/*! 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 227 The uniform power pole of mass \... [FREE SOLUTION] | 91Ó°ÊÓ

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The uniform power pole of mass \(m\) and length \(L\) is hoisted into a vertical position with its lower end supported by a fixed pivot at \(O .\) The guy wires supporting the pole are accidentally released, and the pole falls to the ground. Plot the \(x\) - and \(y\) components of the force exerted on the pole at \(O\) in terms of \(\theta\) from 0 to \(90^{\circ}\). Can you explain why \(O_{y}\) increases again after going to zero?

Short Answer

Expert verified
\(O_y\) decreases to zero as \(\theta\) increases to \(90^\circ\), then increases due to the pendulum effect.

Step by step solution

01

Understanding the Scenario

The problem describes a uniform power pole of mass \(m\) and length \(L\) pivoting about a fixed point \(O\) as it falls to the ground. Initially, it is upright, and then it rotates downward due to gravity when the supporting guy wires are released.
02

Identifying Forces Exerted

The force at the pivot point \(O\) has two components: the horizontal force \(O_x\) and the vertical force \(O_y\). As the pole falls, these components will change according to the angle \(\theta\), which is the angle between the pole and the horizontal ground.
03

Equation for Angular Position

As the pole falls, its angular position \(\theta\) varies from \(0^{\circ}\) (upright) to \(90^{\circ}\) (horizontal on the ground). The torque about point \(O\) is due to the gravitational force acting on the center of mass of the pole, located at \(L/2\) from \(O\).
04

Calculating the Force Components

For a uniform rod pivoting about the end, the force components can be expressed as:1. \(O_x = - \frac{mg}{2} \sin \theta\)2. \(O_y = mg \cos \theta\)These come from the conditions of rotational equilibrium (though briefly disrupted by the fall) and Newton's laws.
05

Explanation of \(O_y\) Behavior

Initially, \(O_y = mg\) when \(\theta = 0\) because the pole is upright. As \(\theta\) increases, \(O_y\) decreases because \(\cos \theta\) decreases. \(O_y\) goes to zero when \(\theta = 90^{\circ}\), meaning the pole is fully horizontal and undergoing free fall in the y-direction. \(O_y\) increases because the pole acts like a pendulum; after reaching horizontal, it starts rotating back (due to momentum) and \(O_y\) reflects the centripetal force.

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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 fundamental concept in physics, especially when discussing objects in rotational motion. For a rotating body like our power pole, angular momentum can be thought of as an object's rotational version of linear momentum. It determines how much rotation an object has when pivoting around a point.

The angular momentum (\( L \)) of a rod rotating about a fixed pivot point is given by:
  • \( L = I \cdot \omega \)
Where:
  • \( I \) is the moment of inertia, which depends on the mass distribution of the rod.
  • \( \omega \) is the angular velocity.
When the guy wires are released, the pole starts falling, causing changes in its angular velocity. However, due to conservation laws, the angular momentum about point \( O \) remains constant unless acted upon by an external torque. Understanding this helps explain the pole's motion as it falls.
Rotational Motion
Rotational motion involves an object spinning around a central point or axis, much like a wheel turning around an axle. The power pole in our exercise exhibits this type of motion as it pivots around point \( O \). When discussing rotational motion, several key elements come into play:
  • Angular Displacement: Describes how far the object rotates, measured in radians or degrees, like \( \theta \).
  • Angular Velocity: It is the rate of change of angular displacement, denoted as \( \omega \).
  • Angular Acceleration: How quickly the angular velocity is changing.
As the pole falls from an upright to a horizontal position, its angular velocity and displacement change. Initially, its angular velocity is zero, but it picks up speed as it falls under gravity's influence until reaching maximum velocity at the horizontal position. Understanding rotational motion helps to explain why the pole swings rather than just dropping straight downward.
Torque
Torque is the measure of force that causes an object to rotate around an axis. In simpler terms, it's what makes things spin. To calculate the torque exerted by the gravitational force on our power pole, we consider:
  • \( \tau = r \times F = rF \sin \theta \)
Where:
  • \( \tau \) is torque.
  • \( r \) is the lever arm distance (half the pole's length since the center of mass is at \( L/2 \)).
  • \( F \) is the force due to gravity, \( mg \).
  • \( \theta \) is the angle between the force vector and the lever arm.
As the angle \( \theta \) changes, so does the effective torque, which in turn alters the pole's rotation speed. This is why understanding torque is crucial for analyzing the dynamics of falling objects with a pivot point, like our fallen power pole.
Pivot Point
The pivot point, in this case, point \( O \), serves as the fixed point around which all rotation occurs. It plays a significant role in dictating the motion of the pole. In physics, the pivot point is crucial because it influences:
  • The distribution of forces.
  • The calculation of torque.
  • The determination of angular momentum.
Since the pole is balanced on point \( O \), this point acts as the axis of rotation, allowing the gravitational force to create torque that causes the pole to fall. Notably, forces exerted at the pivot can be decomposed into horizontal (\( O_x \)) and vertical (\( O_y \)) components.

The behavior of these force components during the fall is a direct result of their interaction with the pivot point, and why understanding this point is an imperative part of solving dynamics problems involving rotation.

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Most popular questions from this chapter

The uniform cylinder is rolling without slip with a velocity \(v\) along the horizontal surface when it overtakes a ramp traveling with speed \(v_{0} .\) Determine an expression for the speed \(v^{\prime}\) which the cylinder has relative to the ramp immediately after it rolls up onto the ramp. Finally, determine the percentage \(n\) of cylinder kinetic energy lost if \((a) \theta=10^{\circ}\) and \(v_{0}=0.25 v\) and (b) \(\theta=10^{\circ}\) and \(v_{0}=\) \(0.5 v .\) Assume that the clearance between the ramp and the ground is essentially zero, that the mass of the ramp is very large, and that the cylinder does not slip on the ramp.

A small experimental vehicle has a total mass \(m\) of \(500 \mathrm{kg}\) including wheels and driver. Each of the four wheels has a mass of \(40 \mathrm{kg}\) and a centroidal radius of gyration of \(400 \mathrm{mm} .\) Total frictional resistance \(R\) to motion is \(400 \mathrm{N}\) and is measured by towing the vehicle at a constant speed on a level road with engine disengaged. Determine the power output of the engine for a speed of \(72 \mathrm{km} / \mathrm{h}\) up the 10-percent grade (a) with zero acceleration and (b) with an acceleration of \(3 \mathrm{m} / \mathrm{s}^{2}\). (Hint: Power equals the time rate of increase of the total energy of the vehicle plus the rate at which frictional work is overcome.)

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