/*! 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 92 The uniform slender rod of lengt... [FREE SOLUTION] | 91Ó°ÊÓ

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The uniform slender rod of length \(L,\) mass \(m,\) and cross-sectional area \(A\) is rotating in a horizontal plane about the vertical central axis \(0-0\) at a constant high angular velocity \(\omega .\) By analyzing the horizontal forces on the accelerating differential element shown, derive an expression for the tensile stress \(\sigma\) in the rod as a function of \(r .\) The stress, commonly referred to as centrifugal stress, equals the tensile force divided by the cross-sectional area \(A\).

Short Answer

Expert verified
The tensile stress \( \sigma \) in the rod is \( \sigma = \frac{m \cdot \omega^2 \cdot r^2}{2LA} \).

Step by step solution

01

Establish Differential Element

Consider a small differential element of length \( dr \) at a distance \( r \) from the axis of rotation \( O-O \). This element is part of the rod experiencing angular motion.
02

Calculate Centrifugal Force

The centrifugal force acting on the differential element due to its rotation is \( dF = dm \cdot \omega^2 \cdot r \), where \( dm \) is the differential mass of the element. Mass per unit length is \( \frac{m}{L} \), hence \( dm = \frac{m}{L} \cdot dr \). So, \( dF = \frac{m}{L} \cdot \omega^2 \cdot r \cdot dr \).
03

Set Up Force Balance

For equilibrium of the differential element, the internal tensile force \( T \) at \( r \) and \( T + dT \) at \( r + dr \) balance the centrifugal force. Therefore, \( dT = dF = \frac{m}{L} \cdot \omega^2 \cdot r \cdot dr \).
04

Integrate Force Equation

Integrate the force balance equation \( dT = \frac{m}{L} \cdot \omega^2 \cdot r \cdot dr \) with respect to \( r \) from \( 0 \) to \( r \): \( T = \int_0^r \frac{m}{L} \cdot \omega^2 \cdot r \, dr = \frac{m}{L} \cdot \omega^2 \cdot \frac{r^2}{2} \).
05

Calculate Tensile Stress \( \sigma \)

The tensile stress \( \sigma \) is the tensile force \( T \) divided by the cross-sectional area \( A \). Substituting for \( T \) results in \( \sigma = \frac{\frac{m}{L} \cdot \omega^2 \cdot \frac{r^2}{2}}{A} = \frac{m \cdot \omega^2 \cdot r^2}{2LA} \).
06

Final Expression for Tensile Stress

The tensile stress in the rod, which arises from the centrifugal effect, is given by \( \sigma = \frac{m \cdot \omega^2 \cdot r^2}{2LA} \).

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Key Concepts

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

Centrifugal Force
Centrifugal force is an apparent force experienced by an object moving in a circular path, directed away from the center of rotation. This force arises due to the object's inertia, which makes the object appear to be pushed outward. When considering a rotating rod, such as in our exercise, each element of the rod feels a centrifugal force due to its circular motion.
The centrifugal force acting on a differential element of the rod is dependent on the distance from the axis of rotation and the angular velocity. It is calculated using the formula \( dF = dm \cdot \omega^2 \cdot r \), where \( dm \) is the mass of the element and \( r \) is the distance from the rotation axis. This force effectively causes the rod to experience tension or stretch.
Differential Element
A differential element is a small segment of a larger object, used in calculus to help break down a problem into manageable pieces. In the context of the physical rod, we consider a small length segment of the rod as a differential element. The beauty of this approach is that it allows for the application of fundamental physics laws to complex, variable systems like a continuously rotating rod.
In this exercise, the differential element is described as a small part of length \( dr \) taken at a distance \( r \) from the axis of rotation. By analyzing the forces on this small element, we can derive expressions that describe the behavior of the entire system through integration.
Angular Velocity
Angular velocity is a measure of the rate of rotation of an object, defining how fast the object spins around a central point. It is usually denoted by the symbol \( \omega \), and has units of radians per second.
In the case of our exercise, the angular velocity is constant, meaning the rod rotates at a steady rate. This steady rotation is crucial in ensuring that the centrifugal force experienced by each part of the rod remains uniform, simplifying the analysis and allowing for accurate stress calculations.
Force Balance
The concept of force balance involves ensuring that all the forces acting on an object are in equilibrium, meaning that the sum of all forces results in zero acceleration. This concept is pivotal in analyzing the differential element in this problem.
For the differential element in the rotating rod, we consider two forces: the tensile force across the element and the centrifugal force experienced due to rotation. The tensile force at one end of the element is \( T \), and at the other end is \( T + dT \). The force balance ensures that the net force on the element results in \( dT = dF = \frac{m}{L} \cdot \omega^2 \cdot r \cdot dr \), facilitating integration to find the internal forces throughout the rod.
Stress Analysis
Stress analysis is a method used to determine the stresses present in an object due to external forces. In this scenario, we focus on tensile stress, which arises from the centrifugal force in the rotating rod. Tensile stress is defined as the tensile force per unit area, symbolized by \( \sigma \).
The tensile stress in a rotating rod can be expressed as \( \sigma = \frac{m \cdot \omega^2 \cdot r^2}{2LA} \). This result is derived by first calculating the tensile force using integration and then dividing by the cross-sectional area of the rod \( A \). Stress analysis is crucial for understanding material behavior under specific conditions and ensuring safe design applications.

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Most popular questions from this chapter

Each tire on the 1350 -kg car can support a maximum friction force parallel to the road surface of 2500 N. This force limit is nearly constant over all possible rectilinear and curvilinear car motions and is attainable only if the car does not skid. Under this maximum braking, determine the total stopping distance \(s\) if the brakes are first applied at point \(A\) when the car speed is \(25 \mathrm{m} / \mathrm{s}\) and if the car follows the centerline of the road.

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