/*! 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 168 The 75 -kg flywheel has a radius... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 168

The 75 -kg flywheel has a radius of gyration about its shaft axis of \(\bar{k}=0.50 \mathrm{m}\) and is subjected to the torque \(M=10\left(1-e^{-t}\right) \mathrm{N} \cdot \mathrm{m},\) where \(t\) is in seconds. If the flywheel is at rest at time \(t=0\) determine its angular velocity \(\omega\) at \(t=3\) s.

Short Answer

Expert verified
The angular velocity at \( t=3 \) s is approximately \( 1.093 \text{ rad/s} \).

Step by step solution

01

Determine Moment of Inertia

First, we calculate the moment of inertia \( I \) using the formula \( I = m \bar{k}^2 \), where \( m \) is the mass of the flywheel and \( \bar{k} \) is the radius of gyration. Substitute \( m = 75 \text{ kg} \) and \( \bar{k} = 0.50 \text{ m} \) to get:\[ I = 75 \times (0.50)^2 = 18.75 \text{ kg}\cdot\text{m}^2.\]
02

Set up the Torque Equation

The torque \( M \) applied to the flywheel is \( M = 10(1 - e^{-t}) \text{ N}\cdot\text{m} \). Torque is related to angular acceleration \( \alpha \) as \( M = I \alpha \). Rearrange to find angular acceleration: \[ \alpha = \frac{M}{I} = \frac{10(1 - e^{-t})}{18.75}. \]
03

Integrate Angular Acceleration to Find Angular Velocity

We know \( \alpha = \frac{d\omega}{dt} \). Integrate \( \alpha \) with respect to time \( t \) to find \( \omega \):\[ \omega = \int_0^{3} \frac{10(1 - e^{-t})}{18.75} \, dt. \]Split the integral:\[ \omega = \frac{10}{18.75} \int_0^3 (1 - e^{-t}) \,dt. \]
04

Evaluate the Integral

Perform the integration:\( \int (1 - e^{-t}) \, dt = \int 1 \, dt - \int e^{-t} \, dt \). This evaluates to:\( t + e^{-t} \). So:\[ \omega = \frac{10}{18.75} \left[ t + e^{-t} \right]_0^{3}. \]
05

Calculate the Angular Velocity

Substitute the limits into the integrated function:\[ \left. t + e^{-t} \right|_0^{3} = \left(3 + e^{-3}\right) - \left(0 + e^{0}\right). \]Evaluating gives:\[ 3 + \frac{1}{e^3} - 1 \approx 3 + 0.0498 - 1 = 2.0498. \]So:\[ \omega = \frac{10}{18.75} \times 2.0498 \approx 1.093 \text{ rad/s}. \]

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Key Concepts

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

Flywheel Dynamics
Flywheel dynamics involves understanding how rotational objects, like flywheels, behave when subjected to external forces or moments, such as torque. A flywheel is a mechanical device specifically designed to efficiently store rotational energy. It is often used in machines to ensure a smooth delivery of power.
In the given exercise, we see the relationship between torque and rotational motion of a flywheel. Torque, measured in Newton-meters (Nm), is the force that causes an object to rotate about an axis. In essence, it is the rotational equivalent of linear force.
A flywheel's behavior under torque depends on its mass distribution, quantified by the moment of inertia, and the radius of gyration. The radius of gyration is a measure of how far the mass of an object is spread out from its axis of rotation. Knowing these properties helps predict how the flywheel will react to applied torque, ultimately allowing calculation of angular velocity, as seen in this exercise.
Moment of Inertia
The moment of inertia (\( I \)) plays a critical role in analyzing rotational motion. It is a measure of an object's resistance to changes in its rotational state. In other words, it quantifies how difficult it is to change the rotational speed of an object.
For a flywheel, the moment of inertia depends on its mass and how the mass is distributed relative to the axis of rotation. The formula \( I = m \bar{k}^2 \) is used, where \( m \) is the mass and \( \bar{k} \) is the radius of gyration. The radius of gyration simplifies calculations by treating the mass distribution as if it were concentrated at a single radius.
This exercise demonstrates calculating the moment of inertia using the given mass of 75 kg and a radius of gyration of 0.50 meters, leading to a moment of inertia of 18.75 kg路m虏. This value is vital because it is used to determine how the applied torque will affect the flywheel's angular acceleration.
Angular Velocity Calculation
Calculating angular velocity (\( \omega \)) involves understanding the relationship between torque and angular acceleration. According to Newton's second law for rotation, the torque (\( M \)) results in angular acceleration (\( \alpha \)) in the system, given by the equation \( M = I \alpha \). Rearranging gives \( \alpha = \frac{M}{I} \).
In this problem, the torque applied is time-dependent: \( M = 10(1 - e^{-t}) \), and \( \alpha = \frac{10(1 - e^{-t})}{18.75} \). This equation shows angular acceleration as a function of time. To find angular velocity, we integrate \( \alpha \) with respect to time over the period from 0 to 3 seconds.
Through integration, the angular velocity is determined by evaluating the function \( \omega = \int_0^3 \frac{10(1 - e^{-t})}{18.75} \, dt \), resulting in a final angular velocity of approximately 1.093 rad/s. This indicates how fast the flywheel rotates after 3 seconds under the given torque.

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Most popular questions from this chapter

The uniform 5 -kg portion of a circular hoop is released from rest while in the position shown where the torsional spring of stiffness \(k_{T}=15 \mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\) has been twisted \(90^{\circ}\) clockwise from its undeformed position. Determine the magnitude of the pin force at \(O\) at the instant of release. Motion takes place in a vertical plane and the hoop radius is \(r=150 \mathrm{mm}\)

The 30 -kg spool of outer radius \(r_{o}=450 \mathrm{mm}\) has a centroidal radius of gyration \(\bar{k}=275 \mathrm{mm}\) and a central shaft of radius \(r_{i}=200 \mathrm{mm} .\) The spool is at rest on the incline when a tension \(T=300 \mathrm{N}\) is applied to the end of a cable which is wrapped securely around the central shaft as shown. Determine the acceleration of the spool center \(G\) and the magnitude and direction of the friction force acting at the interface of the spool and incline. The friction coefficients there are \(\mu_{s}=0.45\) and \(\mu_{k}=0.30 .\) The tension \(T\) is applied parallel to the incline and the angle \(\theta=20^{\circ}\)

The uniform 72 -ft mast weighs 600 lb and is hinged at its lower end to a fixed support at \(O .\) If the winch \(C\) develops a starting torque of 900 lb- ft, calculate the total force supported by the pin at \(O\) as the mast begins to lift off its support at \(B\). Also find the corresponding angular acceleration \(\alpha\) of the mast. The cable at \(A\) is horizontal, and the mass of the pulleys and winch is negligible.

The 100 -lb platform rolls without slipping along the \(10^{\circ}\) incline on two pairs of 16 -in.-diameter wheels. Each pair of wheels with attached axle weighs 25 lb and has a centroidal radius of gyration of 5.5 in. The platform has an initial speed of \(3 \mathrm{ft} / \mathrm{sec}\) down the incline when a tension \(T\) is applied through a cable attached to the platform. If the platform acquires a speed of \(3 \mathrm{ft} / \mathrm{sec}\) up the incline after the tension has been applied for 8 seconds, what is the average value of the tension in the cable?

Two small variable-thrust jets are actuated to keep the spacecraft angular velocity about the \(z\) -axis constant at \(\omega_{0}=1.25 \mathrm{rad} / \mathrm{s}\) as the two telescoping booms are extended from \(r_{1}=1.2 \mathrm{m}\) to \(r_{2}=4.5 \mathrm{m}\) at a constant rate over a 2 -min period. Determine the necessary thrust \(T\) for each jet as a function of time where \(t=0\) is the time when the telescoping action is begun. The small \(10-\mathrm{kg}\) experiment modules at the ends of the booms may be treated as particles, and the mass of the rigid booms is negligible.
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