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Chapter 18: Problem 1

A thin, homogeneous disk of mass \(m\) and radius \(r\) spins at the constant rate \(\omega_{1}\) about an axle held by a fork-ended vertical rod that rotates at the constant rate \(\omega_{2}\). Determine the angular momentum \(\mathbf{H}_{G}\) of the disk about its mass center \(G .\)

Short Answer

Expert verified
\(\mathbf{H}_G = \frac{1}{2}mr^2\omega_1\hat{k} + \frac{1}{2}mr^2\omega_2\hat{j}\).

Step by step solution

01

Understand the System

The problem describes a disk spinning around its own axis with angular velocity \(\omega_1\) and also rotating around a vertical rod with angular velocity \(\omega_2\). Since the rod produces precession, we need to consider the overall motion of the disk to determine the total angular momentum.
02

Calculate the Moment of Inertia of the Disk

The moment of inertia \(I\) for a homogeneous disk about its center of mass is given by the formula \(I = \frac{1}{2}mr^2\), where \(m\) is the mass and \(r\) is the radius of the disk.
03

Determine Angular Velocities

The disk has two angular velocities: \(\omega_1\) about its own axis (spin) and \(\omega_2\) due to the rotation around the vertical rod (precession). We need to consider the vector sum of these two angular velocities to determine the total angular velocity \(\vec{\omega}\) of the disk.
04

Express the Total Angular Velocity

The total angular velocity \(\vec{\omega}\) can be expressed as the vector sum of the axial spin \(\omega_1\hat{k}\) and the precession angular velocity. Since the disk is spinning about an axle parallel to the \(z\)-axis, and assuming that precession occurs in the \(xy\)-plane, \(\vec{\omega} = \omega_1\hat{k} + \omega_2\hat{j}\) or some equivalent vector, depending on the orientation.
05

Calculate the Angular Momentum About the Mass Center

The angular momentum \(\mathbf{H}_{G}\) of the disk about its mass center \(G\) is given by \(\mathbf{H}_{G} = I \cdot \vec{\omega}\). Substituting the expressions, we have \(\mathbf{H}_{G} = \frac{1}{2}mr^2(\omega_1\hat{k} + \omega_2\hat{j})\).
06

Simplify and State the Result

Substitute and simplify: \(\mathbf{H}_{G} = \frac{1}{2}mr^2\omega_1\hat{k} + \frac{1}{2}mr^2\omega_2\hat{j}\). This expression represents the angular momentum of the disk about its mass center \(G\) in terms of its angular velocities and geometric properties.

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Key Concepts

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

Moment of Inertia
Imagine you're holding a spinning disk. What makes it hard to stop or start it spinning is something called the "moment of inertia." It tells us how much "resistance" the disk gives when we're trying to change its spinning motion. The moment of inertia depends on how mass is distributed in the object.

For a disk, this can be calculated using the formula:
  • \( I = \frac{1}{2}mr^2 \)
where:
  • \( I \) is the moment of inertia.
  • \( m \) is the mass of the disk.
  • \( r \) is the radius of the disk.
This formula tells us that if you double the radius while keeping the mass the same, the moment of inertia (and hence the resistance to change its spin) increases by four times. So, the further away the mass is from the rotating axis (the edge of the disk), the harder it is to spin or stop it.
Angular Velocity
Angular velocity is all about how fast something spins or rotates. If you imagine a spinning wheel or a merry-go-round, the angular velocity describes how quickly it goes round and round. It's about speed, but not in a straight line.

In our exercise, the disk has two angular velocities:
  • \( \omega_1 \), the spin around its own axle. This is like turning around on its own without moving anywhere else.
  • \( \omega_2 \), the rotation with the vertical rod. This is more about moving in a circular path because of the precession.
To get the total angular velocity, we combine these two using something called vector addition. This means if you look at it from above, the disk spins along one path while also circling around another! The total angular velocity becomes a combination of these motions:
\( \vec{\omega} = \omega_1\hat{k} + \omega_2\hat{j} \). This ensures that we consider both the spin and the precession of the disk.
Precession
Now, precession is a fascinating phenomenon. It's the wobbling motion you see in a spinning object, like a top when it isn't perfectly upright. For the disk in our problem, precession happens because the axle is rotating with the vertical rod.

Imagine holding a spinning disk by a rod. If you start rotating the rod, the disk doesn't just spin in place; it also follows the path created by the movement of the rod. This leads to a combined motion where the disk spins around its axis and rotates with the axle's movement. This kind of circular path is known as precession.
  • This movement is crucial for understanding the overall angular momentum, as it adds a layer of direction and motion.
  • Incorporating precession when calculating the total angular momentum ensures that we account for all possible movements of the disk, not just the simple spin around its center.

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Most popular questions from this chapter

A uniform thin disk with a 6 -in. diameter is attached to the end of a rod \(A B\) of negligible mass that is supported by a ball-and-socket joint at point \(A\). Knowing that the disk is spinning about its axis of symmetry \(A B\) at the rate of 2100 rpm in the sense indicated and that \(A B\) forms an angle \(\beta=45^{\circ}\) with the vertical axis \(A C\), determine the two possible rates of steady precession of the disk about the axis \(A C\).

A coin is tossed into the air. It is observed to spin at the rate of 600 rpm about an axis \(G C\) perpendicular to the coin and to precess about the vertical direction \(G D .\) Knowing that \(G C\) forms an angle of \(15^{\circ}\) with \(G D,\) determine \((a)\) the angle that the angular velocity \(\omega\) of the coin forms with \(G D,(b)\) the rate of precession of the coin about \(G D .\)

A solid aluminum sphere of radius 4 in. is welded to the end of a 10 -in-long rod \(A B\) of negligible mass that is supported by a ball- and-socket joint at \(A\). Knowing that the sphere is observed to precess about a vertical axis at the constant rate of \(60 \mathrm{rpm}\) in the sense indicated and that rod \(A B\) forms an angle \(\beta=20^{\circ}\) with the vertical, determine the rate of spin of the sphere about line \(A B .\)

A 6 -lb homogeneous disk of radius 3 in. spins as shown at the constant rate \(\omega_{1}=60\) rad/s. The disk is supported by the fork-ended rod \(A B,\) which is welded to the vertical shaft \(C B D .\) The system is at rest when a couple \(M_{0}=(0.25 \mathrm{ft}-16)\) ) is applied to the shaft for \(2 \mathrm{s}\) and then removed. Determine the dynamic reactions at \(C\) and \(D\) after the couple has been removed.

A solid cube of side \(c=120 \mathrm{mm}\) is attached as shown to a cord \(A B\) of length \(240 \mathrm{mm}\). The cube spins about its diagonal \(B C\) and precesses about the vertical axis \(A D\). Knowing that \(\theta=25^{\circ}\) and \(\beta=40^{\circ},\) determine \((a)\) the rate of spin of the cube, \((b)\) its rate of precession. (See hint of Prob. 18.115 .)
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