \):\[ 0 = 52.36 + (-1.396)t \]. Solving gives:\[ t \approx \frac{52.36}{1.396} \approx 37.5 \, \text{seconds} \].
05

Calculate Total Revolutions Until Stop

Total angular displacement from the power failure start is calculated using:\[ \theta = \omega_i \, t + \frac{1}{2} \alpha \, t^2 \]For \( t = 37.5 \, \text{seconds} \):\[ \theta = 52.36(37.5) + \frac{1}{2}(-1.396)(37.5)^2 \].This calculates to approximately \( \theta \approx 491.5 \, \text{radians} \). Convert radians to revolutions:\[ \text{revolutions} = \frac{491.5}{2\pi} \approx 78.2 \text{ revolutions} \].

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Key Concepts

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

Angular Speed
Angular speed is a measure of the rate at which an object rotates or spins around a central point. It is often given in units of radians per second (rad/s), which provides a clear understanding of how fast an angle is swept out in a given time. In the context of the exercise, the initial angular speed of the flywheel was given in revolutions per minute (RPM). To use it in calculations, it was converted to radians per second. This is done using the conversion factor: 1 revolution equals \(2\pi\) radians, and 1 minute equals 60 seconds.

In our exercise, the flywheel’s initial speed was 500 RPM. Converting this to radians per second using the relation: \[ \omega_i = 500 \times \frac{2\pi}{60} \approx 52.36 \, \text{rad/s}\] This conversion is crucial to analyze rotational systems, as radians is the SI unit for angular measurements.

Another point to highlight is that angular speed provides insight into how fast an object is spinning at a given moment. It's a fundamental concept when studying rotational motion, as it helps understand kinetic energies or torques involved in a system.
Angular Acceleration
Angular acceleration is defined as the rate of change of angular velocity over time. It tells us how quickly an object is speeding up or slowing down its rotation. In this problem, the flywheel's angular acceleration was determined using the angular kinematic equation:\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \] Here, \(\theta\) is the angular displacement, \(\omega_i\) is the initial angular speed, and \(\alpha\) is the angular acceleration.

During the time the power was off, the flywheel's angular acceleration was calculated to be negative, which indicates that the flywheel was decelerating. By substituting the given values:\[ \alpha = \frac{2(400\pi - 1570.8)}{900} \approx -1.396 \, \text{rad/s}^2\]This negative value indicates a decline in speed due to friction. The concept of angular acceleration is important in rotational dynamics as it influences the rotational inertia and can significantly affect the motion depending on the friction and external torques involved.
Angular Displacement
Angular displacement is the angle through which an object rotates about a fixed point. It is measured in radians or degrees and gives an indication of how far the object has turned. The calculation of angular displacement for the flywheel was:\[ \theta = 200 \, \text{revolutions} \times 2\pi \approx 400\pi \approx 1256.64 \, \text{radians}\] This shows that during the power failure, the flywheel made 200 full turns.

In understanding rotational dynamics, angular displacement is similar to linear displacement in linear motion—it's a measure of the distance traveled, albeit in a circular path. It helps us understand how much motion takes place in a given task and is pivotal to solving problems involving rotational kinematics.

It is important to remember that angular displacement is distinct from arc length, though they are related concepts. While arc length measures the actual distance along a circular path, angular displacement is strictly about the angle through which the object has rotated.
Revolutions per Minute (RPM)
Revolutions per minute (RPM) is a unit of angular velocity commonly used to express the speed at which an object, like a flywheel, is rotating. In this unit, the rotation speed is expressed as the number of complete revolutions an object makes in one minute.

In contexts such as this exercise, RPM is often the initial or observed state of rotating systems. The initial RPM of the flywheel was 500, which needed conversion to more universally applicable SI units for calculations involving physics. The formula used for conversion was:\[ \omega = \text{RPM} \times \frac{2\pi}{60}\]This is important because while RPM is intuitive and regularly used in engineering and everyday contexts, understanding and working with radian measures gives a more mathematical and scientific perspective.

The final speed result, after the power failure, was converted back to RPM for practicality and ease of understanding:\[ \omega_f \approx 100 \, \text{RPM} \]Understanding RPM is not only helpful for engineers but also for anyone dealing with machines where rotation is integral. It's a concept that connects daily use with more complex mathematical descriptions of motion.

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Most popular questions from this chapter

You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm 1rev/min2. (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

A computer disk drive is turned on starting from rest and has constant angular acceleration. If it took 0.0865 s for the drive to make its \(second\) complete revolution, (a) how long did it take to make the first complete revolution, and (b) what is its angular acceleration, in rad/s\(^2\)?

If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \\( \frac{1}{2} \\)\(MV^2\). However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13% of a person's mass, while the legs and feet together account for 37%. For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm30^\circ\) (a total of 60\(^\circ\)) from the vertical in approximately 1 second. Assume that they are held straight, rather than being bent, which is not quite true. Consider a 75-kg person walking at 5.0 km/h, having arms 70 cm long and legs 90 cm long. (a) What is the average angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

At \(t\) \(=\) 3.00 s a point on the rim of a 0.200-m-radius wheel has a tangential speed of 50.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.0 m/s\(^2\). (a) Calculate the wheel’s constant angular acceleration. (b) Calculate the angular velocities at \(t\) \(=\) 3.00 s and \(t\) \(=\) 0. (c) Through what angle did the wheel turn between \(t\) \(=\) 0 and \(t\) \(=\) 3.00 s? (d) At what time will the radial acceleration equal g?

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s. (a) Find the angular acceleration in rev/s\(^2\) and the number of revolutions made by the motor in the 4.00-s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

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