Show that the moments of inertia of any body satisfy the triangle inequalities $$ I_{3} \leq I_{2}+I_{1} \quad I_{2} \leq I_{1}+I_{3} \text { and } I_{1} \leq I_{2}+I_{3} \text {. } $$ and that equality holds only for a planar body.

Short Answer

Expert verified
Answer: Yes, the moments of inertia of any three-dimensional object satisfy the triangle inequalities (I₃ ≤ I₂ + I₁, I₂ ≤ I₁ + I₃, and I₁ ≤ I₂ + I₃). Equality holds only for a planar body, in which case I₃ = I₁ + I₂.

Step by step solution

01

Define Moment of Inertia

Moments of inertia are scalar values representing how an object's mass is distributed relative to a particular axis of rotation. The three moments of inertia (I₁, I₂, and I₃) correspond to three orthogonal axes (x, y, and z) passing through the center of mass of the object.
02

Recall Inertia Tensor

The inertia tensor is a 3x3 matrix that relates the angular velocity and angular momentum vectors. For any body, the principal moments of inertia are the eigenvalues of the inertia tensor. The inertia tensor (I) can be written as: $$ I = \begin{bmatrix} I_{1} & -P_{23} & -P_{31} \\ -P_{23} & I_{2} & -P_{12} \\ -P_{31} & -P_{12} & I_{3} \end{bmatrix} $$ where P23, P12, and P31 are the products of inertia, which are equal to zero when the coordinate system is aligned with the principal axes of inertia.
03

Analyze the Determinant

The determinant of the inertia tensor should be non-negative, as it is related to the volume in the inertia space. We can compute the determinant of the inertia tensor: $$ \begin{vmatrix} I_{1} & -P_{23} & -P_{31} \\ -P_{23} & I_{2} & -P_{12} \\ -P_{31} & -P_{12} & I_{3} \end{vmatrix} $$ = I₁I₂I₃ + 2P12P23P31 - I₁P12² - I₂P23² - I₃P31² Since the products of inertia are zero for the principal axes, the determinant becomes: det(I) = I₁I₂I₃ ≥ 0
04

Prove Triangle Inequalities

Based on the non-negativity of the determinant, we can derive the triangle inequalities as follows: det(I) = I₁I₂I₃ ≥ 0 Divide both sides by I₁I₂I₃ (assuming none of them are zero) 1 ≥ (I₂I₃) / I₁ + (I₁I₃) / I₂ + (I₁I₂) / I₃ Rewrite the inequality in terms of moments of inertia: I₃ ≤ I₂ + I₁, I₂ ≤ I₁ + I₃, and I₁ ≤ I₂ + I₃ This proves that the moments of inertia of any three-dimensional object satisfy these triangle inequalities.
05

Show Equality for Planar Body

Suppose the body is a planar body. This means the body lies entirely in one plane, say the xy-plane. In this case, the moment of inertia about the z-axis (I₃) will be maximum, and there will be no contribution to the moment of inertia about the x-axis (I₁) and the y-axis (I₂) from the z-direction. So, I₁ and I₂ will only depend on the distribution of mass within the plane itself. From this, it can be inferred that I₁ = I₂ for symmetric distribution, and I₃ = I₁ + I₂ for a planar body. This demonstrates that equality in the triangle inequalities holds only for a planar body, as required.

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Key Concepts

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

Inertia Tensor
The concept of the inertia tensor is integral to understanding how an object's mass distribution affects its rotational properties. Imagine a rigid body as an assortment of masses scattered in space, each particle contributing to how difficult it is to spin the object around any given axis.

The inertia tensor is a mathematical concept that takes all these contributions and organizes them into a matrix that links the angular velocity of an object to its angular momentum. It is crucial in physics because it offers a comprehensive description of an object's inertia in all directions. The tensor is expressed as a 3x3 matrix, embodying not only the individual moments of inertia along three perpendicular axes but also the product of inertia which accounts for mass distribution in mixed planes.

Understanding and calculating the inertia tensor enable physicists and engineers to predict the rotational motion of bodies more accurately, crucial for applications ranging from mechanical engineering to aerospace dynamics.
Principal Moments of Inertia
The principal moments of inertia are special values that are derived from the inertia tensor. They represent the moments of inertia along three mutually perpendicular axes called principal axes. These axes are unique to the body's geometry and mass distribution and align with the directions in which the mass is most and least spread out.

To find the principal moments of inertia, one typically computes the eigenvalues of the inertia tensor. These eigenvalues, denoted as I₁, I₂, and I₃, have a profound physical meaning: they are the moments of inertia for rotations around the principal axes, where the products of inertia are zero. These values are fundamental since they provide the simplest picture of the object's rotational dynamics, free from the complications of off-diagonal terms (products of inertia) in the tensor matrix.

The step-by-step solution provided describes how these principal moments are not arbitrary but satisfy an important set of inequalities, akin to the triangle inequalities found in geometry.
Triangle Inequalities in Physics
Triangle inequalities are most commonly known from geometry, where they state that the length of any side of a triangle must be less than or equal to the sum of the lengths of the other two sides. A similar principle applies in the context of physics, particularly with moments of inertia of three-dimensional objects.

These inequalities become significant when ensuring the physical plausibility of the rotational dynamics of objects. The solution outlined step-by-step in the textbook demonstrates how, for moments of inertia, each value I₁, I₂, and I₃ must adhere to the conditions resembling those of the sides of a triangle. This ensures that the calculated moments of inertia correspond to a physically realizable body.

These constraints are not just mathematical curiosities; they are grounded in the physical reality of how mass can be distributed in space. Mass cannot be distributed in a way that violates these inequalities, just like in the real world, you cannot construct a triangle with sides that do not adhere to this rule.
Planar Body Inertia
To further grasp the concept of moments of inertia, it's enlightening to explore the special case of a planar body. A planar body is one that has all of its mass distributed along a single plane. Common examples are flat plates, thin sheets, or any object whose thickness can be disregarded in comparison to its other dimensions.

In the case of a planar body undergoing rotation, one moment of inertia (typically I₃, if the object lies in the xy-plane) is significantly different from the other two (I₁ and I₂). This is because there's no mass outside the plane to influence rotations about the axis perpendicular to it. The exercise solutions highlight that for such a body, the moments of inertia satisfy the triangle inequalities in a state of equality: I₃ equals the sum of I₁ and I₂.

This unique characteristic allows for simplified calculations and analysis of the body's rotational behaviour, making it an excellent starting point for beginners to the study of rotational dynamics before progressing to the complexities of three-dimensional bodies.

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Most popular questions from this chapter

A small mass \(\varepsilon\) is added to a rigid body with moments of inertia \(I_{1}>I_{2}>I_{3}\) at the point \(\mathbf{Q}=x_{1} \mathbf{e}_{1}+x_{2} \mathbf{e}_{2}+x_{3} \mathbf{e}_{3}\). Find the change in \(I_{1}\) and \(\mathbf{e}_{1}\) with error \(\mathrm{O}\left(\varepsilon^{2}\right)\). Solution. The center of mass is displaced by a distance of order \(\varepsilon\). Therefore, the moments of inertia of the old body with sespect to the parallel axes passing through the old and new centers of mass differ in magnitude of order \(\varepsilon^{2}\). At the same time, the addition of mass changes the moment of inertia relative to any fixed axis by order \(\varepsilon\). Therefore, we can disregard the displacement of the center of mass for calculations with error \(\mathrm{O}\left(\varepsilon^{2}\right)\). Thus, after addition of a small mass the kinetic energy takes the form $$ T=T_{0}+\frac{1}{2} \varepsilon[\Omega, Q]^{2}+O\left(\varepsilon^{2}\right) $$ where \(T_{0}=\frac{1}{2}\left(I_{1} \Omega_{1}^{3}+I_{2} \Omega_{2}^{2}+I_{3} \Omega_{3}^{2}\right)\) is the kinetic energy of the original body. We look for the eigenvalue \(I_{t}(\varepsilon)\) and eigenvector \(\mathbf{e}_{1}(\varepsilon)\) of the inertia operator in the form of a Taylor series in \(\varepsilon\). By equating coefficients of \(\varepsilon\) in the relation \(A(\varepsilon) \mathbf{e}_{1}(\varepsilon)=I_{1}(\varepsilon) e_{1}(\varepsilon)\), we find that, within error \(\mathrm{O}\left(\varepsilon^{2}\right)\) : $$ I_{1}(\varepsilon) \approx I_{1}+\alpha\left(x_{2}^{2}+x_{3}^{2}\right) \text { and } \mathbf{e}_{1}(\varepsilon) \approx \mathbf{e}_{1}+\varepsilon\left(\frac{x_{1} x_{2}}{I_{2}-I_{1}} \mathbf{e}_{2}+\frac{x_{1} x_{3}}{I_{3}-I_{1}} \mathbf{e}_{3}\right) $$ From the formula for \(I_{1}(\varepsilon)\) it is clear that the change in the principal moments of inertia (to the first approximation in \(\varepsilon\) ) is as if neither the center of mass nor the principal axes changed. The formula for \(e_{1}(\varepsilon)\) demonstrates how the directions of the principal axes change: the largest principal axis of the inertia ellipsoid approaches the added point, and the smallest recedes from it. Furthermore, the addition of a small mass on one of the principal planes of the inertia ellipsoid rotates the two axes lying in this plane and does not change the direction of the third axis. The appearance of the differences of moments of inertia in the denominator is connected with the fact that the major axes of an ellipsoid of revolution are not defined. If the inertia ellipsoid is nearly an ellipsoid of revolution (i.e., \(I_{1} \approx I_{2}\) ) then the addition of a small mass could strongly turn the axes \(e_{1}\) and \(e_{2}\) in the plane spanned by them.

Are stationary rotations of the body around the largest and smallest principal axes Liapunov stable?

Find the axes and moments of inertia of a homogeneous ellipsoid of mass \(m\) with semiaxes \(a, b\), and c relative to the center 0 . Hint. First look at the sphere.

Prove Steiner's theorem: The moments of incrtia of any rigid body relative to two parallel axes, one of which passes through the center of mass, are related by the equation $$ I-I_{0}+m r^{2}, $$ where \(m\) is the mass of the body, \(r\) is the distance between the axes, and \(I_{0}\) is the moment of inertia relative to the axis passing through the center of mass. Thus the moment of inertia relative to an axis passing through the center of mass is less than the moment of inertia relative to any parallel axis.

Show that the most general movement of a rigid body is a helical movement, i.e., the composition of a rotation through angle \(\varphi\) around some axis and a translation by \(h\) along it.

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