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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.

Short Answer

Expert verified
The inertia tensor is [10ma20006ma25ma205ma26ma2]. Principal moments are 5ma2,11ma2,10ma2. Principal axes: [1,0,0], [0,12,12], [0,12,12].

Step by step solution

01

Write the coordinates and mass of each point

We have three masses: m located at (a,0,0), 2m located at (0,a,a), and 3m located at (0,a,a). These coordinates will be used to calculate the inertia tensor.
02

Calculate moments of inertia

The moments of inertia for each axis Ixx,Iyy,Izz need to be calculated. For Ixx, sum y2+z2 for each mass:Ixx=m(02+02)+2m(a2+a2)+3m(a2+(a)2)Ixx=0+4ma2+6ma2=10ma2For Iyy, sum x2+z2 for each mass:Iyy=m(a2+02)+2m(02+a2)+3m(02+(a)2)Iyy=ma2+2ma2+3ma2=6ma2For Izz, sum x2+y2 for each mass:Izz=m(a2+02)+2m(02+a2)+3m(02+a2)Izz=ma2+2ma2+3ma2=6ma2
03

Calculate products of inertia

The products of inertia Ixy,Ixz,Iyz are calculated by summing the product of two different coordinates for each mass.For Ixy:Ixy=m(a0)+(2m)(0a)+(3m)(0a)=0For Ixz:Ixz=m(a0)+2m(0a)+3m(0(a))=0For Iyz:Iyz=m(00)+(2m)(aa)+3m(a(a))=2ma23ma2=5ma2
04

Construct the inertia tensor

Combine the moments and products of inertia into the inertia matrix I=[IxxIxyIxz IyxIyyIyz IzxIzyIzz]=[10ma200 06ma25ma2 05ma26ma2]
05

Find the principal moments (eigenvalues)

To find the principal moments, solve the characteristic equation for the inertia tensor, which is given by:det(IλI)=0Solving this for λ gives the eigenvalues of the inertia tensor: \[\lambda_1 = 5ma^2, \quad \lambda_2 = 11ma^2, \quad \lambda_3 = 10ma^2\]
06

Determine the principal axes (eigenvectors)

Calculate the eigenvectors corresponding to each eigenvalue determined in the previous step. The eigenvectors provide the directions of the principal axes. Using an eigenvalue solver on the inertia tensor yields: Principal axis 1: [1,0,0] Principal axis 2: [0,12,12] Principal axis 3: [0,12,12]

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Key Concepts

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

Principal Moments
Principal moments are specific scalar values associated with the inertia tensor of an object. They are derived as the eigenvalues from that tensor. These values are crucial because they represent the moments of inertia about the principal axes of the object, which are directions where the inertia tensor is diagonalized.
The three principal moments obtained as eigenvalues give insight into how a rigid body behaves under rotation. In this exercise, the calculated principal moments were:
  • λ1=5ma2
  • λ2=11ma2
  • λ3=10ma2
These correspond to the distinct rotational characteristics of the body relative to each principal axis. They are integral in determining stability and dynamics of the system as each principal moment indicates how resistant the body is to changes in rotational velocity around one of its axes.
Products of Inertia
Products of inertia are off-diagonal terms in the inertia tensor, and they represent interactions between different rotational axes. These terms are generally present in systems where the axes are not principal axes and include contributions from cross products of coordinate values.
In the given exercise, the products of inertia were calculated to be:
  • Ixy=0
  • Ixz=0
  • Iyz=5ma2
These values provide information on how the component interacts across different planes. A zero indicates no interaction, while a non-zero value, like in Iyz, shows that rotations about the y- and z-axes influence each other. In practice, minimizing or eliminating products of inertia through proper alignment of the principal axes can simplify calculations and understanding of rotational dynamics.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors hold a key role in characterizing the inertia tensor. When dealing with rigid body rotations, finding these components allows for the simplification of rotational dynamics.
  • Eigenvalues: They correspond to the principal moments of inertia. The eigenvalues, λ1, λ2, and λ3, were calculated as 5ma2, 11ma2, and 10ma2 respectively. These values help to understand how mass distribution affects rotational resistance.
  • Eigenvectors: These vectors give the directions of the principal axes. For the inertia tensor, eigenvectors are derived through solving linear equations associated with their respective eigenvalues. In this exercise, the principal axes were calculated as:
    • First axis: [1,0,0]
    • Second axis: [0,12,12]
    • Third axis: [0,12,12]
    These eigenvectors indicate the coordinate directions in which the moments of inertia are measured, ultimately simplifying down the dynamics of the system by aligning them with principal axes.

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Most popular questions from this chapter

(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 x2μdx where μ is the linear mass density, mass/length.] (b) What if the rod's center is at the origin?

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).

The moment of inertia of a continuous mass distribution with density ϱ is obtained by converting the sum of (10.25) into the volume integral ρ2dm=ρ2ϱdV. (Note the two forms of the Greek "rho": ρ= distance from z axis, ϱ= 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.

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.

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.

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