/*! 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 10 (a) A thin uniform rod of mass \... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 10

(a) A thin uniform rod of mass \(M\) and length \(L\) lies on the \(x\) axis with one end at the origin. Find its moment of inertia for rotation about the \(z\) axis. [Here the sum of (10.25) must be replaced by an integral of the form \(\int x^{2} \mu d x\) where \(\mu\) is the linear mass density, mass/length.] (b) What if the rod's center is at the origin?

Short Answer

Expert verified
(a) \( \frac{ML^2}{3} \); (b) \( \frac{ML^2}{12} \).

Step by step solution

01

Understand the Problem

We need to calculate the moment of inertia of a thin rod about the z-axis in two scenarios. First, with one end at the origin, and second, with the rod's center at the origin.
02

Set Up the Integral for Case (a)

For a rod extending from 0 to L, the linear mass density is given by \( \mu = \frac{M}{L} \). The element's mass, \( dm \), is \( \mu \cdot dx \). We use the formula for moment of inertia: \[ I = \int x^2 dm = \int x^2 \mu \, dx \] Substitute \( \mu \) to get \[ I = \frac{M}{L} \int_{0}^{L} x^2 \, dx \].
03

Solve the Integral for Case (a)

Evaluate the integral:\[ I = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{0}^{L} = \frac{M}{L} \cdot \frac{L^3}{3} = \frac{ML^2}{3} \].Thus, the moment of inertia of the rod with one end at the origin is \( \frac{ML^2}{3} \).
04

Set Up for Case (b)

Now, consider the rod's center at the origin. The rod extends from \(-\frac{L}{2}\) to \(\frac{L}{2}\). The integral becomes:\[ I = \frac{M}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 \, dx \].
05

Solve the Integral for Case (b)

Evaluate the integral:\[ I = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{M}{L} \left[ \frac{(\frac{L}{2})^3}{3} - \frac{(-\frac{L}{2})^3}{3} \right] = \frac{M}{L} \cdot \frac{L^3}{12} \cdot 2 = \frac{ML^2}{12} \cdot 2 = \frac{ML^2}{6} \].Thus, the moment of inertia of the rod with its center at the origin is \( \frac{ML^2}{12} \).

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Key Concepts

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

Linear Mass Density
Linear mass density is a way to describe how mass is distributed along a rod. Imagine stretching a rod along the x-axis, where its mass seems evenly spread across its length. The concept of linear mass density, \( \mu \), will help us understand how to calculate properties like moment of inertia for such objects.

A uniform rod鈥攐ne with the same thickness and material throughout鈥攈as a constant linear mass density. It is defined by the formula:
  • \( \mu = \frac{M}{L} \)
where \( M \) is the total mass of the rod and \( L \) is its length.

Knowing \( \mu \) allows us to find the mass of any small segment of the rod. For a tiny piece of length \( dx \), located at position \( x \), the mass \( dm \) is given by:

\[ dm = \mu \cdot dx \]

This calculation is pivotal in setting up the integral used in finding the moment of inertia, as each little piece \( dm \) contributes a small amount to the total inertia.
Integral Calculus
Integral calculus is a mathematical tool that helps us sum an infinite number of tiny parts to find a whole. When measuring the moment of inertia, we integrate the squared distance of each tiny mass segment from the axis of rotation, reflected in its use within our solution.

The formula we use is:

\[ I = \int x^2 dm \]

This integral takes every small part of the rod, represented by \( dm \), and multiplies it by the square of its distance \( x^2 \) from the rotation axis. This operation tells us how much each part "contributes" to the rotational inertia.

To solve for \( I \) through calculus, we transform it using the linear mass density, converting \( dm \) into \( \mu \, dx \). Our integral setup for case (a) involves multiplying the density \( \mu \) with \( x^2 \) and integrating over the rod's length, from 0 to \( L \). The complete form integrates all small sections of the rod, combining them to give a final inertia value.
Rotation About an Axis
When calculating the moment of inertia, it's crucial to understand the axis of rotation. In physics, the axis of rotation is an imaginary line around which an object turns. For this problem, we consider the rod turning about the z-axis.

A change in the rod's position relative to this axis changes how we calculate the moment of inertia. If the rod is anchored at one end, as in case (a), our integration bounds span from 0 to \( L \). However, when the center of the rod is placed over the origin, as in case (b), the rotation affects parts of the rod equally left and right from this axis.

In this situation, our new limits of integration become \(-\frac{L}{2}\) to \(\frac{L}{2}\). This change correctly accounts for the symmetrical distance contributing to the inertia. The moment of inertia values demonstrate how the distribution and position relative to the axis significantly impact the rotational dynamics.

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Most popular questions from this chapter

The moment of inertia of a continuous mass distribution with density \(\varrho\) is obtained by converting the sum of (10.25) into the volume integral \(\int \rho^{2} d m=\int \rho^{2} \varrho d V\). (Note the two forms of the Greek "rho": \(\rho=\) distance from \(z\) axis, \(\varrho=\) mass density.) Find the moment of inertia of a uniform circular cylinder of radius \(R\) and mass \(M\) for rotation about its axis. Explain why the products of inertia are zero.

Consider a rigid plane body or "lamina," such as a flat piece of sheet metal, rotating about a point \(O\) in the body. If we choose axes so that the lamina lies in the \(x y\) plane, which elements of the inertia tensor \(\mathbf{I}\) are automatically zero? Prove that \(I_{z z}=I_{x x}+I_{y y}\).

A rigid body consists of three equal masses \((m)\) fastened at the positions \((a, 0,0),(0, a, 2 a)\) and \((0,2 a, a) .\) (a) Find the inertia tensor \(\mathbf{I}\). (b) Find the principal moments and a set of orthogonal principal axes.

A thin rod (of width zero, but not necessarily uniform) is pivoted freely at one end about the horizontal \(z\) axis, being free to swing in the \(x y\) plane ( \(x\) horizontal, \(y\) vertically down). Its mass is \(m,\) its \(\mathrm{CM}\) is a distance \(a\) from the pivot, and its moment of inertia (about the \(z\) axis) is \(I\). (a) Write down the equation of motion \(\dot{L}_{z}=\Gamma_{z}\) and, assuming the motion is confined to small angles (measured from the downward vertical), find the period of this compound pendulum. ("Compound pendulum" is traditionally used to mean any pendulum whose mass is distributed \(-\) as contrasted with a "simple pendulum," whose mass is concentrated at a single point on a massless arm.) (b) What is the length of the "equivalent" simple pendulum, that is, the simple pendulum with the same period?

Here is a good exercise in vector identities and matrices, leading to some important general results: (a) For a rigid body made up of particles of mass \(m_{\alpha},\) spinning about an axis through the origin with angular velocity \(\omega,\) prove that its total kinetic energy can be written as $$ T=\frac{1}{2} \sum m_{\alpha}\left[\left(\omega r_{\alpha}\right)^{2}-\left(\omega \cdot \mathbf{r}_{\alpha}\right)^{2}\right] $$ Remember that \(\mathbf{v}_{\alpha}=\boldsymbol{\omega} \times \mathbf{r}_{\alpha} .\) You may find the following vector identity useful: For any two vectors a and b, $$ (\mathbf{a} \times \mathbf{b})^{2}=a^{2} b^{2}-(\mathbf{a} \cdot \mathbf{b})^{2} $$ (If you use the identity, please prove it.) (b) Prove that the angular momentum \(\mathbf{L}\) of the body can be written as $$ \mathbf{L}=\sum m_{\alpha}\left[\omega r_{\alpha}^{2}-\mathbf{r}_{\alpha}\left(\omega \cdot \mathbf{r}_{\alpha}\right)\right] $$ For this you will need the so-called \(B A C-C A B\) rule, that \(\mathbf{A} \times(\mathbf{B} \times \mathbf{C})=\mathbf{B}(\mathbf{A} \cdot \mathbf{C})-\mathbf{C}(\mathbf{A} \cdot \mathbf{B})\) (c) Combine the results of parts (a) and (b) to prove that $$ T=\frac{1}{2} \omega \cdot \mathbf{L}=\frac{1}{2} \tilde{\omega} \mathbf{I} \omega $$ Prove both equalities. The last expression is a matrix product; \(\omega\) denotes the \(3 \times 1\) column of numbers \(\omega_{x}, \omega_{y}, \omega_{z},\) the tilde on \(\tilde{\omega}\) denotes the matrix transpose (in this case a row), and I is the moment of inertia tensor. This result is actually quite important; it corresponds to the much more obvious result that for a particle, \(T=\frac{1}{2} \mathbf{v} \cdot \mathbf{p} .\) (d) Show that with respect to the principal axes, \(T=\frac{1}{2}\left(\lambda_{1} \omega_{1}^{2}+\lambda_{2} \omega_{2}^{2}+\lambda_{3} \omega_{3}^{2}\right)\) as in Equation (10.68).
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