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

Short Answer

Expert verified
Inertia tensor: m[5a20006a24a204a27a2] Principal moments: 11a2,7a2,0, with corresponding eigenvectors as principal axes.

Step by step solution

01

Understand the Problem

We need to calculate the inertia tensor for a system of three equal masses located at specified positions and then find the principal moments and axes of inertia for the system. The inertia tensor helps describe how the mass is distributed relative to an axis.
02

Recall the Inertia Tensor Formula

The inertia tensor for a set of point masses is given by: I=imi[yi2+zi2xiyixizi yixixi2+zi2yizi zixiziyixi2+yi2],where the summation is taken over all point masses of the rigid body.
03

Calculate Moments of Inertia for Each Mass

Define the position vectors for the masses: - Mass at (a,0,0)- Mass at (0,a,2a)- Mass at (0,2a,a)Calculate the moment of inertia for each position and sum them up:- For the first mass at (a,0,0), I1=m[02+020a00 0aa2+0200 0000a2+02]=m[000 0a20 00a2].- For the second mass at (0,a,2a), I2=m[a2+(2a)20a02a 0a02+(2a)2a2a 02aa2a02+a2]=m[5a200 04a22a2 02a22a2].- For the third mass at (0,2a,a), I3=m[(2a)2+a202a0a 02a02+a22aa 0a2aa02+4a2]=m[5a200 0a22a2 02a24a2].
04

Sum the Components to Get the Inertia Tensor

Sum all three inertia matrices to obtain the total inertia tensor:I=I1+I2+I3=m[5a200 0(a2+4a2+a2)(2a22a2) 0(2a22a2)(a2+2a2+4a2)],Resulting in:I=m[5a200 06a24a2 04a27a2].
05

Diagonalize the Inertia Tensor

To find the principal moments, we need to diagonalize the inertia tensor. Solve the characteristic equation det(IλI3)=0, where I3 is the identity matrix.m[5a2λ00 06a2λ4a2 04a27a2λ]The characteristic equation is:λ318a2λ2+69a4λ104a6=0Solve for λ to find the principal moments.
06

Solve for Principal Moments and Axes

Solving the characteristic equation, we find the principal moments: λ1=11a2,λ2=7a2,λ3=0. Corresponding eigenvectors give the principal axes. The orthogonal principal axes are derived using the standard procedure for finding eigenvectors.

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Key Concepts

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

Principal Moments of Inertia
The principal moments of inertia are special values that show how a rigid body resists angular acceleration around specific axes, known as the principal axes. These axes are perpendicular to each other and pass through the body's center of mass. Essentially, they are a particular set of axes for which the inertia tensor becomes diagonal, meaning all off-diagonal components become zero. This simplification makes it easier to analyze the rotational dynamics of the body.

For the given problem, we find the principal moments by solving the characteristic equation derived from the inertia tensor. The values of these principal moments (λ1=11a2,λ2=7a2,λ3=0λ), tell us about the rotational inertia of the body around these principal axes. If one of these moments is zero, it means the axis through the center of mass is an axis of rotation without resistance, indicating a principal axis lies along it.
Principal Axes
Principal axes are critical to understanding the rotational behavior of a rigid body. They are the axes around which the body can rotate without experiencing any torque due to inertia. These axes correspond to the eigenvectors of the inertia tensor. To find them, one must solve the eigenvector equation for the tensor.

In practical terms, these axes are the natural directions a body will rotate around if perturbed. For symmetric objects, like a cube or a sphere, the principal axes align with the object's geometric symmetry axes. In the calculation, by determining the principal moments first, we then find the orientation of these axes by solving the system that arises from substituting each principal moment into the inertia tensor equation. This gives us a set of orthogonal vectors corresponding to the principal axes.
Rigid Body Dynamics
Rigid body dynamics involves analyzing how solid objects move when subjected to forces. Unlike fluid or elastic bodies, a rigid body is assumed to have no deformation under force, maintaining its size and shape. The core elements of rigid body dynamics include translational and rotational motion, described by Newton's laws applied to mass and inertia.

The inertia tensor plays a critical role in describing how a rigid body resists angular motion. It quantitatively describes how mass distribution affects the body's rotation about an axis. In our specific exercise, we calculated this tensor for three point masses. By simplifying the dynamics into principal moments and axes, we make the analysis straightforward by reducing the inertia's complex interaction to simpler, more intuitive terms.
Inertia Tensor Calculation
The inertia tensor is a matrix that provides detailed information about how a body's mass is distributed relative to an axis of rotation. The formula for calculating the inertia tensor involves summing contributions from all point masses or particles composing the rigid body.

In the exercise, we calculated the inertia tensor by using the positions of the three masses. Each position contributed to the final tensor through its squared distances from each axis, as well as the products of its coordinates which fill the off-diagonal elements of the tensor. Once all contributions are summed, we can examine this matrix to extract physical properties like principal moments and principal axes, simplifying further analyses of the body's rotational dynamics. This process is crucial in many areas such as mechanical engineering and robotics, where understanding and controlling motion is essential.

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Most popular questions from this chapter

Five equal point masses are placed at the five corners of a square pyramid whose square base is centered on the origin in the xy plane, with side L, and whose apex is on the z axis at a height H above the origin. Find the CM of the five-mass system.

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 xy plane, which elements of the inertia tensor I are automatically zero? Prove that Izz=Ixx+Iyy.

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 mα, spinning about an axis through the origin with angular velocity ω, prove that its total kinetic energy can be written as T=12mα[(ωrα)2(ωrα)2] Remember that vα=ω×rα. You may find the following vector identity useful: For any two vectors a and b, (a×b)2=a2b2(ab)2 (If you use the identity, please prove it.) (b) Prove that the angular momentum L of the body can be written as L=mα[ωrα2rα(ωrα)] For this you will need the so-called BACCAB rule, that A×(B×C)=B(AC)C(AB) (c) Combine the results of parts (a) and (b) to prove that T=12ωL=12ω~Iω Prove both equalities. The last expression is a matrix product; ω denotes the 3×1 column of numbers ωx,ωy,ωz, the tilde on ω~ 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=12vp. (d) Show that with respect to the principal axes, T=12(λ1ω12+λ2ω22+λ3ω32) as in Equation (10.68).

A rigid body consists of three masses fastened as follows: m at (a,0,0),2m at (0,a,a), and 3m at (0,a,a). (a) Find the inertia tensor I. (b) Find the principal moments and a set of orthogonal principal axes.

An axially symmetric space station (principal axis e3, and λ1=λ2 ) is floating in free space. It has rockets mounted symmetrically on either side that are firing and exert a constant torque Γ about the symmetry axis. Solve Euler's equations exactly for ω (relative to the body axis) and describe the motion. At t=0 take ω=(ω10,0,ω30)

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