/*! 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 102 A demonstration gyroscope wheel ... [FREE SOLUTION] | 91Ó°ÊÓ

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A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 \(\mathrm{m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 \(\mathrm{m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 \(\mathrm{kg}\) ; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.00 rev \(/ \mathrm{s}\) . Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at 0.050 rev \(/ \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at 0.300 \(\mathrm{rev} / \mathrm{s} .\) (d) At what rate must the shaft rotate in order that it may be supported at one end only?

Short Answer

Expert verified
(a) 39.24 N per hand up. (b) Solve \( F_{precession} \) using \( \omega_p = 0.050 \) rad/s. (c) Solve \( F_{precession} \) using \( \omega_p = 0.300 \) rad/s. (d) Solve \( \omega_p \) for one-hand support.

Step by step solution

01

Understanding the Gyroscope

The gyroscope consists of a wheel whose mass is concentrated at the rim. The shaft, where the force is applied, extends horizontally from the wheel. The wheel is spinning about the shaft axis, causing precession around the shaft. We need to consider these spinning and precession motions to solve the problem.
02

Calculating Linear Parameters

The diameter of the wheel is 0.650 m, so the radius is \( r = \frac{0.650}{2} = 0.325 \) m. The mass of the system is \( m = 8.00 \) kg. The wheel spins at \( \omega_s = 5 \) rev/s, which we convert to radians per second: \( \omega_s = 5 \times 2\pi \) rad/s.
03

Applying Force Equilibrium when Shaft is Stationary (a)

When the shaft is at rest (no precession), each hand supports half the weight of the system. The gravitational force is \( F_g = mg = 8.00 \cdot 9.81 \) N. Each hand exerts a force of \( \frac{F_g}{2} = \frac{8.00 \cdot 9.81}{2} = 39.24 \) N vertically upward.
04

Calculating Precession (b & c)

Precession rate \( \Omega \) is given by \( \Omega L = I \omega_s \omega_p \), where \( I = m r^2 \), \( L \) is the angular momentum, and \( \omega_p \) is precession angular speed. For (b), \( \Omega \) is \( 0.050 \times 2\pi \). For (c), \( \Omega \) is \( 0.300 \times 2\pi \).
05

Determining Torque and Force (b)

Calculate the torque from precession rate for \( \omega_p = 0.050 \): \( \tau = I \omega_s \omega_p = m r^2 \omega_s \omega_p \). In this case, \( \tau = 8 \cdot (0.325)^2 \cdot (10\pi) \cdot (0.1\pi) \). Each hand exerts a torque balancing force, \( F_{precession} = \frac{\tau}{0.2} \), adding vertically.
06

Determining Torque and Force (c)

For \( \omega_p = 0.300 \): repeat torque calculation using \( \omega_p = 0.6\pi \), \( \tau = 8 \cdot (0.325)^2 \cdot (10\pi) \cdot (0.6\pi) \). Solve \( F_{precession} \) as \( \frac{\tau}{0.2} \) to find precession force per hand.
07

Constraining for Single Supporting Hand (d)

For supporting with one hand, all torque must balance, \( \tau = F_g \times 0.2 \). Set \( I \omega_s \omega_p = mg \times 0.2 \). Solve for \( \omega_p = \frac{mg \times 0.2}{I \omega_s} \) given wheel radius and system mass, yielding the necessary precession rate.

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Key Concepts

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

Angular Momentum
Angular momentum is a crucial concept in understanding gyroscopes. It measures the quantity of rotation a body has. The gyroscope wheel, in this problem, is spinning, and its angular momentum can be calculated using the formula: \[ L = I \cdot \omega_s \] where \( L \) is the angular momentum, \( I \) is the moment of inertia, and \( \omega_s \) is the angular velocity of the spinning wheel. Since the mass of the wheel is concentrated on its rim, the moment of inertia can be simplified to \( I = m r^2 \), with \( m \) being the mass and \( r \) the radius of the wheel. The gyroscope maintains a stable rotation because of its angular momentum, resisting changes in its orientation. This stability is what makes gyroscopes so useful in navigation systems. Understanding how angular momentum works is key to grasping the behavior of gyroscopes. It helps explain both the balance and the persistence of spin, essential for the calculations carried out in this exercise.
Precession
Precession occurs when the axis of a spinning gyroscope slowly changes its orientation due to an applied external force. It doesn't stop the spin but causes the axis to move around. This exercise demonstrates that when the shaft of the gyroscope begins to rotate in a horizontal plane, the precession angular speed \( \omega_p \) is introduced. The rate of precession is influenced by the angular momentum and external torque acting on the gyroscope.The formula to determine the precession rate is given by: \[ \Omega = \frac{r \times F_g}{L} \] where \( \Omega \) represents the rate of precession, \( r \) is the radius, and \( F_g \) is the gravitational force. In context, precession is why the gravitational forces acting on a rotating object can cause its axis to rotate. This phenomenon can be counterintuitive, as the axis shifts perpendicularly to the direction of applied forces.Students need to understand this property, as it explains why the shaft moves differently when rotated at various speeds.
Torque
Torque is the measure of the force that can cause an object to rotate about an axis. In this gyroscope problem, torque plays a significant role when the system is both at rest and during precession. When the gyroscope is stationary, the hands exerting forces must balance the torque to prevent rotation. This is calculated using the gravitational force. However, as soon as the gyroscope begins to precess, additional torque from centrifugal force must be considered. The torque due to precession is given by:\[ \tau = I \cdot \omega_s \cdot \omega_p \] where \( \tau \) is the torque, \( I \) is the moment of inertia, \( \omega_s \) is the spin rate, and \( \omega_p \) is the precession rate. Torque ensures that forces are applied in such a way that they cause the desired motion, be it balanced stationary or steady precession. Understanding how torque interacts with rotational dynamics is crucial to accurately compute the forces required in gyroscope mechanics.
Rotational Dynamics
Rotational dynamics explores how torques and angular momenta interact with forces to create motion. For gyroscopes, this includes the spin of the wheel, the shaft’s precession, and external forces exerted by hands. These aspects come together to illustrate the overall dynamic behavior of the system. Dependence on spin speed is seen in how changing velocities alter torque effects and the rate of precession. In the given exercise, the rotational dynamics of the gyroscope are shown through:
  • The static force balance when the shaft is at rest.
  • The precession response when the shaft begins to rotate.
  • The calculation required for single-end hand support.
The central theme of rotational dynamics is to analyze how changes in angular momentum cause precession and necessitate torque adjustments. Understanding these interactions allows for predicting rotational outcomes accurately. This foundation is key for more advanced topics in rotational motion and complex systems involving multiple rotational axes.

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Most popular questions from this chapter

The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Fig. 10.49 ) When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled, hollow cylinder. His hands and arms have a combined mass 8.0 \(\mathrm{kg}\) . When outstretched, they span 1.8 \(\mathrm{m}\) ; when wrapped, they form a cylinder of radius 25 \(\mathrm{cm}\) . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . If his original angular speed is 0.40 \(\mathrm{rev} / \mathrm{s}\) , what is his final angular speed?

A uniform drawbridge 8.00 \(\mathrm{m}\) long is attached to the road- way by a frictionless hinge at one end, and it can be raised by a cable attached to the other end. The bridge is at rest, suspended at \(60.0^{\circ}\) above the horizontal, when the cable suddenly breaks. (a) Find the angular acceleration of the drawbridge just after the cable breaks. (Gravity behaves as though it all acts at the center of mass. 6 ) Could you use the equation \(\omega=\omega_{0}+\alpha t\) to calculate the angular speed of the drawbridge at a later time? Explain why. (c) What is the angular speed of the drawbridge as it becomes horizontal?

A solid, uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and \(\mathrm{rim}\) is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

A \(55-\mathrm{kg}\) runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.8 \(\mathrm{m} / \mathrm{s}\) . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 \(\mathrm{rad} / \mathrm{s}\) relative to the earth. The radius of the turntable is \(3.0 \mathrm{m},\) and its moment of inertia about the axis of rotation is 80 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

A gyroscope is precessing about a vertical axis. Describe what happens to the precession angular speed if the following changes in the variables are made, with all other variables remaining the same: (a) the angular speed of the spinning flywheel is doubled; (b) the total weight is doubled; (c) the moment of inertia about the axis of the spinning flywheel is doubled; \((\mathrm{d})\) the distance from the pivot to the center of gravity is doubled. (e) What happens if all four of the variables in parts (a) through (d) are doubled?

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