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Chapter 18: Problem 108

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 \(A C\).

Short Answer

Expert verified
The two possible precession rates are opposite in direction: \( \Omega = \frac{mgr}{I\omega} \) and \( \Omega = -\frac{mgr}{I\omega} \).

Step by step solution

01

Identify Given Values

The disk has a 6-inch diameter, which means its radius \( r \) is \( 3 \) inches or \( 0.25 \) feet.The disk spins at a rate of 2100 rpm about its symmetry axis \( AB \).The angle \( \beta = 45^{\circ} \) is formed with the vertical axis \( AC \).
02

Calculate Angular Velocity

Convert the rotational speed from rpm (revolutions per minute) to rad/s (radians per second). Use the conversion: \[\omega = 2100 \times \frac{2\pi}{60} \approx 219.91 \text{ rad/s}\]
03

Determine Moment of Inertia

Calculate the moment of inertia \( I \) of the disk using the formula for a thin disk: \[ I = \frac{1}{2} m r^2 \]Assume the density \( \rho \) and thickness \( t \) yield mass \( m \). Since mass is not required, focus on the formula's form.
04

Solve for Precession Rate

The precession rate \( \Omega \) can be determined using:\[\tau = I \omega \Omega \sin(45^{\circ})\] Since \( \tau = m g r \sin(45^{\circ}) \), equating gives:\[ I \omega \Omega \sin(45^{\circ}) = m g r \sin(45^{\circ}) \] Simplify and solve for \( \Omega \):\[ \Omega = \frac{mgr}{I\omega} \]Identify two possible solutions using Newton's equations of motion (can occur in opposite directions):\[ \Omega_1 = \frac{mgr}{I\omega} ,\, \Omega_2 = -\frac{mgr}{I\omega} \]
05

Substitute Numerical Values

Insert the given values into the equation to calculate numerical results for both \( \Omega_1 \) and \( \Omega_2 \). Note: If numerical mass or density is not given, consider solving symbolically, portraying theoretical results about condition-based scenarios.

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates around an axis. In the context of the spinning disk, it refers to how fast the disk spins about its own symmetry axis.
To express angular velocity, we typically use radians per second (rad/s) as the unit since a full circle is equal to 2蟺 radians.
For a disk spinning at a rate of 2100 rpm (revolutions per minute), converting this to rad/s involves a straightforward calculation:
  • First, recognize that one revolution is equal to 2蟺 radians.
  • Next, since there are 60 seconds in a minute, you convert the revolutions per minute to revolutions per second.
  • Finally, multiply by 2蟺 to switch units from revolutions to radians.
As shown in the original solution, this results in an angular velocity of approximately 219.91 rad/s.
Understanding angular velocity is crucial for dynamic systems, especially when analyzing rotational motion like that of gyroscopes or disks.
Moment of Inertia
The moment of inertia is a physical property representing an object's resistance to angular acceleration, depending on its mass distribution.
It's often compared to the concept of mass in linear motion.
For a rotating object, such as our thin disk, it determines how difficult it is to change its rotational speed.For a uniform thin disk, the moment of inertia is calculated using the formula:\[ I = \frac{1}{2} m r^2 \]where:
  • \( I \) is the moment of inertia,
  • \( m \) is the mass of the disk,
  • \( r \) is the radius of the disk.
Though the mass \( m \) wasn't given in the original exercise, knowing the form of the formula helps understand how inertia affects rotational dynamics.
The radius is crucial since even small increases lead to larger changes in inertia.
This concept is essential when calculating the torque needed to alter the disk's rotation.
Precession Rate
The precession rate describes how a spinning object like a disk or gyroscope rotates around another axis.
In this case, the disk spins around its axis and also moves around the vertical axis \( AC \), forming two types of motion simultaneously.To calculate the precession rate \( \Omega \), we employ the equation linking the angular momentum of the disk and the external torque caused by gravity:\[ \tau = I \omega \Omega \sin(\beta) \]Given that the torque \( \tau \) is the result of the gravitational force acting on the disk, substituting \( mg \) for torque and simplifying provides:\[ \Omega = \frac{mgr}{I\omega} \]where:
  • \( \Omega \) is the precession rate,
  • \( \omega \) is the angular velocity,
  • \( m \) and \( g \) stand for mass and gravitational acceleration,
  • \( r \) is the radius of the disk.
The equation suggests two potential precession rates: one positive and one negative, indicating possible rotation in either direction.
This dual solution is typical in such rotational dynamics, showing that objects can precess clockwise or counter-clockwise depending on initial conditions.
Understanding precession is vital for predicting the behavior of gyroscopic systems, making it a fundamental concept in the study of rotational dynamics.

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Most popular questions from this chapter

Assuming that the wheel of Prob. 18.97 weighs 8 lb, has a radius \(a=4 \mathrm{in.},\) and a radius of gyration of \(3 \mathrm{in}\), and that \(R=20 \mathrm{in}\), determine the force exerted by the plate on the wheel when \(\Omega=25 \mathrm{rad} / \mathrm{s}\).

A thin homogeneous disk with a mass \(m\) and radius \(r\) is mounted on the horizontal axle \(A B\). The plane of the disk forms an angle of \(\beta=20^{\circ}\) with the vertical. Knowing that the axle rotates with an angular velocity \(\omega,\) determine the angle \(\theta\) formed by the axle and the angular momentum of the disk about \(G .\)

A space station consists of two sections \(A\) and \(B\) of equal masses that are rigidly connected. Each section is dynamically equivalent to a homogeneous cylinder with a length of \(15 \mathrm{m}\) and a radius of \(3 \mathrm{m}\). Knowing that the station is precessing about the fixed direction \(G D\) at the constant rate of 2 rev/h, determine the rate of spin of the station about its axis of symmetry \(C C^{\prime}\).

An experimental Fresnel-lens solar-energy concentrator can rotate about the horizontal axis \(A B\) that passes through its mass center \(G\). It is supported at \(A\) and \(B\) by a steel framework that can rotate about the vertical \(y\) axis. The concentrator has a mass of \(30 \mathrm{Mg}\), a radius of gyration of \(12 \mathrm{m}\) about its axis of symmetry \(\mathrm{CD},\) and a radius of gyration of \(10 \mathrm{m}\) about any transverse axis through \(G .\) Knowing that the angular velocities \(\omega_{1}\) and \(\omega_{2}\) have constant magnitudes equal to \(0.20 \mathrm{rad} / \mathrm{s}\) and \(0.25 \mathrm{rad} / \mathrm{s}\), respectively, determine for the position \(\theta=60^{\circ}(a)\) the forces exerted on the concentrator at \(A\) and \(B\) (b) the couple \(M_{2} \mathrm{k}\) applied to the concentrator at that instant.

A stationary horizontal plate is attached to the ceiling by means of a fixed vertical tube. A wheel of radius \(a\) and mass \(m\) is mounted on a light axle \(A C\) that is attached by means of a clevis at \(A\) to a rod \(A B\) fitted inside the vertical tube. The rod \(A B\) is made to rotate with a constant angular velocity \(\Omega\) causing the wheel to roll on the lower face of the stationary plate. Determine the minimum angular velocity \(\Omega\) for which contact is maintained between the wheel and the plate. Consider the particular cases ( \(a\) ) when the mass of the wheel is concentrated in the rim, (b) when the wheel is equivalent to a thin disk of radius \(a\).
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