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Chapter 10: Q6P (page 508)

Point masses 1 at (1, 1, -2) and 2 at (1, 1, 1).

Short Answer

Expert verified

Answer:

Inertia tensor is l=9-30-390006.

Step by step solution

01

Given Information.

The given information is mentioned below:

m1=1m2=2r1=1,1,-2r2=1,1,1

02

Definition of Point mass.

The concept of a physical object with non-zero mass but explicitly and particularly minuscule volume or linear dimension is known as point mass.

03

Find the eigenvalues.

Solve for eigenvalues.

lij=6δij--2δ3i+δ3i+23δij-1l=600060006-11-211-2-2-24+22-1-1-12-1-1-12=5-12-152222+4-2-2-24-2-2-24l=9-30-390006

04

Find principal axes and corresponding moment of inertia.

Find the moment of inertia.

v1=-1,1,0v2=0,0,1v3=1,1,0

Simplify further.

l1=12l2=6l3=6

Hence, inertia tensor is, l=9-30-390006

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Most popular questions from this chapter

Verify equations(2.6).

The square matrix in equation (10.3)is called the Jacobian matrix J; the determinant of this matrix is the Jacobian J=detJ which we used in Chapter 5 , Section 4 to find volume elements in multiple integrals. (Note that as in Chapter 3, J represents a matrix; J in italics is its determinant.) For the transformation to spherical coordinates in localid="1659266126385" (10.1)and (10.2) show that J=detJ=r2sinθ . Recall that the spherical coordinate volume element is r2sinθdrdθdϕ . Hint: Find JTJ and note that det(JTJ)=(detJ)2.

Write the transformation equations forW×S to verify the results of Example 3.

Show that, in polar coordinates, theθcontravariant component of dsis which is unitless, the dθphysical component of ds is rdθwhich has units of length, and theθcovariant component of ds isr2dθwhich has units role="math" localid="1659265070715" (length)2 .

XiAij=Bj.
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