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Chapter 6: Problem 137

Under active development is the storage of energy in high-speed rotating disks where friction is effectively eliminated by encasing the rotor in an evacuated enclosure and by using magnetic bearings. For a 10 -kg rotor with a radius of gyration of \(90 \mathrm{mm}\) rotating initially at \(80000 \mathrm{rev} / \mathrm{min},\) calculate the power \(P\) which can be extracted from the rotor by applying a constant \(2.10-\mathrm{N} \cdot \mathrm{m}\) retarding torque \((a)\) when the torque is first applied and ( \(b\) ) at the instant when the torque has been applied for 120 seconds.

Short Answer

Expert verified
(a) 17,591 W when first applied; (b) 11,052 W after 120 seconds.

Step by step solution

01

Convert Units

First, convert the given rotational speed from revolutions per minute to radians per second. We have: \[80,000 \text{ rev/min} = 80,000 \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1}{60 \text{ s}} \approx 8,377 \text{ rad/s}\]
02

Calculate Moment of Inertia

Using the radius of gyration \( k = 90 \text{ mm} = 0.09 \text{ m} \), calculate the moment of inertia \( I \) for the rotor.\[ I = m \cdot k^2 = 10 \cdot (0.09)^2 = 0.081 \text{ kg} \cdot \text{m}^2\]
03

Calculate Initial Angular Deceleration

The angular deceleration \( \alpha \) can be calculated from the applied torque \( \tau = 2.10 \text{ Nm} \) using the formula \( \tau = I \alpha \).\[ \alpha = \frac{\tau}{I} = \frac{2.10}{0.081} \approx 25.93 \text{ rad/s}^2 \]
04

Compute Initial Extractable Power

The power \( P \) can be calculated as the product of the torque and the initial angular velocity.\[ P = \tau \omega = 2.10 \times 8,377 \approx 17,591 \text{ W} \]
05

Calculate Angular Velocity after 120 Seconds

Use the relation \( \omega = \omega_0 - \alpha t \) to find the angular velocity after 120 seconds:\[ \omega = 8,377 - 25.93 \times 120 = 5,263 \text{ rad/s} \]
06

Compute Extractable Power after 120 Seconds

Calculate the power after 120 seconds using the new angular velocity.\[ P = \tau \omega = 2.10 \times 5,263 \approx 11,052 \text{ W} \]

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Key Concepts

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

Energy Storage
Energy storage in rotational dynamics involves the use of rotating disks to store kinetic energy. High-speed rotating disks, like the one described in the exercise, leverage rotational inertia to store energy efficiently. This method is especially effective when friction is minimized by enclosing the rotor in a vacuum and using magnetic bearings.
The stored energy in a rotating object is given by its rotational kinetic energy, which can be expressed as:
\[ KE = \frac{1}{2} I \omega^2 \] where
  • \(KE\) is the kinetic energy
  • \(I\) is the moment of inertia
  • \(\omega\) is the angular velocity
This formula demonstrates how energy is stored in the form of rotational motion, relying on both the rotational speed and the distribution of mass within the rotor.
Moment of Inertia
The moment of inertia is a fundamental concept in understanding how easy or difficult it is to change the rotational motion of an object. It reflects the distribution of mass relative to the axis of rotation. For a rotor, as in the exercise, the moment of inertia can be calculated using the formula:
\[ I = m \cdot k^2 \] where
  • \( m \) is the mass of the rotor
  • \( k \) is the radius of gyration
The radius of gyration \( k \) is a measure of how far the mass is spread from the axis; a larger \( k \) means the mass is further out, increasing the moment of inertia. A higher moment of inertia implies that the rotor will be more resistant to changes in rotational speed when a torque is applied.
Angular Velocity
Angular velocity refers to the rate of rotation of an object and is measured in radians per second. It indicates how quickly the rotor is spinning. For instance, in the original problem, the rotor's initial angular velocity was calculated from revolutions per minute to radians per second:
\[ \omega = 8,377 \text{ rad/s} \] This conversion is crucial because angular velocity is used in calculating both the initial energy stored in the rotor and the power that can be extracted.

Angular velocity can change under the action of a torque. This change is governed by the relationship between torque, moment of inertia, and angular acceleration, highlighting the dynamic nature of rotational systems.
Angular Deceleration
Angular deceleration describes the decrease in angular velocity over time. It is caused by an opposing torque and is crucial in determining how quickly a rotor slows down. In rotational dynamics, the angular deceleration \(\alpha\) can be predicted using the formula:
\[ \alpha = \frac{\tau}{I} \] where
  • \( \tau \) is the applied torque
  • \( I \) is the moment of inertia
In the solution, the rotor began decelerating at a rate of \(25.93 \text{ rad/s}^2\). This rate determines how the angular velocity decreases over time.
After a specified time, as shown in the exercise, you can use \( \omega = \omega_0 - \alpha t \) to find the new angular velocity.

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Most popular questions from this chapter

The sector and attached wheels are released from rest in the position shown in the vertical plane. Each wheel is a solid circular disk weighing \(12 \mathrm{lb}\) and rolls on the fixed circular path without slipping. The sector weighs 18 lb and is closely approximated by one-fourth of a solid circular disk of 16-in. radius. Determine the initial angular acceleration \(\alpha\) of the sector.

The uniform 225 -lb crate is supported by the thin homogeneous 40 -lb platform \(B F\) and light support links whose motion is controlled by the hydraulic cylinder \(C D .\) If the cylinder is extending at a constant rate of 6 in./sec when \(\theta=75^{\circ},\) determine the magnitudes of the forces supported by the pins at \(B\) and \(F\). Additionally, determine the total friction force acting on the crate. The crate is centered on the 6 -ft platform, and friction is sufficient to keep the crate motionless relative to the platform. (Hint: Be careful with the location of the resultant normal force beneath the crate.)

The gear train shown starts from rest and reaches an output speed of \(\omega_{C}=240\) rev/min in 2.25 s. Rotation of the train is resisted by a constant \(150 \mathrm{N} \cdot \mathrm{m}\) moment at the output gear \(C .\) Determine the required input power to the \(86 \%\) efficient motor at \(A\) just before the final speed is reached. The gears have masses \(m_{A}=6 \mathrm{kg}, m_{B}=10 \mathrm{kg},\) and \(m_{C}=24 \mathrm{kg}\) pitch diameters \(d_{A}=120 \mathrm{mm}, d_{B}=160 \mathrm{mm},\) and \(d_{C}=240 \mathrm{mm},\) and centroidal radii of gyration \(k_{A}=\) \(48 \mathrm{mm}, k_{B}=64 \mathrm{mm},\) and \(k_{C}=96 \mathrm{mm}\)

The 50 -kg flywheel has a radius of gyration \(\bar{k}=0.4 \mathrm{m}\) about its shaft axis and is subjected to the torque \(M=2\left(1-e^{-0.1 \theta}\right) \mathrm{N} \cdot \mathrm{m},\) where \(\theta\) is in radians. If the flywheel is at rest when \(\theta=0,\) determine its angular velocity after 5 revolutions.

A space telescope is shown in the figure. One of the reaction wheels of its attitude-control system is spinning as shown at \(10 \mathrm{rad} / \mathrm{s}\), and at this speed the friction in the wheel bearing causes an internal moment of \(10^{-6} \mathrm{N} \cdot \mathrm{m}\). Both the wheel speed and the friction moment may be considered constant over a time span of several hours. If the mass moment of inertia of the entire spacecraft about the \(x\) -axis is \(150\left(10^{3}\right) \mathrm{kg} \cdot \mathrm{m}^{2},\) determine how much time passes before the line of sight of the initially stationary spacecraft drifts by 1 arcsecond, which is \(1 / 3600\) degree. All other elements are fixed relative to the spacecraft, and no torquing of the reaction wheel shown is performed to correct the attitude drift. Neglect external torques.
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