Moment of inertia unit mm4

ENGR 516 Computational Methods for Graduate Students Catholic University of America

Assignment #6 EigenValue Problems

1.) The moment of inertia, Ix, Iy, and the product of inertia Ixy of the cross sectional area shown in the figure are:

2.) The structure of an acetylene molecule may be idealized as four masses connected by two springs as shown in the figure below. Applying the equation of motion; we can generate a system model for the amplitudes of vibration of each atom:

1

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4.23 The moment of inertia , , and the product of inertia of the cross-sectional area shown in the figure are:

mm4, mm4, and mm4

The principal moments of inertia are the eigenvalues of the matrix

, and the principal axes are in the direction of the eigen-

vectors. Determine the principal moments of inertia by solving the char- acteristic equation. Determine the orientation of the principal axes of inertia (unit vectors in the directions of the eigenvectors).

Solution

The eigenvalues of the matrix are the principal moments of inertia. They are determined from

the roots of the characteristic equation:

which simplifies to . The equation is solved by using the quadratic formula:

or and

By definition, the eigenvectors corresponding to each eigenvalue must satisfy:

Starting with the first eigenvalue:

These two equations are redundant and yield .

Since the eigenvector is a unit vector: .

4 mm

O

y

x

24 mm 2 mm

20 mm

3 mm

2 mm

Ix Iy Ixy

Ix 5286= Iy 4331= Ixy 2914=

5286 2914 2914 4331

5286 2914 2914 4331

5286 λ–( ) 4331 λ–( ) 2914( ) 2914( )– 0=

14402270 9617λ– λ2+ 0=

λ1 2 9617 9617( )2 4 1( ) 14402270( )–

2 ----------------------------------------------------------------------------------------= λ1 7761.36= λ2 1855.64=

5286 2914 2914 4331

u1 i( )

u2 i( )

λi u1 i( )

u2 i( )

=

5286 2914 2914 4331

u1 1( )

u2 1( )

λ1 u1

1( )

u2 1( )

7761.36 u1 1( )

u2 1( )

= =

u1 1( ) 1.1772u2

1( )=

u1 1( )( )

2 u2

1( )( ) 2

+ 1=

1

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

4.23 The moment of inertia , , and the product of inertia of the cross-sectional area shown in the figure are:

mm4, mm4, and mm4

The principal moments of inertia are the eigenvalues of the matrix

, and the principal axes are in the direction of the eigen-

vectors. Determine the principal moments of inertia by solving the char- acteristic equation. Determine the orientation of the principal axes of inertia (unit vectors in the directions of the eigenvectors).

Solution

The eigenvalues of the matrix are the principal moments of inertia. They are determined from

the roots of the characteristic equation:

which simplifies to . The equation is solved by using the quadratic formula:

or and

By definition, the eigenvectors corresponding to each eigenvalue must satisfy:

Starting with the first eigenvalue:

These two equations are redundant and yield .

Since the eigenvector is a unit vector: .

4 mm

O

y

x

24 mm 2 mm

20 mm

3 mm

2 mm

Ix Iy Ixy

Ix 5286= Iy 4331= Ixy 2914=

5286 2914 2914 4331

5286 2914 2914 4331

5286 λ–( ) 4331 λ–( ) 2914( ) 2914( )– 0=

14402270 9617λ– λ2+ 0=

λ1 2 9617 9617( )2 4 1( ) 14402270( )–

2 ----------------------------------------------------------------------------------------= λ1 7761.36= λ2 1855.64=

5286 2914 2914 4331

u1 i( )

u2 i( )

λi u1 i( )

u2 i( )

=

5286 2914 2914 4331

u1 1( )

u2 1( )

λ1 u1

1( )

u2 1( )

7761.36 u1 1( )

u2 1( )

= =

u1 1( ) 1.1772u2

1( )=

u1 1( )( )

2 u2

1( )( ) 2

+ 1=

1

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

4.50 The structure of the C2H2 (acetylene) molecule may be ide- alized as four masses connected by two springs (see discussion in Problem 4.25). By applying the equation of motion, the following system of equations can be written for the amplitudes of vibration of each atom:

where is the frequency, kg/s2 and kg/s2 are the restoring force spring constants representing the C–H and C–C bonds, respectively, and and are the masses of the atoms ( kg). (a) Determine the eigenvalues (frequencies), and the corresponding wavelengths (where

m/s is the speed of light). (b) Determine the eigenvectors corresponding to the eigenvalues found in part (a). From the eigenvectors,

deduce the relative motion of the atoms (i.e., are they moving toward, or away from each other?).

Solution

The following script file solves this problem:

clear, clc kCH=5.92e2; kCC=15.8e2; mH=1.6605e-27; mC=12*1.6605e-27; c=3e8; M=zeros(4); M(1,1)=kCH/mH; M(1,2)=-M(1,1); M(2,1)=-kCH/mC; M(2,2)=(kCH+kCC)/mC; M(2,3)=-kCC/mC; M(3,2)=M(2,3); M(3,3)=M(2,2); M(3,4)=M(2,1); M(4,3)=M(1,2); M(4,4)=-M(4,3);

kCH mH --------- ω2–

kCH mH ---------– 0 0

kCH mC ---------–

kCH kCC+( ) mC

----------------------------- ω2– kCC mC --------– 0

0 kCC mC --------–

kCH kCC+( ) mC

----------------------------- ω2– kCH mC ---------–

0 0 kCH mH ---------–

kCH mH --------- ω2–

A1 A2 A3 A4

0 0 0 0

=

ω kCH 5.92 10 2×= kCC 15.8 10

2×= mH 1amu= mC 12amu=

1amu 1.6605 10 27–×= ω λ 2πc

ω ---------=

c 3 108×=

1

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

4.50 The structure of the C2H2 (acetylene) molecule may be ide- alized as four masses connected by two springs (see discussion in Problem 4.25). By applying the equation of motion, the following system of equations can be written for the amplitudes of vibration of each atom:

where is the frequency, kg/s2 and kg/s2 are the restoring force spring constants representing the C–H and C–C bonds, respectively, and and are the masses of the atoms ( kg). (a) Determine the eigenvalues (frequencies), and the corresponding wavelengths (where

m/s is the speed of light). (b) Determine the eigenvectors corresponding to the eigenvalues found in part (a). From the eigenvectors,

deduce the relative motion of the atoms (i.e., are they moving toward, or away from each other?).

Solution

The following script file solves this problem:

clear, clc kCH=5.92e2; kCC=15.8e2; mH=1.6605e-27; mC=12*1.6605e-27; c=3e8; M=zeros(4); M(1,1)=kCH/mH; M(1,2)=-M(1,1); M(2,1)=-kCH/mC; M(2,2)=(kCH+kCC)/mC; M(2,3)=-kCC/mC; M(3,2)=M(2,3); M(3,3)=M(2,2); M(3,4)=M(2,1); M(4,3)=M(1,2); M(4,4)=-M(4,3);

kCH mH --------- ω2–

kCH mH ---------– 0 0

kCH mC ---------–

kCH kCC+( ) mC

----------------------------- ω2– kCC mC --------– 0

0 kCC mC --------–

kCH kCC+( ) mC

----------------------------- ω2– kCH mC ---------–

0 0 kCH mH ---------–

kCH mH --------- ω2–

A1 A2 A3 A4

0 0 0 0

=

ω kCH 5.92 10 2×= kCC 15.8 10

2×= mH 1amu= mC 12amu=

1amu 1.6605 10 27–×= ω λ 2πc

ω ---------=

c 3 108×=

1

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

4.50 The structure of the C2H2 (acetylene) molecule may be ide- alized as four masses connected by two springs (see discussion in Problem 4.25). By applying the equation of motion, the following system of equations can be written for the amplitudes of vibration of each atom:

where is the frequency, kg/s2 and kg/s2 are the restoring force spring constants representing the C–H and C–C bonds, respectively, and and are the masses of the atoms ( kg). (a) Determine the eigenvalues (frequencies), and the corresponding wavelengths (where

m/s is the speed of light). (b) Determine the eigenvectors corresponding to the eigenvalues found in part (a). From the eigenvectors,

deduce the relative motion of the atoms (i.e., are they moving toward, or away from each other?).

Solution

The following script file solves this problem:

clear, clc kCH=5.92e2; kCC=15.8e2; mH=1.6605e-27; mC=12*1.6605e-27; c=3e8; M=zeros(4); M(1,1)=kCH/mH; M(1,2)=-M(1,1); M(2,1)=-kCH/mC; M(2,2)=(kCH+kCC)/mC; M(2,3)=-kCC/mC; M(3,2)=M(2,3); M(3,3)=M(2,2); M(3,4)=M(2,1); M(4,3)=M(1,2); M(4,4)=-M(4,3);

kCH mH --------- ω2–

kCH mH ---------– 0 0

kCH mC ---------–

kCH kCC+( ) mC

----------------------------- ω2– kCC mC --------– 0

0 kCC mC --------–

kCH kCC+( ) mC

----------------------------- ω2– kCH mC ---------–

0 0 kCH mH ---------–

kCH mH --------- ω2–

A1 A2 A3 A4

0 0 0 0

=

ω kCH 5.92 10 2×= kCC 15.8 10

2×= mH 1amu= mC 12amu=

1amu 1.6605 10 27–×= ω λ 2πc

ω ---------=

c 3 108×=