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Chapter 7: Problem 55

The aircraft landing gear viewed from the front is being retracted immediately after takeoff, and the wheel is spinning at the rate corresponding to the takeoff speed of \(200 \mathrm{km} / \mathrm{h}\). The 45 -kg wheel has a radius of gyration about its \(z\) -axis of \(370 \mathrm{mm} .\) Neglect the thickness of the wheel and calculate the angular momentum of the wheel about \(G\) and about \(A\) for the position where \(\theta\) is increasing at the rate of \(30^{\circ}\) per second.

Short Answer

Expert verified
The angular momentum about both G and A is approximately 922.03 kg鈰卪虏/s.

Step by step solution

01

Convert Units

First, we convert the speed of the wheel from kilometers per hour to meters per second:\[200 \text{ km/h} = \frac{200}{3.6} \text{ m/s} \approx 55.56 \text{ m/s}\]
02

Calculate Angular Velocity

The angular velocity \(\omega\) of the wheel can be found using the relationship between linear and angular speed: \[v = r\omega \Rightarrow \omega = \frac{v}{r}\]where \(v = 55.56 \text{ m/s}\) and the radius of gyration \(r = 0.37 \text{ m}\). Thus, \[\omega = \frac{55.56}{0.37} \approx 150 \text{ rad/s}\]
03

Calculate Moment of Inertia

The moment of inertia \(I_G\) of the wheel about the axis through its center of gravity \(G\) is given by:\[I_G = m k^2\]where \(m = 45 \text{ kg}\) and \(k = 0.37 \text{ m}\). Therefore,\[I_G = 45 \times (0.37)^2 \approx 6.14685 \text{ kg} \cdot \text{m}^2\]
04

Calculate Angular Momentum about G

The angular momentum \(H_G\) about the axis through the center of gravity \(G\) is:\[H_G = I_G \omega\]Substituting the values from the previous steps:\[H_G = 6.14685 \cdot 150 \approx 922.0275 \text{ kg} \cdot \text{m}^2/\text{s}\]
05

Determine Angular Momentum about A

The angular momentum about any point \(A\) is found by considering the parallel axis theorem if \(A\) is not along the axis of rotation. Here assume \(A\) is along z-axis through G so \(H_A = H_G\).\[H_A = 922.0275 \text{ kg} \cdot \text{m}^2/\text{s}\]

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Key Concepts

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

Moment of Inertia
The moment of inertia is a critical concept in rotational dynamics. It is the rotational equivalent of mass in linear motion.
When calculating how difficult it is to change the rotational speed (angular velocity) of an object, we use the moment of inertia. It depends on the distribution of mass in relation to the axis of rotation.
For a wheel, the formula is given by \[ I = mr^2 \]where:
  • \( I \) is the moment of inertia,
  • \( m \) is the mass of the wheel, and
  • \( r \) is the radius of gyration. This is the distance from the axis of rotation at which the total mass could be concentrated to have the same moment of inertia.
By concentrating on the radius of gyration, which is 370 mm or 0.37 meters in this problem, you can simplify calculations. The formula indicates that the further mass is distributed from the axis of rotation, the larger the moment of inertia. It directly affects how much torque is needed to change the rotational state of the object.
Angular Velocity
Angular velocity refers to how fast an object is rotating, similar to how linear velocity measures how fast an object is moving in a straight line. It is described by the symbol \( \omega \) and is measured in radians per second (rad/s).
This type of velocity relates the linear speed of a point on the edge of the wheel to the radius.
The formula to determine angular velocity is:\[ \omega = \frac{v}{r} \]where:
  • \( v \) is the linear speed, and
  • \( r \) is the distance from the axis of rotation (radius in this context).
In the problem, we calculated \( \omega \) by converting the initial linear speed from kilometers per hour to meters per second and using the radius of gyration (\( 0.37 \) m) in the formula. Knowing \( \omega \) helps understand how fast the wheel spins, which is crucial for figuring out its behavior during landing gear retraction in the aircraft.
Radius of Gyration
The radius of gyration is a key term when exploring rotational motion, especially when it comes to understanding how mass is distributed relative to the axis of rotation.
It simplifies the computation of moment of inertia by providing a distance where the entire mass of a body could be concentrated, maintaining the same rotational characteristics.
Typically denoted by \( k \), the radius of gyration can be seen as:
  • A representation of the distribution of mass,
  • A factor to easily compute the moment of inertia using the formula \( I = mk^2 \),
  • In this exercise, a given value that determines how far the mass appears 'spread out' from the axis of rotation.
For our wheel, considering its radius of gyration about the axis is 0.37 m gives significant insight into how mass affects its rotational inertia. The larger the radius of gyration, the greater the moment of inertia will be, thus requiring more torque to change its rotational velocity. This understanding is pivotal for solving angular momentum problems efficiently.

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Most popular questions from this chapter

The wheel of radius \(r\) is free to rotate about the bent axle \(C O\) which turns about the vertical axis at the constant rate \(p\) rad/s. If the wheel rolls without slipping on the horizontal circle of radius \(R,\) determine the expressions for the angular velocity \(\omega\) and angular acceleration \(\alpha\) of the wheel. The \(x\) -axis is always horizontal.

The primary structure of a proposed space station consists of five spherical shells connected by tubular spokes. The moment of inertia of the structure about its geometric axis \(A-A\) is twice as much as that about any axis through \(O\) normal to \(A-A\). The station is designed to rotate about its geometric axis at the constant rate of 3 rev/min. If the spin axis \(A-A\) precesses about the \(Z\) -axis of fixed orientation and makes a very small angle with it, calculate the rate \(\psi\) at which the station wobbles. The mass center \(O\) has negligible acceleration.

The motor turns the disk at the constant speed \(p=30\) rad/sec. The motor is also swiveling about the horizontal axis \(B-O(y\) -axis ) at the constant speed \(\dot{\theta}=2\) rad/sec. Simultaneously, the entire assembly is rotating about the vertical axis \(C-C\) at the constant rate \(q=8\) rad/sec. For the instant when \(\theta=30^{\circ},\) determine the angular acceleration \(\alpha\) of the disk and the acceleration a of point \(A\) at the bottom of the disk. Axes \(x-y-z\) are attached to the motor housing, and plane \(O-x_{0}-y\) is horizontal.

Each of the identical wheels has a mass of \(4 \mathrm{kg}\) and a radius of gyration \(k_{z}=120 \mathrm{mm}\) and is mounted on a horizontal shaft \(A B\) secured to the vertical shaft at \(O .\) In case \((a),\) the horizontal shaft is fixed to a collar at \(O\) which is free to rotate about the vertical \(y\) -axis. In case \((b),\) the shaft is secured by a yoke hinged about the \(x\) -axis to the collar. If the wheel has a large angular velocity \(p=3600\) rev/min about its \(z\) -axis in the position shown, determine any precession which occurs and the bending moment \(M_{A}\) in the shaft at \(A\) for each case. Neglect the small mass of the shaft and fitting at \(O\)

The collar and clevis \(A\) are given a constant upward velocity of 8 in./sec for an interval of motion and cause the ball end of the bar to slide in the radial slot in the rotating disk. Determine the angular acceleration of the bar when the bar passes the position for which \(z=3\) in. The disk turns at the constant rate of 2 rad/sec.
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