/*! 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 137 The boom \(O A B\) pivots about ... [FREE SOLUTION] | 91Ó°ÊÓ

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The boom \(O A B\) pivots about point \(O,\) while section \(A B\) simultaneously extends from within section OA. Determine the velocity and acceleration of the center \(B\) of the pulley for the following conditions: \(\theta=20^{\circ}, \dot{\theta}=5\) deg/sec, \(\ddot{\theta}=2\) deg/sec \(^{2}, l=7 \mathrm{ft}\) \(i=1.5 \mathrm{ft} / \mathrm{sec}, \ddot{l}=-4 \mathrm{ft} / \mathrm{sec}^{2} .\) The quantities \(l\) and \(l\) are the first and second time derivatives, respectively, of the length \(l\) of section \(A B\).

Short Answer

Expert verified
Velocity of B: 2.20 ft/sec; acceleration of B: 3.81 ft/sec².

Step by step solution

01

Convert Angular and Linear Velocities to Radians

The given angular velocities and accelerations are in degrees per second. First, convert them to radians per second. \( \dot{\theta} = 5 \text{ deg/sec} = \frac{5\pi}{180} \text{ rad/sec} \approx 0.0873 \text{ rad/sec} \)\( \ddot{\theta} = 2 \text{ deg/sec}^2 = \frac{2\pi}{180} \text{ rad/sec}^2 \approx 0.0349 \text{ rad/sec}^2 \)
02

Identify Position of Point B

The position of point B can be described as:\( \mathbf{r}_B = l \mathbf{e}_r = (l+7) \mathbf{e}_r\).Since section AB extends within OA, we account for both the length \(l\) and the segment OA (7 ft).
03

Express Velocity of Point B in Polar Form

The velocity \(v_B\) in polar coordinates is given by:\[ v_B = \left( \dot{l}+7 \dot{\theta} \right)\mathbf{e}_r + (l+7) \dot{\theta} \mathbf{e}_\theta \]Substitute the given values:\( \dot{l} = 1.5 \text{ ft/sec}, \dot{\theta} = 0.0873 \text{ rad/sec} \)\[ v_B = \left(1.5 + 0.6111\right) \mathbf{e}_r + (7+0) \cdot 0.0873 \mathbf{e}_\theta = 2.1111 \, \mathbf{e}_r + 0.6111 \mathbf{e}_\theta \]Thus, the magnitude of the velocity \(v_B\) is \(\sqrt{(2.1111)^2 + (0.6111)^2} \approx 2.2006 \, \text{ft/sec} \).
04

Express Acceleration of Point B in Polar Form

The acceleration \(a_B\) includes both radial and tangential accelerations. It can be calculated as:\[ a_B = (\ddot{l} + 7\ddot{\theta} - 7\dot{\theta}^2 )\mathbf{e}_r + ((l+7)\ddot{\theta} + 2i \dot{\theta})\mathbf{e}_\theta \]Given, \(\ddot{l} = -4 \text{ ft/sec}^2 \) and substituting \(\dot{\theta}, \ddot{\theta}\) from earlier calculations:Radial component:\[ \ddot{l} + 7\ddot{\theta} - 7\dot{\theta}^2 = -4 + 7\cdot0.0349 - 7\cdot(0.0873)^2 \approx -3.7581 \, \text{ft/sec}^2\]Transversal component:\[ (l + 7)\ddot{\theta} + 2i \dot{\theta} = 7\cdot0.0349 + 2\cdot1.5\cdot0.0873 \approx 0.6103 \, \text{ft/sec}^2\]The acceleration vector becomes: \(-3.7581 \, \mathbf{e}_r + 0.6103 \, \mathbf{e}_\theta\). The magnitude is \(\sqrt{(-3.7581)^2 + (0.6103)^2} \approx 3.8073 \, \text{ft/sec}^2 \).

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Key Concepts

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

Polar Coordinates
In kinematics, polar coordinates are used to describe the position of a point in a two-dimensional plane, methodically using a radius and angle. This system is particularly useful when dealing with problems involving rotating bodies or systems with radial symmetry, as it allows us to separate the motion into radial and angular components.

To locate a point using polar coordinates, we use two factors:
  • The radial distance (r), which tells us how far the point is from a fixed origin.
  • The angular position (θ), which measures the angle from a fixed line, often the positive x-axis.
In this exercise, we consider the point B, whose position can be expressed in polar coordinates using the radial distance (abbreviated 'l' for the extension within section OA) and the angle θ about the pivot point O.
Velocity Calculation
Velocity in polar coordinates is a combination of radial and angular components. Let's break it down:

When a point moves, it can experience changes in both its distance from the origin and its angular position. Therefore, the velocity vector can be expressed as:
  • The radial velocity, which is the rate of change of the radius with respect to time. Expressed as \( \dot{l} \), it corresponds to how fast point B is moving in the radial direction.
  • The angular velocity, expressed as \( (l+7) \cdot \dot{\theta} \), which indicates how fast the point is moving around the pivot in the angular direction.
By calculating the magnitudes and combining these components, we find the velocity of the center B in this exercise to be approximately 2.2006 ft/sec.
Angular Motion
Angular motion refers to the change in the orientation of a line segment around a fixed point. When discussing kinematics problems involving rotation, it's essential to understand how angular velocity and angular acceleration contribute.

Here, \( \dot{\theta} = 5 \text{ deg/sec} \) is converted to radians per second as angular velocity. This conversion is necessary because most mathematical models use radians for accuracy and simplicity.

Angular acceleration, \( \ddot{\theta} \), is the rate of change of angular velocity, which in this context is 2 degrees per second squared or approximately 0.0349 rad/sec² after conversion. Knowing these values helps determine how quickly point B changes its angular position as the segment pivots around point O.
Acceleration Analysis
Acceleration involves the rate of change of velocity and can also be split into radial and tangential components in polar coordinates.

For radial acceleration, we consider both the rate at which the length changes, \( \ddot{l} \), and the term \(-7\dot{\theta}^2 \), which accounts for changes due to the movement along the arc itself. Tangential acceleration, on the other hand, focuses on the change in angle and how fast that angle is increasing or decreasing, represented by \((l+7)\ddot{\theta} + 2\dot{l}\dot{\theta}\).

The combined effects of these accelerations provide a comprehensive view of how point B's motion evolves, with a computed total acceleration magnitude of approximately 3.8073 ft/sec². An understanding of these two types of acceleration is crucial for analyzing the dynamics of any rotational system.

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Most popular questions from this chapter

The car \(A\) is ascending a parking-garage ramp in the form of a cylindrical helix of 24 -ft radius rising 10 ft for each half turn. At the position shown the car has a speed of \(15 \mathrm{mi} / \mathrm{hr}\), which is decreasing at the rate of \(2 \mathrm{mi} / \mathrm{hr}\) per second. Determine the \(r, \theta\) and \(z\) -components of the acceleration of the car.

At a football tryout, a player runs a 40 -yard dash in 4.25 seconds. If he reaches his maximum speed at the 16 -yard mark with a constant acceleration and then maintains that speed for the remainder of the run, determine his acceleration over the first 16 yards, his maximum speed, and the time duration of the acceleration.

A baseball is dropped from an altitude \(h=200 \mathrm{ft}\) and is found to be traveling at \(85 \mathrm{ft} / \mathrm{sec}\) when it strikes the ground. In addition to gravitational acceleration, which may be assumed constant, air resistance causes a deceleration component of magnitude \(k v^{2},\) where \(v\) is the speed and \(k\) is a constant. Determine the value of the coefficient \(k .\) Plot the speed of the baseball as a function of altitude \(y .\) If the baseball were dropped from a high altitude, but one at which \(g\) may still be assumed constant, what would be the terminal velocity \(v_{t} ?\) (The terminal velocity is that speed at which the acceleration of gravity and that due to air resistance are equal and opposite, so that the baseball drops at a constant speed.) If the baseball were dropped from \(h=200 \mathrm{ft},\) at what speed \(v^{\prime}\) would it strike the ground if air resistance were neglected?

An electric motor \(M\) is used to reel in cable and hoist a bicycle into the ceiling space of a garage. Pulleys are fastened to the bicycle frame with hooks at locations \(A\) and \(B\), and the motor can reel in cable at a steady rate of 12 in./sec. At this rate, how long will it take to hoist the bicycle 5 feet into the air? Assume that the bicycle remains level.

The base structure of the firetruck ladder rotates about a vertical axis through \(O\) with a constant angular velocity \(\Omega=10\) deg/s. At the same time, the ladder unit \(O B\) elevates at a constant rate \(\dot{\phi}=7 \mathrm{deg} / \mathrm{s},\) and section \(A B\) of the ladder extends from within section \(O A\) at the constant rate of \(0.5 \mathrm{m} / \mathrm{s}\). At the instant under consideration, \(\phi=30^{\circ}, \overline{O A}=9 \mathrm{m},\) and \(\overline{A B}=6 \mathrm{m} .\) Determine the magnitudes of the velocity and acceleration of the end \(B\) of the ladder.

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