/*! 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 12 The \(100-\mathrm{kg}\) projecti... [FREE SOLUTION] | 91Ó°ÊÓ

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The \(100-\mathrm{kg}\) projectile shown has a radius of gyration of \(100 \mathrm{mm}\) about its axis of symmetry \(G x\) and a radius of gyration of \(250 \mathrm{mm}\) about the transverse axis \(G y .\) Its angular velocity \(\omega\) can be resolved into two components; one component, directed along \(G x,\) measures the rate of spin of the projectile, while the other component, directed along \(G D,\) measures its rate of precession. Knowing that \(\theta=6^{\circ}\) and that the angular momentum of the projectile about its mass center \(G\) is \(\mathbf{H}_{G}=\left(500 \mathrm{g} \cdot \mathrm{m}^{2} / \mathrm{s}\right) \mathrm{i}-\left(10 \mathrm{g} \cdot \mathrm{m}^{2} / \mathrm{s}\right) \mathrm{j},\) determine \((a)\) the rate of spin, \((b)\) the rate of precession.

Short Answer

Expert verified
The rate of spin is 0.5 rad/s and the rate of precession is -0.0016 rad/s.

Step by step solution

01

Understanding the Inertial Properties

First, we calculate the moments of inertia. For spin about the symmetry axis, use the given radius of gyration about the symmetry axis: \( I_x = m k_x^2 = 100 \times (0.1)^2 \). For precession, use the transverse radius of gyration: \( I_y = m k_y^2 = 100 \times (0.25)^2 \).
02

Calculation of the Moments of Inertia

Compute the values from the formulas in Step 1: \( I_x = 100 \times 0.01 = 1 \text{ kg} \cdot \text{m}^2 \) \( I_y = 100 \times 0.0625 = 6.25 \text{ kg} \cdot \text{m}^2 \).
03

Resolving Angular Momentum Components

Angular momentum about the mass center \( G \) is given as \( \mathbf{H}_G = (500 \text{ g} \cdot \text{m}^2/s) \mathbf{i} - (10 \text{ g} \cdot \text{m}^2/s) \mathbf{j} \). Convert this to \( \text{kg} \cdot \text{m}^2/s \): \[ \mathbf{H}_G = (0.5 \text{ kg} \cdot \text{m}^2/s) \mathbf{i} - (0.01 \text{ kg} \cdot \text{m}^2/s) \mathbf{j} \]
04

Calculating the Rate of Spin

The rate of spin \( \omega_x \) is determined using the component along \( \mathbf{i} \): \[ \omega_x = \frac{H_{Gx}}{I_x} = \frac{0.5}{1} = 0.5 \text{ rad/s} \]
05

Calculating the Rate of Precession

The rate of precession \( \omega_{GD} \) is determined using the component along \( \mathbf{j} \): \[ \omega_{GD} = \frac{H_{Gy}}{I_y} = \frac{-0.01}{6.25} = -0.0016 \text{ rad/s} \]

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Key Concepts

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

Radius of Gyration
The radius of gyration is a concept in engineering and physics that basically measures how the mass of an object is distributed relative to a specific axis. It's a handy way of expressing the distribution of an object's mass. For an object rotating about an axis, the radius of gyration helps to simplify its mass distribution to a single point. This point is assumed to be at a specific distance from the axis, allowing us to calculate moments of inertia in a straightforward manner.

When it comes to our projectile, there are two radii of gyration: one along its symmetry axis (G x) and another perpendicular to that axis (G y). The symmetry axis radius of gyration is 100 mm, indicating that for rotation about the symmetry axis, the mass appears as if it is concentrated at a distance of 100 mm from the axis. The other radius of gyration, 250 mm, describes rotation along the transverse axis.

Knowing the radius of gyration enables us to calculate the moment of inertia using the formula: \[ I = m k^2 \] Here, 'm' is the mass of the object, and 'k' represents the radius of gyration. Understanding this concept allows us to efficiently work with the rotational dynamics of the body.
Moments of Inertia
Moments of inertia play a crucial role in rotational dynamics. They are essentially the rotational equivalent of mass in linear motion. Just as mass indicates how much an object resists acceleration in a straight line, moments of inertia tell us how much an object resists rotational acceleration.For the projectile in our exercise, moments of inertia about two axes were calculated:
  • One along the symmetry axis (G x): \( I_x = 1 \text{ kg} \cdot \text{m}^2 \).
  • One transverse to it (G y): \( I_y = 6.25 \text{ kg} \cdot \text{m}^2 \).
These calculations used the radius of gyration we discussed earlier. Moments of inertia are particularly important when analyzing how the projectile spins and processes, which ultimately affect its path and behavior during motion.

For practical calculations, if you have one moment of inertia and the corresponding radius of gyration, you can determine everything else quite easily. This is why understanding moments of inertia is crucial for properly engaging with any rotational dynamics problems.
Precession
Precession is an interesting phenomenon in rotational dynamics, often seen in spinning wheels or tops. It refers to the change in orientation of the rotational axis. In simpler terms, while something spins, its axis itself can also move in a circular motion.

For our projectile, precession is represented by the rate at which the axis moves around, in addition to actually rotating. In this exercise, the rate of precession, calculated from the angular momentum component along the transverse axis, gave us a value of \( \omega_{GD} = -0.0016 \text{ rad/s} \).

This negative sign indicates the direction of precession, meaning the axis of rotation swings in a particular manner, complementing the spinning motion. Understanding precession is critical for applications relating to gyrocompasses and navigation, as it allows better prediction of how rotating objects will behave over time.
Rate of Spin
The rate of spin is a fundamental aspect that characterizes how fast something rotates. It is measured in radians per second (rad/s) and can be thought of as the angular speed of rotation about an axis.

In our problem with the spinning projectile, the rate of spin was specifically calculated along the symmetry axis (G x). The angular momentum component provided along this axis allowed us to compute the rate of spin as \( \omega_x = 0.5 \text{ rad/s} \).

This value tells us how fast the projectile spins around its own axis. Fast spinning is a key feature in various applications like sports, missile dynamics, and even toys like fidget spinners. The rate of spin influences the stability of the rotating object and thus is critical for performance and control in many systems. Hence, knowing the rate of spin is vital for any accurate modeling of rotational systems.

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Most popular questions from this chapter

One of the sculptures displayed on a university campus consists of a hollow cube made of six aluminum sheets, each \(1.5 \times 1.5 \mathrm{m}\), welded together and reinforced with internal braces of negligible weight. The cube is mounted on a fixed base at \(A\) and can rotate freely about its vertical diagonal \(A B .\) As she passes by this display on the way to a class in mechanics, an engineering student grabs corner \(C\) of the cube and pushes it for \(1.2 \mathrm{s}\) in a direction perpendicular to the plane \(A B C\) with an average force of \(50 \mathrm{N}\). Having observed that it takes \(5 \mathrm{s}\) for the cube to complete one full revolution, she flips out her calculator and proceeds to determine the mass of the cube. What is the result of her calculation? (Hint: The perpendicular distance from the diagonal joining two vertices of a cube to any of its other six vertices can be obtained by multiplying the side of the cube by \(\sqrt{2 / 3}\).)

A homogeneous disk with a radius of 9 in. is welded to a rod \(A G\) with a length of 18 in. and of negligible weight that is connected by a clevis to a vertical shaft \(A B\). The rod and disk can rotate freely about a horizontal axis \(A C,\) and shaft \(A B\) can rotate freely about a vertical axis. Initially, rod \(A G\) is horizontal \(\left(\theta_{0}=90^{\circ}\right)\) and has no angular velocity about \(A C\). Knowing that the smallest value of \(\theta\) in the ensuing motion is \(30^{\circ}\), determine \((a)\) the initial angular velocity of shaft \(A B,(b)\) its maximum angular velocity.

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 \(a\).

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 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.

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