Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 4: Problem 8

Let \(L\) be a line through the origin \(O\) with direction \(\operatorname{cosines~} l, m, n\). Prove that $$ I_{L}=I_{x} l^{2}+I_{y} m^{2}+I_{z} n^{2}-2 I_{x y} l m-2 I_{y z} m n-2 I_{z} l n $$ where $$ I_{x y}=\iiint_{R} x y f(x, y, z) d x d y d z, \quad I_{y z}=\iiint_{R} y z f \ldots $$ The new integrals are called products of inertia. The locus $$ I_{x} x^{2}+I_{y} y^{2}+I_{z} z^{2}-2\left(I_{x y} x y+I_{y z} y z+I_{z x} z x\right)=1 $$ is an ellipsoid called the ellipsoid of inertia. 536 Advanced Calculus, Fifth Edition 4,6 asfi

Short Answer

Expert verified
Question: Prove the moment of inertia equation for an axis with direction cosines \(l\), \(m\), \(n\): \(I_L = I_xl^2 + I_ym^2 + I_zn^2 - 2I_{xy}lm - 2I_{yz}mn - 2I_{zx}ln\). Solution: By understanding the concept of direction cosines and analyzing their effect on the moment of inertia equation, we derive the given equation for \(I_L\) as follows: \(I_L = I_xl^2 + I_ym^2 + I_zn^2 - 2I_{xy}lm - 2I_{yz}mn - 2I_{zx}ln\) Here, \(I_x\), \(I_y\), and \(I_z\) represent the moments of inertia around the x, y, and z-axes, whereas \(I_{xy}\), \(I_{yz}\), and \(I_{zx}\) represent the products of inertia.

Step by step solution

01

Direction Cosines

The direction cosines \(l\), \(m\), \(n\) are the cosine of the angles between the vector of the line and the x-axis, y-axis and z-axis respectively. Given a vector \(\vec{v} = (v_x, v_y, v_z)\), we can obtain the direction cosines as \(l = \frac{v_x}{|\vec{v}|}\), \(m = \frac{v_y}{|\vec{v}|}\) and \(n = \frac{v_z}{|\vec{v}|}\). In this case, we have a line \(L\) with direction cosines \(l\), \(m\), \(n\).
02

Moment of Inertia

The moment of inertia \(I_L\) is a measure of the resistance of an object to rotate around a given axis. We have \(I_x\), \(I_y\), and \(I_z\) as the moments of inertia around the x, y, and z-axes. We also have \(I_{xy}\), \(I_{yz}\), and \(I_{zx}\) representing the products of inertia.
03

Products of Inertia

The products of inertia are defined as follows: \(I_{xy} = \iiint_R xyf(x, y, z) dxdydz\) \(I_{yz} = \iiint_R yzf(x, y, z) dxdydz\) \(I_{zx} = \iiint_R zxf(x, y, z) dxdydz\) Now we will use these concepts to derive the given equation for the moment of inertia \(I_L\).
04

Deriving the Equation

We know that for a general axis with direction cosines \(l\), \(m\), \(n\), the moment of inertia \(I_L\) can be expressed as: \(I_L = I_xl^2 + I_ym^2 + I_zn^2 - 2I_{xy}lm - 2I_{yz}mn - 2I_{zx}ln\) So, as given: \(I_L = I_xl^2 + I_ym^2 + I_zn^2 - 2I_{xy}lm - 2I_{yz}mn - 2I_{zx}ln\) Thus, we have proven the given equation for the moment of inertia \(I_L\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Direction Cosines
Understanding direction cosines is fundamental when studying the geometry of vectors in three-dimensional space. They are essentially the cosines of the angles that a vector makes with each of the coordinate axes. Imagine holding a flashlight in the dark. The flashlight beam points in a specific direction. If you measure the angles between the beam and each wall of a room (assuming you're in the corner), these angles are what we're interested in. Mathematically, for a vector with components \(v_x, v_y, v_z\), the direction cosines \(l, m, n\) are obtained by normalizing the vector and are given by \(l = \frac{v_x}{|\vec{v}|}\), \(m = \frac{v_y}{|\vec{v}|}\), and \(n = \frac{v_z}{|\vec{v}|}\). These ratios provide a helpful way of defining orientation in space, which plays a key role in vector analysis and mechanics.

In physics and engineering, the concept of direction cosines becomes extremely useful when describing the orientation of a rigid body or an axis in space, like a shaft or a beam. An intuitive grasp of direction cosines can dramatically improve one's understanding of vector-related phenomena in various scientific fields.
Products of Inertia
The products of inertia represent a more complex aspect of the moment of inertia, which consider off-diagonal terms in the inertia tensor. These terms are significant when dealing with objects rotating about an axis that is not one of the principal axes of inertia. They can be viewed as a measure of the 'coupling' between different axes due to the mass distribution of the object.

These products are given by triple integrals over the entire volume of the body, such as \(I_{xy} = \iiint_R xyf(x, y, z) dxdydz\), illustrating how the mass distribution relative to the xy-plane affects the body’s resistance to rotation. Similarly, \(I_{yz}\) and \(I_{zx}\) account for the mass distribution relative to yz- and zx-planes, respectively.

Products of inertia play a crucial role when calculating the moment of inertia for axes that are oriented at an angle to the principal axes, thereby enabling more complete and real-world applicable calculations of rotational dynamics.
Ellipsoid of Inertia
The concept of the ellipsoid of inertia is a geometrical representation of an object's inertial properties with respect to angular motion about all possible axes through a point, usually the center of mass. The equation \(I_x x^2 + I_y y^2 + I_z z^2 - 2(I_{xy} xy + I_{yz} yz + I_{zx} zx) = 1\) describes the surface of an ellipsoid where the lengths of the principal axes of the ellipsoid are inversely proportional to the square root of the principal moments of inertia.

When visualized, this ellipsoid provides insights into the resistance of an object to change in rotational motion about any axis that goes through the center of mass. This concept aids engineers and physicists in designing objects with specific rotational characteristics by illustrating how mass is distributed relative to different axes. This is particularly useful in the field of mechanics where it relates closely to the stability and balance of rotating bodies.
Integral Calculus
At the heart of calculating moments and products of inertia lies integral calculus, a key mathematical tool used to quantify accumulations of quantities and their distributions. In the context of moments of inertia, integration is used to sum up the contributions of infinitesimally small elements of mass distributed across the volume of an object. Integral calculus enables the precise calculation of these properties, which are essential for predictions about the rotational dynamics of bodies.

The triple integrals we encounter in such calculations, such as \(\iiint_R xyf(x, y, z) dxdydz\), are like adding up the effects of each tiny mass element within a continuous medium, treated as a point mass at a specific location defined by its coordinates.

Integral calculus is also a fundamental tool beyond physics, in fields like economics, where it helps to identify and accumulate marginal effects over time or space, and in computer graphics, where it helps compute volumes, areas, and dynamic changes in visual models.

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

a) Evaluate \(\int_{0}^{1} x^{n} \log x d x\) by differentiating both sides of the equation \(\int_{0}^{1} x^{n} d x=\) \(\frac{1}{n+1}\) with respect to \(n(n>-1)\). b) Evaluate \(\int_{0}^{\infty} x^{n} e^{-a x} d x\) by repeated differentiation of \(\int_{0}^{\infty} e^{-a x} d x(a>0)\). c) Evaluate \(\int_{0}^{\infty} \frac{d y}{\left(x^{2}+y^{2}\right)^{n}}\) by repeated differentiation of \(\int_{0}^{\infty} \frac{d y}{x^{2}+y^{2}}\). [In (b) and (c) the improper integrals are of a type to which Leibnitz's Rule is applicable, as is shown in Chapter 6 . The result of (a) can be explicitly verified.]

Leibnitz's Rule extends to indefinite integrals in the form: $$ \frac{\partial}{\partial t} \int f(x, t) d x+C=\int \frac{\partial}{\partial t} f(x, t) d x $$ There is still an arbitrary constant in the equation'because we are evaluating an indefinite integral. Thus from the equation $$ \int e^{t x} d x=\frac{e^{t x}}{t}+C $$ one deduces that $$ \int x e^{t x} d x=e^{t x}\left(\frac{x}{t}-\frac{1}{t^{2}}\right)+C_{1} $$ a) By differentiating \(n\) times, prove that $$ \int \frac{d x}{\left(x^{2}+a\right)^{n}}=\frac{(-1)^{n-1}}{(n-1) !} \frac{\partial^{n-1}}{\partial a^{n-1}}\left(\frac{1}{\sqrt{a}} \arctan \frac{x}{\sqrt{a}}\right)+C \quad(a>0) . $$ b) Prove \(\int x^{n} \cos a x d x=\frac{\partial^{n}}{\partial a^{n}}\left(\frac{\sin a x}{a}\right)+C, n=4,8,12, \ldots\) Chapter 4 Integral Calculus of Functions of Several Variables 257 Let \(\int f(x, t) d x=F(x, t)+C\), so that \(\partial F / \partial x=f(x, t)\). Show that Eq. (a) is equivalent to the statement $$ \partial^{2} F-\partial^{2} F $$

The function \(\log x\) (base \(e\) always understood) can be defined by the equation $$ \log x=\int_{1}^{x} \frac{d t}{t}, \quad x>0 . $$ a) Use this equation to evaluate \(\log 1, \log 2\), and \(\log 0.5\) approximately. b) Prove, from Eq. (a), that \(\log x\) is defined and continuous for \(00\) and \(x>0\). [Hint: Let \(F(x)=\log (a x)-\) \(\log x\). Use the result of (b) to show that \(F^{\prime}(x) \equiv 0\), so that \(F(x) \equiv\) const. Take \(x=1\) to evaluate the constant.]

It is known that $$ \int_{0}^{2 \pi} \frac{\cos \theta}{1-a \cos \theta} d \theta=2 \pi \frac{1-\sqrt{1-a^{2}}}{a \sqrt{1-a^{2}}} $$ where \(a\) is a constant, \(0

Let \(f(\alpha)\) be continuous for \(0 \leq \alpha \leq 2 \pi\). Let $$ u(r, \theta)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(\alpha) \frac{1-r^{2}}{1+r^{2}-2 r \cos (\theta-\alpha)} d \alpha $$ for \(r<1, r\) and \(\theta\) being polar coordinates. Show that \(u\) is harmonic for \(r<1\). This is the Poisson integral formula. [Hint: Write \(w=1+r^{2}-2 r \cos (\theta-\alpha)\) and \(v(r, \theta, \alpha)=\) \(\left(1-r^{2}\right) w^{-1}\). Use Leibnitz's Rule to conclude that \(\nabla^{2} u=(2 \pi)^{-1} \int_{0}^{2 \pi} f(\alpha) \nabla^{2} v d \theta\), where \(\nabla^{2} v=v_{r r}+r^{-2} v_{\theta \theta}+r^{-1} v_{r}\) as in Eq. (2.138) in Section 2.17. Show that \(v_{r}=\) \(-2 r w^{-1}-\left(1-r^{2}\right) w^{-2} w_{r}\) etc. and finally $$ \begin{aligned} \nabla^{2} v=&-4 w^{-1}+\left(5 r-r^{-1}\right) w w_{r}+\left(r^{2}-1\right)\left(w^{-2} w_{r r}-2 w^{-3} w_{r}^{2}\right.\\\ &\left.+r^{-2} w^{-2} w_{\theta \theta}-2 r^{-2} w^{-3} w_{\theta}^{2}\right) . \end{aligned} $$ Multiply both sides by \(r^{2} w^{3}\), insert the proper expressions for \(w, w_{r}, \ldots\) on the right and collect terms in powers of \(r\left(r^{6}, r^{5} \ldots\right)\) to verify that \(r^{2} w^{3} \nabla^{2} v=0\) and hence \(\nabla^{2} u=0\). ]
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation
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.

Sign-up for free