/*! 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 20 The flywheel of a motor has a ma... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 20

The flywheel of a motor has a mass of \(300.0 \mathrm{~kg}\) and a moment of incrtia of \(580 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The motor develops a constant torque of \(2000.0 \mathrm{~N} \cdot \mathrm{m}\), and the flywhecl starts from rest. (a) What is the angular acceleration of the flywheel? (b) What is its angular velocity after it makes 4.00 revolutions? (c) How much work is done by the motor during the first 4.00 revolutions?

Short Answer

Expert verified
(a) \( \alpha = 3.45 \, \mathrm{rad/s^2} \); (b) \( \omega = 13.16 \, \mathrm{rad/s} \); (c) Work = 50260 J.

Step by step solution

01

Calculate Angular Acceleration

To find the angular acceleration \( \alpha \), we use the equation \( \tau = I \alpha \), where \( \tau \) is the torque and \( I \) is the moment of inertia. Substitute \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and \( I = 580 \, \mathrm{kg} \cdot \mathrm{m^2} \):\[ \alpha = \frac{\tau}{I} = \frac{2000.0}{580} \approx 3.45 \, \mathrm{rad/s^2} \]
02

Calculate Angular Velocity After 4 Revolutions

To find the angular velocity \( \omega \) after the flywheel makes 4 revolutions, use the kinematic equation: \( \omega^2 = \omega_0^2 + 2 \alpha \theta \). Here, \( \omega_0 = 0 \), \( \alpha = 3.45 \, \mathrm{rad/s^2} \), and \( \theta = 4 \times 2\pi \approx 25.13 \, \mathrm{rad} \). Substitute these values to solve for \( \omega \):\[ \omega^2 = 0 + 2 \times 3.45 \times 25.13 \]\[ \omega^2 = 173.23 \]\[ \omega = \sqrt{173.23} \approx 13.16 \, \mathrm{rad/s} \]
03

Calculate Work Done by the Motor

The work done \( W \) by the motor can be determined using the equation: \( W = \tau \theta \). Using the torque \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and angle \( \theta = 25.13 \, \mathrm{rad} \):\[ W = 2000.0 \times 25.13 \approx 50260 \, \mathrm{J} \]
04

Conclusion

The solution requires calculating angular acceleration, angular velocity after four revolutions, and the work done by the motor during this time.

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Key Concepts

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

Angular Acceleration
Angular acceleration is essentially how quickly an object speeds up or slows down in rotational motion. In our problem, we calculated the angular acceleration \( \alpha \) using the formula: \[ \tau = I \alpha \] where \( \tau \) represents torque, and \( I \) is the moment of inertia. This formula shows that angular acceleration is directly proportional to torque.
For example, if you apply a greater force (torque) on a flywheel, you'll see it accelerate faster. However, a larger moment of inertia means the object resists rotational changes more, thus needs more torque for the same angular acceleration.
  • In this exercise, the torque was \(2000.0 \, \mathrm{N} \cdot \mathrm{m}\), and the moment of inertia was \(580 \, \mathrm{kg} \cdot \mathrm{m}^2\).
  • Substituting into the equation: \[ \alpha = \frac{2000.0}{580} \approx 3.45 \, \mathrm{rad/s^2} \]
  • So, the flywheel starts its motion with this angular acceleration.
Moment of Inertia
Moment of inertia is a critical factor in rotational motion, akin to mass in linear motion. It describes how the mass is distributed in an object concerning its axis of rotation. In simple terms, it tells us how difficult it is to change the rotation of an object.
For a flywheel, a higher moment of inertia means it will spin steadily without change unless acted on by a force. The values for moment of inertia differ substantially depending on the shape of the object and how its mass is distributed.
  • In our scenario, the flywheel's moment of inertia is \(580 \, \mathrm{kg} \cdot \mathrm{m}^{2}\).
  • It reflects that the flywheel has its mass fairly distributed away from the axis of rotation.
  • This impacts how quickly the flywheel can reach higher angular velocities or be stopped.
Torque
Torque is essentially the rotational equivalent of force in linear motion. It measures how much a force acting on an object causes that object to rotate. The magnitude of torque depends on three factors: the amount of force applied, the distance from the rotational axis, and the angle at which the force is applied.
An easy way to think about torque is using a wrench to tighten a bolt. The longer the wrench, the less force you need to exert to turn the bolt.
  • In this example, the motor applies a constant torque of \(2000.0 \, \mathrm{N} \cdot \mathrm{m}\) to the flywheel.
  • The constant torque leads to a steady increase in the flywheel's angular velocity, as shown by the calculation of angular acceleration.
  • This torque value helps facilitate the determination of both the angular velocity and work done on the system.
Angular Velocity
Angular velocity gives us an idea of how fast an object spins around in circles, rather than moving in a straight line. It's measured in radians per second (rad/s) and defines how fast the angular position or orientation of an object changes over time.
To find the final angular velocity of the flywheel after making 4 revolutions, we used the kinematic equation: \[ \omega^2 = \omega_0^2 + 2 \alpha \theta \] where \( \omega_0 = 0 \), \( \alpha = 3.45 \, \mathrm{rad/s^2} \), and \( \theta = 25.13 \, \mathrm{rad} \). Calculating this gives:
  • \( \omega = \sqrt{173.23} \approx 13.16 \, \mathrm{rad/s}\)
  • This speed is the result of the constant torque acting over the rotational distance the wheel travels.
  • In real-world terms, this angular velocity means the flywheel is spinning 13.16 radians every second after the 4 revolutions.
Work and Energy
In rotational motion, work and energy play a similar role to that in linear motion but with a twist. Work done is the action of force moving something over a distance; whereas in rotation, it's about applying torque to get an object to turn through an angle.
In our flywheel problem, to find the work done by the motor over 4 revolutions, we applied the formula: \[ W = \tau \theta \] where \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and \( \theta = 25.13 \, \mathrm{rad} \). Here's what we find:
  • The work done by the motor is calculated to be about \(50260 \, \mathrm{J}\).
  • This amount of energy is what has been input into the flywheel to reach its current state of motion.
  • Like linear motion, this energy could later be used, say, to perform tasks as the wheel transfers it back into the system or slows down.

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Most popular questions from this chapter

A large turntable rotates about a fixed vertical axis, making one revolution in 6.00 s. The moment of inertia of the turntable about this axis is \(1200 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A child of mass \(40.0 \mathrm{~kg}\). initially standing at the center of the turntable, runs out along a radius. What is the angular speed of the turntable when the child is \(2.00 \mathrm{~m}\) from the center, assuming that you can treat the child as a particle?

For each of the following rotating objects, describe the direction of the angular momentum vector: (a) the minute hand of a clock; (b) the right front tire of a car moving backward; (c) an ice skater spinning clockwise; (d) the earth rotating on its axis.

The moment of inertia of the empty tumtable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m}\). the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s}\). What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. A. \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) B. \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) C. \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) D. \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\)

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of \(5,00 \mathrm{~N} \cdot \mathrm{m}\) is applied to the tire for \(2.00 \mathrm{~s}\), the angular speed of the tire increases from 0 to 100 rev \(/\) min. The external torque is then removed, and the wheel is brought to rest in \(125 \mathrm{~s}\) by friction in its bearings. Compute (a) the moment of inertia of the wheel about the axis of rotation, (b) the friction torque, and (c) the total number of revolutions made by the wheel in the \(125 \mathrm{~s}\) time interval.

A uniform, 7.5 -m-long beam weighing \(9000 \mathrm{~N}\) is hinged to a wall and supported by a thin cable attached \(1.5 \mathrm{~m}\) from the free end of the beam. The cable runs between the beam and the wall and makes a \(40^{\circ}\) angle with the beam. What is the tension in the cable when the beam is at an angle of \(30^{\circ}\) above the horizontal?
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