/*! 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 24 The \(15-\mathrm{kg}\) circular ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 24

The \(15-\mathrm{kg}\) circular disk spins about its axle with a constant angular velocity of \(\omega_{1}=10 \mathrm{rad} / \mathrm{s}\). Simultaneously, the yoke is rotating with a constant angular velocity of \(\omega_{2}=5 \mathrm{rad} / \mathrm{s} .\) Determine the angular momentum of the disk about its center of mass \(O\), and its kinetic energy.

Short Answer

Expert verified
The angular momentum of the disk about its center of mass is \(L_{total} = m \cdot r_{yoke}^2 \cdot \omega_2\) and the total kinetic energy of the disk is \(K.E_{total} = 0.5 \cdot m \cdot r_{yoke}^2 \cdot \omega_2^2\)

Step by step solution

01

Calculate the angular momentum about the center of mass

First, we calculate the angular momentum of the disk about its center of mass. The formula for angular momentum \(L\) of a rotating object is given by \(L = I \cdot \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For a disk of mass \(m\) and radius \(r\) rotating around its center, the moment of inertia \(I\) is given by \(\frac{1}{2}mr^2\). However, we don't have the radius of the disk in the question, and since we are asked to find the angular momentum about the center of mass, we can simplify the moment of inertia to \(I = m \cdot r^2\), where \(r = 0\) (since the rotation is right at the center). This simplifies our calculation to \(L = 0 \cdot \omega_1\)
02

Compute the overall motion's angular momentum

Next, we calculate the angular momentum caused by the overall motion of the center of mass. Since we don't have the radius for this rotation either, we'll assume it to be \(r_{yoke}\). So the angular momentum due to this motion is \(L = m \cdot r_{yoke}^2 \cdot \omega_2\)
03

Calculate the total angular momentum

Now, we add the two components of angular momentum to get the total angular momentum of the disk. So, \(L_{total} = 0 + m \cdot r_{yoke}^2 \cdot \omega_2\)
04

Compute the rotational kinetic energy

The formula for rotational kinetic energy is \(K.E = 0.5 \cdot I \cdot \omega^2\). Now finding the rotational kinetic energy about its own axis, we substitute \(I = m \cdot r^2\) and \(r = 0\) (since rotation is at the center), so \(K.E = 0\).
05

Compute the kinetic energy due to overall motion

The kinetic energy due to the overall motion, or translational kinetic energy, is given by \(K.E = 0.5 \cdot I \cdot \omega^2\). Substituting \(I = m \cdot r^2\), we get \(K.E = 0.5 \cdot m \cdot r_{yoke}^2 \cdot \omega_2^2\).
06

Calculate the total kinetic energy

Finally, to calculate the total kinetic energy of the disk, we add the rotational kinetic energy and the translational kinetic energy, so \(K.E_{total} = 0 + 0.5 \cdot m \cdot r_{yoke}^2 \cdot \omega_2^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A wheel of mass \(m\) and radius \(r\) rolls with constant spin \(\omega\) about a circular path having a radius \(a\). If the angle of inclination is \(\theta\), determine the rate of precession. Treat the wheel as a thin ring. No slipping occurs.

The football has a mass of \(450 \mathrm{~g}\) and radii of gyration about its axis of symmetry ( \(z\) axis) and its transverse axes ( \(x\) or \(y\) axis) of \(k_{z}=30 \mathrm{~mm}\) and \(k_{x}=k_{y}=50 \mathrm{~mm}\), respectively. If the football has an angular momentum of \(H_{G}=0.02 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), determine its precession \(\dot{\phi}\) and spin \(\dot{\psi}\). Also, find the angle \(\beta\) that the angular velocity vector makes with the \(z\) axis

The 1-lb top has a center of gravity at point \(G\). If it spins about its axis of symmetry and precesses about the vertical axis at constant rates of \(\omega_{s}=60 \mathrm{rad} / \mathrm{s}\) and \(\omega_{p}=10 \mathrm{rad} / \mathrm{s}\), respectively, determine the steady state angle \(\theta .\) The radius of gyration of the top about the \(z\) axis is \(k_{z}=1\) in., and about the \(x\) and \(y\) axes it is \(k_{x}=k_{y}=4\) in.

Show that if the angular momentum of a body is determined with respect to an arbitrary point \(A\), then \(\mathbf{H}_{A}\) can be expressed by Eq. \(21-9 .\) This requires substituting \(\boldsymbol{\rho}_{A}=\boldsymbol{\rho}_{G}+\boldsymbol{\rho}_{G / A}\) into Eq. \(21-6\) and expanding, noting that \(\int \boldsymbol{\rho}_{G} d m=\mathbf{0}\) by definition of the mass center and \(\mathbf{v}_{G}=\mathbf{v}_{A}+\omega \times \boldsymbol{\rho}_{G / A}\)

The propeller on a single-engine airplane has a mass of \(15 \mathrm{~kg}\) and a centroidal radius of gyration of \(0.3 \mathrm{~m}\) computed about the axis of spin. When viewed from the front of the airplane, the propeller is turning clockwise at \(350 \mathrm{rad} / \mathrm{s}\) about the spin axis. If the airplane enters a vertical curve having a radius of \(80 \mathrm{~m}\) and is traveling at \(200 \mathrm{~km} / \mathrm{h}\), determine the gyroscopic bending moment which the propeller exerts on the bearings of the engine when the airplane is in its lowest position.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Physical Quantities And Units

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Turning Points in Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.