/*! 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 222 The 6 -lb pendulum with mass cen... [FREE SOLUTION] | 91Ó°ÊÓ

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The 6 -lb pendulum with mass center at \(G\) is pivoted at \(A\) to the fixed support \(C A .\) It has a radius of gyration of 17 in. about \(O-O\) and swings through an amplitude \(\theta=60^{\circ} .\) For the instant when the pendulum is in the extreme position, calculate the moments \(M_{x}, M_{y},\) and \(M_{z}\) applied by the base support to the column at \(C\)

Short Answer

Expert verified
\(M_z \approx 240.3\ \text{ft}\cdot \text{lb}, M_x = 0, M_y = 0.\)

Step by step solution

01

Understand the Problem

We are given a pendulum with a 6-lb weight and a radius of gyration of 17 in. We need to find the moments \(M_x\), \(M_y\), and \(M_z\) applied by the base support to the column at point \(C\) when the pendulum is at the extreme position (\(\theta=60^\circ\)).
02

Identify Forces

Consider the forces acting on the pendulum. At the extreme position, the swing is purely in the plane, and the tension in the rod acts as a centripetal force. Gravitational force \(mg\) acts vertically downward at the center of mass \(G\).
03

Calculate Moment Contributions

The moment \(M_z\) is typically computed due to gravitational forces, while \(M_x\) and \(M_y\) are due to forces in the perpendicular directions. Since the pendulum swings on a vertical plane, calculate the horizontal distance from \(A\) to \(G\) at \(\theta=60^\circ\) and use \(r=17\) in.
04

Compute Moments

1. Compute \(M_z = mgh\sin(\theta)\): - \(h = 6\times 17^2/386.4 \approx 1.5\ ft\). - \(M_z = 6 \times 32.2 \times 1.5 \times \sin(60^\circ)\) ft\cdot lb.2. Moments \(M_x\) and \(M_y\) are zero as pendulum swings in the plane.
05

Solve for Numerical Values

Calculate \(M_z\) by plugging in the values: \[M_z = 6 \cdot 32.2 \cdot 1.5 \cdot \sin \left( \frac{\pi}{3} \right) \approx 240.3\ \text{ft}\cdot \text{lb}\].Since the pendulum is not accelerating laterally in the extreme positions, \(M_x = 0\ \text{and}\ M_y = 0\).

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Key Concepts

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

Pendulum Motion
Pendulum motion refers to the movement of a weight suspended from a pivot, which swings due to the force of gravity. The pendulum in the given problem is a physical pendulum, which means its mass is distributed along its length rather than concentrated at a point. When it's released from a raised position, gravity pulls it downward, creating a swinging motion. The point of suspension is pivotal in determining its range and motion dynamics. A key aspect of pendulum motion is the force of gravity, which always acts downwards. At each extreme of the swing, the pendulum momentarily stops before reversing direction. This happens because all its kinetic energy is converted into potential energy at this point. The extreme positions are critical because they define the amplitude of the swing. In the problem, the pendulum reaches an amplitude of 60°, which indicates it swings 60° away from its resting vertical position. Understanding these dynamics is crucial in analyzing the forces at play when it is at the extreme position.
Moments of Forces
The moment of a force, often referred to as torque, is a measure of its tendency to rotate an object about an axis or point. In physics, moments are calculated as the product of the force and the perpendicular distance from the axis or point to the line of action of the force. For the pendulum in this problem, we calculate the moments when it is at the extreme position. As the pendulum reaches its maximum displacement angle, we consider how gravitational force (acting at the pendulum's center of mass) generates a moment about the pivot. At - The moment about the z-axis ( M_z ) is due to the gravitational forces acting at the center of mass. - M_x and M_y are zero because the pendulum swings in a single vertical plane and thus has no perpendicular force components relative to those axes.
Radius of Gyration
The radius of gyration is a property that quantifies the distribution of an object's mass relative to a specific axis. It's defined as the distance from the axis at which the entire mass of a body can be assumed to be concentrated, if it were a point mass. In the context of a pendulum, the radius of gyration ( r ) is pivotal for understanding how its mass is spread out along its length. This property contributes to calculating the moment of inertia, which measures an object's resistance to rotational motion about an axis. A larger radius of gyration indicates that the mass is further from the axis of rotation, which affects the object's dynamics during motion. In the specific exercise, the pendulum has a radius of gyration of 17 inches about the axis O-O. This value is crucial when calculating the pendulum's moment of inertia, allowing us to comprehend how easily it rotates around the pivot and how that contributes to the moments calculated.
Gravitational Forces
Gravitational forces play a crucial role in pendulum motion. They act on the pendulum's center of mass, pulling it towards the Earth's center. This downward force is the driving factor behind the pendulum's swinging motion. Gravitational force is calculated as the product of mass ( m ) and the gravitational acceleration ( g , approximately 32.2 ft/s² for Earth). This force influences both the potential and kinetic energy at various points of the pendulum's path. For the pendulum in question, gravitational forces determine the torque or moment created about the pivot. These forces affect how the pendulum moves through its arcs and how moments are calculated at the extreme positions. The potential energy due to gravity is highest at these points, and this energy conversion governs the pendulum's behavior, making an understanding of gravitational effects essential to solving dynamic problems like this one.

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Most popular questions from this chapter

The 20 -kg wheel has an eccentric mass which places the center of mass \(G\) a distance \(\bar{r}=70 \mathrm{mm}\) away from the geometric center \(0 .\) A constant couple \(M=6 \mathrm{N} \cdot \mathrm{m}\) is applied to the initially stationary wheel, which rolls without slipping along the horizontal surface and enters the curve of radius \(R=600 \mathrm{mm} .\) Determine the normal force under the wheel just before it exits the curve at \(C .\) The wheel has a rolling radius \(r=100 \mathrm{mm}\) and a radius of gyration \(k_{O}=65 \mathrm{mm}\)

The thin hoop of negligible mass and radius \(r\) contains a homogeneous semicylinder of mass \(m\) which is rigidly attached to the hoop and positioned such that its diametral face is vertical. The assembly is centered on the top of a cart of mass \(M\) which rolls freely on the horizontal surface. If the system is released from rest, what \(x\) -directed force \(P\) must be applied to the cart to keep the hoop and semicylinder stationary with respect to the cart, and what is the resulting acceleration \(a\) of the cart? Motion takes place in the \(x\) -y plane. Neglect the mass of the cart wheels and any friction in the wheel bearings. What is the requirement on the coefficient of static friction between the hoop and cart?

The 30 -kg spool of outer radius \(r_{o}=450 \mathrm{mm}\) has a centroidal radius of gyration \(\bar{k}=275 \mathrm{mm}\) and a central shaft of radius \(r_{i}=200 \mathrm{mm} .\) The spool is at rest on the incline when a tension \(T=300 \mathrm{N}\) is applied to the end of a cable which is wrapped securely around the central shaft as shown. Determine the acceleration of the spool center \(G\) and the magnitude and direction of the friction force acting at the interface of the spool and incline. The friction coefficients there are \(\mu_{s}=0.45\) and \(\mu_{k}=0.30 .\) The tension \(T\) is applied parallel to the incline and the angle \(\theta=20^{\circ}\)

The nose-wheel assembly is raised by the application of a torque \(M\) to link \(B C\) through the shaft at \(B\) The arm and wheel \(A O\) have a combined weight of 100 lb with center of mass at \(G\), and a centroidal radius of gyration of 14 in. If the angle \(\theta=30^{\circ},\) determine the torque \(M\) necessary to rotate link \(A O\) with a counterclockwise angular velocity of 10 deg/sec that is increasing at the rate of 5 deg/sec every second. Additionally, determine the total force supported by the pin at \(A\). The mass of links \(B C\) and \(C D\) may be neglected for this analysis.

The uniform 20 -kg slender bar is pivoted at \(O\) and swings freely in the vertical plane. If the bar is released from rest in the horizontal position, calculate the initial value of the force \(R\) exerted by the bearing on the bar an instant after release.

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