/*! 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 42 The 30 -in. slender bar weighs 2... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The 30 -in. slender bar weighs 20 lb and is mounted on a vertical shaft at \(O .\) If a torque \(M=100 \mathrm{lb}\) -in. is applied to the bar through its shaft, calculate the horizontal force \(R\) on the bearing as the bar starts to rotate.

Short Answer

Expert verified
The horizontal force R on the bearing is approximately 10.04 lb.

Step by step solution

01

Define the Problem

We have a slender bar weighing 20 lb that is 30 inches long. A torque of 100 lb-in is applied at the shaft located at point O. We need to find the horizontal force R on the bearing as the bar starts to rotate.
02

Determine the Moment of Inertia

The moment of inertia ( \(I\) ) for a slender bar of length \(L\) and mass \(m\) about one end is given by \[I = \frac{1}{3}mL^2\] where \(m = \frac{20}{32.2} = 0.62 \text{ slugs}\) (using \(g = 32.2 \text{ ft/s}^2\)). The length \(L = 2.5 \text{ ft} = 30 \text{ in.}/12 \text{ in/ft}\). Thus, \[I = \frac{1}{3} \times 0.62 \times (2.5)^2 = 1.29 \text{ slug-ft}^2\].
03

Calculate Angular Acceleration

The angular acceleration \(\alpha\) can be found from the torque equation \(M = I \alpha\) . Solving for \(\alpha\): \[\alpha = \frac{M}{I} = \frac{100/12}{1.29} = 6.48 \text{ rad/s}^2\]
04

Determine the Resultant Force on the Bar

The horizontal force \(R\) at the bearing is due to the inertia of the bar. Using \(F = ma\), where \(a = \alpha \cdot L\) , we find: \[a = 6.48 \times 2.5 = 16.2 \text{ ft/s}^2\]. Now, \[R = m \cdot a = 0.62 \times 16.2 = 10.04 \text{ lb}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Torque
Torque is a measure of how force applied at a distance causes an object to rotate. Imagine pushing a door open. The force applied at the edge of the door creates a torque around the hinges, causing it to swing open. The formula for torque is given as \( \tau = r \times F \), where \( r \) is the distance from the pivot point to where the force is applied, and \( F \) is the force. This product results in units such as pound-inches or Newton-meters.
This exercise involves applying a torque of 100 lb-in to a slender bar mounted on a shaft. The problem demonstrates how torque initiates rotational movement, which is fundamentally different from how linear forces move objects. Torque's role is crucial in systems like engines, gears, and levers, making it an important aspect of dynamics.
The concept of torque helps us understand not only that a force is applied but also where it is applied and its ability to cause rotation. It is essential to calculate torque accurately to understand and predict the mechanical behavior in applications ranging from simple door mechanisms to complex machinery.
Moment of Inertia
Moment of Inertia, often symbolized by \( I \), is an object's resistance to changes in its rotation. This is akin to mass in linear movement—the greater the moment of inertia, the more torque is needed to change its rotational speed. The formula for the moment of inertia for a slender bar about one end is \( I = \frac{1}{3}mL^2 \), where \( m \) is mass, and \( L \) is length.
In our exercise's example, the bar's calculated moment of inertia is 1.29 slug-ft\(^2\). This value illustrates how the bar's mass distribution affects its rotation about the shaft. A larger moment of inertia would resist change more effectively, necessitating greater torque for the same angular acceleration.
Understanding the moment of inertia is vital when designing rotating systems. It impacts everything from the efficiency of flywheels in engines to the stability of spinning satellite components. By knowing an object's moment of inertia, engineers can optimize torque requirements and energy consumption.
Angular Acceleration
Angular acceleration, denoted by \( \alpha \), describes how quickly an object's rotation is speeding up or slowing down. It is tied to the torque applied and the object's moment of inertia, as described by the equation \( M = I\alpha \). This relationship means angular acceleration is directly proportional to the torque and inversely proportional to the moment of inertia.
In the problem, the bar experienced an angular acceleration of 6.48 rad/s\(^2\), calculated using the torque and its moment of inertia. This acceleration tells us how rapidly the bar is beginning to rotate once the torque is applied.
Angular acceleration is a fundamental concept in dynamics, providing insights into how rotative systems respond to applied forces. It is essential in understanding everything from the startup of electric motors to how wheels accelerate in vehicles. By calculating angular acceleration, engineers can predict motion patterns and control rotational systems effectively for safe and efficient operation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The 165 -lb ice skater with arms extended horizontally spins about a vertical axis with a rotational speed of 1 rev/sec. Estimate his rotational speed \(N\) if he fully retracts his arms, bringing his hands very close to the centerline of his body. As a reasonable approximation, model the extended arms as uniform slender rods, each of which is 27 in. long and weighs 15 lb. Model the torso as a solid 135-lb cylinder 13 in. in diameter. Treat the man with arms retracted as a solid 165 -lb cylinder of 13-in. diameter. Neglect friction at the skate-ice interface.

The uniform slender bar \(A B\) has a mass of \(8 \mathrm{kg}\) and swings in a vertical plane about the pivot at \(A\). If \(\dot{\theta}=2 \mathrm{rad} / \mathrm{s}\) when \(\theta=30^{\circ},\) compute the force supported by the pin at \(A\) at that instant.

The 50 -kg flywheel has a radius of gyration \(\bar{k}=0.4 \mathrm{m}\) about its shaft axis and is subjected to the torque \(M=2\left(1-e^{-0.1 \theta}\right) \mathrm{N} \cdot \mathrm{m},\) where \(\theta\) is in radians. If the flywheel is at rest when \(\theta=0,\) determine its angular velocity after 5 revolutions.

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 75 -kg flywheel has a radius of gyration about its shaft axis of \(\bar{k}=0.50 \mathrm{m}\) and is subjected to the torque \(M=10\left(1-e^{-t}\right) \mathrm{N} \cdot \mathrm{m},\) where \(t\) is in seconds. If the flywheel is at rest at time \(t=0\) determine its angular velocity \(\omega\) at \(t=3\) s.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.