/*! 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

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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, \[R = m \cdot a = 0.62 \times 16.2 = 10.04 \text{ lb}\]

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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.

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Most popular questions from this chapter

A uniform slender bar of mass \(M\) and length \(L\) is translating on the smooth horizontal \(x\) -y plane with a velocity \(v_{M}\) when a particle of mass \(m\) traveling with a velocity \(v_{m}\) as shown strikes and becomes embedded in the bar. Determine the final linear and angular velocities of the bar with its embedded particle.

The experimental Formula One race car is traveling at \(300 \mathrm{km} / \mathrm{h}\) when the driver begins braking to investigate the behavior of the extreme-grip tires. An accelerometer in the car records a maximum deceleration of \(4 g\) when both the front and rear tires are on the verge of slipping. The car and driver have a combined mass of \(690 \mathrm{kg}\) with mass center \(G .\) The horizontal drag acting on the car at this speed is \(4 \mathrm{kN}\) and may be assumed to pass through the mass center \(G .\) The downforce acting over the body of the car at this speed is \(13 \mathrm{kN}\). For simplicity, assume that \(35 \%\) of this force acts directly over the front wheels, \(40 \%\) acts directly over the rear wheels, and the remaining portion acts at the mass center. What is the necessary coefficient of friction \(\mu\) between the tires and the road for this condition? Compare your results with those for passenger car tires.

A person who walks through the revolving door exerts a 90 -N horizontal force on one of the four door panels and keeps the \(15^{\circ}\) angle constant relative to a line which is normal to the panel. If each panel is modeled by a 60 -kg uniform rectangular plate which is \(1.2 \mathrm{m}\) in length as viewed from above, determine the final angular velocity \(\omega\) of the door if the person exerts the force for 3 seconds. The door is initially at rest and friction may be neglected.

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 planetary gear system is shown, where the gear teeth are omitted from the figure. Each of the three identical planet gears \(A, B,\) and \(C\) has a mass of \(0.8 \mathrm{kg},\) a radius \(r=50 \mathrm{mm},\) and a radius of gyration of \(30 \mathrm{mm}\) about its center. The spider \(E\) has a mass of \(1.2 \mathrm{kg}\) and a radius of gyration about \(O\) of \(60 \mathrm{mm} .\) The ring gear \(D\) has a radius \(R=150 \mathrm{mm}\) and is fixed. If a torque \(M=5 \mathrm{N} \cdot \mathrm{m}\) is applied to the shaft of the spider at \(O,\) determine the initial angular acceleration \(\alpha\) of the spider.

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