Ellipsoid moment of inertia matrix












2












$begingroup$


Some background info: torque $tau$ is defined as $$tau = I*domega$$



Where $I$ is the moment of inertia matrix and $domega$ is an object's rotational acceleration. As I understand it, the inertia matrix acts just like mass in that it counteracts the torque (for example, if an object is spinning around the x axis, a big value of $I_{xx}$ means that the object needs more torque around the x axis in order to spin).



However, angular momentum $M$ can be defined as $$M=I * omega$$



Where $omega$ is the rotational velocity. So it seems that torque is the time derivative of angular momentum.



Using these facts, how would I find the moment of inertia matrix for an ellipsoid with uniform density of the form $$frac{x^2}{a}+frac{y^2}{b}+frac{z^2}{c}≤9$$ with $a≠b≠c≠0$? Would I have to use spherical coordinates somehow? I'm not given any torque or angular velocity information. Any guidance is appreciated.






multivariable-calculus physics classical-mechanics elliptic-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
    $endgroup$
    – Paul
    May 9 '16 at 20:26
















2












$begingroup$


Some background info: torque $tau$ is defined as $$tau = I*domega$$



Where $I$ is the moment of inertia matrix and $domega$ is an object's rotational acceleration. As I understand it, the inertia matrix acts just like mass in that it counteracts the torque (for example, if an object is spinning around the x axis, a big value of $I_{xx}$ means that the object needs more torque around the x axis in order to spin).



However, angular momentum $M$ can be defined as $$M=I * omega$$



Where $omega$ is the rotational velocity. So it seems that torque is the time derivative of angular momentum.



Using these facts, how would I find the moment of inertia matrix for an ellipsoid with uniform density of the form $$frac{x^2}{a}+frac{y^2}{b}+frac{z^2}{c}≤9$$ with $a≠b≠c≠0$? Would I have to use spherical coordinates somehow? I'm not given any torque or angular velocity information. Any guidance is appreciated.






multivariable-calculus physics classical-mechanics elliptic-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
    $endgroup$
    – Paul
    May 9 '16 at 20:26














2












2








2


1



$begingroup$


Some background info: torque $tau$ is defined as $$tau = I*domega$$



Where $I$ is the moment of inertia matrix and $domega$ is an object's rotational acceleration. As I understand it, the inertia matrix acts just like mass in that it counteracts the torque (for example, if an object is spinning around the x axis, a big value of $I_{xx}$ means that the object needs more torque around the x axis in order to spin).



However, angular momentum $M$ can be defined as $$M=I * omega$$



Where $omega$ is the rotational velocity. So it seems that torque is the time derivative of angular momentum.



Using these facts, how would I find the moment of inertia matrix for an ellipsoid with uniform density of the form $$frac{x^2}{a}+frac{y^2}{b}+frac{z^2}{c}≤9$$ with $a≠b≠c≠0$? Would I have to use spherical coordinates somehow? I'm not given any torque or angular velocity information. Any guidance is appreciated.






multivariable-calculus physics classical-mechanics elliptic-integrals







share|cite|improve this question











$endgroup$




Some background info: torque $tau$ is defined as $$tau = I*domega$$



Where $I$ is the moment of inertia matrix and $domega$ is an object's rotational acceleration. As I understand it, the inertia matrix acts just like mass in that it counteracts the torque (for example, if an object is spinning around the x axis, a big value of $I_{xx}$ means that the object needs more torque around the x axis in order to spin).



However, angular momentum $M$ can be defined as $$M=I * omega$$



Where $omega$ is the rotational velocity. So it seems that torque is the time derivative of angular momentum.



Using these facts, how would I find the moment of inertia matrix for an ellipsoid with uniform density of the form $$frac{x^2}{a}+frac{y^2}{b}+frac{z^2}{c}≤9$$ with $a≠b≠c≠0$? Would I have to use spherical coordinates somehow? I'm not given any torque or angular velocity information. Any guidance is appreciated.






multivariable-calculus physics classical-mechanics elliptic-integrals




multivariable-calculus physics classical-mechanics elliptic-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '16 at 0:27







user137731

















asked May 9 '16 at 20:22









Jim MJim M

777




777












  • $begingroup$
    farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
    $endgroup$
    – Paul
    May 9 '16 at 20:26


















  • $begingroup$
    farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
    $endgroup$
    – Paul
    May 9 '16 at 20:26
















$begingroup$
farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
$endgroup$
– Paul
May 9 '16 at 20:26




$begingroup$
farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
$endgroup$
– Paul
May 9 '16 at 20:26










1 Answer
1






active

oldest

votes


















2












$begingroup$

In usual notation,
begin{align*}
frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} &= 1 \
I_{ij} &=
iiint rho(mathbf{r}) (r^{2} delta_{ij}-x_{i} x_{j}) d^{3} , mathbf{r} \
I &= frac{m}{5}
begin{pmatrix}
b^2+c^2 & 0 & 0 \
0 & c^2+a^2 & 0 \
0 & 0 & a^2+b^2 \
end{pmatrix}
end{align*}



In your case: $$I=frac{3m}{5}
begin{pmatrix}
b+c & 0 & 0 \
0 & c+a & 0 \
0 & 0 & a+b \
end{pmatrix}$$



See also the link here.



P.S.:



By symmetry, $ineq j implies I_{ij}=0$



Let $x=a X$, $y=b Y$ and $z=c Z$



begin{align*}
iiint_{x^2/a^2+y^2/b^2+z^2/c^2<1} x^{2} dV &=
abc iiint_{X^2+Y^2+Z^2<1} a^{2} X^{2} dX dY dZ \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
left( int_{-sqrt{1-X^2-Y^2}}^{sqrt{1-X^2-Y^2}} dZ right) dY
right] a^2X^2 dX \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
2sqrt{1-X^2-Y^2} dY
right] a^2X^2 dX \
&= abcint_{-1}^{1} pi (1-X^2) a^2X^2 dX \
&= pi abc times frac{4a^2}{15} \
rho &= frac{3m}{4pi abc} \
I_{11} &= rho iiint (y^2+z^2) dV \
&= frac{m}{5} (b^2+c^2)
end{align*}



Finish the rest by symmetry.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $z^2$ for the third term?
    $endgroup$
    – Jim M
    May 12 '16 at 3:19










  • $begingroup$
    @JimM Typo corrected
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 3:24










  • $begingroup$
    How do you evaluate this triple integral?
    $endgroup$
    – Chill2Macht
    May 12 '16 at 4:50












  • $begingroup$
    See PS in answer
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 5:55










  • $begingroup$
    Also, once you do the substitution that reduced the integral to a sphere, you can finish with spherical coordinates and eliminate most of the work (instead of a triple horror of square roots), or just "steal" the known answer from the sphere.
    $endgroup$
    – orion
    May 12 '16 at 6:40












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2












$begingroup$

In usual notation,
begin{align*}
frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} &= 1 \
I_{ij} &=
iiint rho(mathbf{r}) (r^{2} delta_{ij}-x_{i} x_{j}) d^{3} , mathbf{r} \
I &= frac{m}{5}
begin{pmatrix}
b^2+c^2 & 0 & 0 \
0 & c^2+a^2 & 0 \
0 & 0 & a^2+b^2 \
end{pmatrix}
end{align*}



In your case: $$I=frac{3m}{5}
begin{pmatrix}
b+c & 0 & 0 \
0 & c+a & 0 \
0 & 0 & a+b \
end{pmatrix}$$



See also the link here.



P.S.:



By symmetry, $ineq j implies I_{ij}=0$



Let $x=a X$, $y=b Y$ and $z=c Z$



begin{align*}
iiint_{x^2/a^2+y^2/b^2+z^2/c^2<1} x^{2} dV &=
abc iiint_{X^2+Y^2+Z^2<1} a^{2} X^{2} dX dY dZ \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
left( int_{-sqrt{1-X^2-Y^2}}^{sqrt{1-X^2-Y^2}} dZ right) dY
right] a^2X^2 dX \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
2sqrt{1-X^2-Y^2} dY
right] a^2X^2 dX \
&= abcint_{-1}^{1} pi (1-X^2) a^2X^2 dX \
&= pi abc times frac{4a^2}{15} \
rho &= frac{3m}{4pi abc} \
I_{11} &= rho iiint (y^2+z^2) dV \
&= frac{m}{5} (b^2+c^2)
end{align*}



Finish the rest by symmetry.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $z^2$ for the third term?
    $endgroup$
    – Jim M
    May 12 '16 at 3:19










  • $begingroup$
    @JimM Typo corrected
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 3:24










  • $begingroup$
    How do you evaluate this triple integral?
    $endgroup$
    – Chill2Macht
    May 12 '16 at 4:50












  • $begingroup$
    See PS in answer
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 5:55










  • $begingroup$
    Also, once you do the substitution that reduced the integral to a sphere, you can finish with spherical coordinates and eliminate most of the work (instead of a triple horror of square roots), or just "steal" the known answer from the sphere.
    $endgroup$
    – orion
    May 12 '16 at 6:40
















2












$begingroup$

In usual notation,
begin{align*}
frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} &= 1 \
I_{ij} &=
iiint rho(mathbf{r}) (r^{2} delta_{ij}-x_{i} x_{j}) d^{3} , mathbf{r} \
I &= frac{m}{5}
begin{pmatrix}
b^2+c^2 & 0 & 0 \
0 & c^2+a^2 & 0 \
0 & 0 & a^2+b^2 \
end{pmatrix}
end{align*}



In your case: $$I=frac{3m}{5}
begin{pmatrix}
b+c & 0 & 0 \
0 & c+a & 0 \
0 & 0 & a+b \
end{pmatrix}$$



See also the link here.



P.S.:



By symmetry, $ineq j implies I_{ij}=0$



Let $x=a X$, $y=b Y$ and $z=c Z$



begin{align*}
iiint_{x^2/a^2+y^2/b^2+z^2/c^2<1} x^{2} dV &=
abc iiint_{X^2+Y^2+Z^2<1} a^{2} X^{2} dX dY dZ \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
left( int_{-sqrt{1-X^2-Y^2}}^{sqrt{1-X^2-Y^2}} dZ right) dY
right] a^2X^2 dX \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
2sqrt{1-X^2-Y^2} dY
right] a^2X^2 dX \
&= abcint_{-1}^{1} pi (1-X^2) a^2X^2 dX \
&= pi abc times frac{4a^2}{15} \
rho &= frac{3m}{4pi abc} \
I_{11} &= rho iiint (y^2+z^2) dV \
&= frac{m}{5} (b^2+c^2)
end{align*}



Finish the rest by symmetry.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $z^2$ for the third term?
    $endgroup$
    – Jim M
    May 12 '16 at 3:19










  • $begingroup$
    @JimM Typo corrected
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 3:24










  • $begingroup$
    How do you evaluate this triple integral?
    $endgroup$
    – Chill2Macht
    May 12 '16 at 4:50












  • $begingroup$
    See PS in answer
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 5:55










  • $begingroup$
    Also, once you do the substitution that reduced the integral to a sphere, you can finish with spherical coordinates and eliminate most of the work (instead of a triple horror of square roots), or just "steal" the known answer from the sphere.
    $endgroup$
    – orion
    May 12 '16 at 6:40














2












2








2





$begingroup$

In usual notation,
begin{align*}
frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} &= 1 \
I_{ij} &=
iiint rho(mathbf{r}) (r^{2} delta_{ij}-x_{i} x_{j}) d^{3} , mathbf{r} \
I &= frac{m}{5}
begin{pmatrix}
b^2+c^2 & 0 & 0 \
0 & c^2+a^2 & 0 \
0 & 0 & a^2+b^2 \
end{pmatrix}
end{align*}



In your case: $$I=frac{3m}{5}
begin{pmatrix}
b+c & 0 & 0 \
0 & c+a & 0 \
0 & 0 & a+b \
end{pmatrix}$$



See also the link here.



P.S.:



By symmetry, $ineq j implies I_{ij}=0$



Let $x=a X$, $y=b Y$ and $z=c Z$



begin{align*}
iiint_{x^2/a^2+y^2/b^2+z^2/c^2<1} x^{2} dV &=
abc iiint_{X^2+Y^2+Z^2<1} a^{2} X^{2} dX dY dZ \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
left( int_{-sqrt{1-X^2-Y^2}}^{sqrt{1-X^2-Y^2}} dZ right) dY
right] a^2X^2 dX \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
2sqrt{1-X^2-Y^2} dY
right] a^2X^2 dX \
&= abcint_{-1}^{1} pi (1-X^2) a^2X^2 dX \
&= pi abc times frac{4a^2}{15} \
rho &= frac{3m}{4pi abc} \
I_{11} &= rho iiint (y^2+z^2) dV \
&= frac{m}{5} (b^2+c^2)
end{align*}



Finish the rest by symmetry.






share|cite|improve this answer











$endgroup$



In usual notation,
begin{align*}
frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} &= 1 \
I_{ij} &=
iiint rho(mathbf{r}) (r^{2} delta_{ij}-x_{i} x_{j}) d^{3} , mathbf{r} \
I &= frac{m}{5}
begin{pmatrix}
b^2+c^2 & 0 & 0 \
0 & c^2+a^2 & 0 \
0 & 0 & a^2+b^2 \
end{pmatrix}
end{align*}



In your case: $$I=frac{3m}{5}
begin{pmatrix}
b+c & 0 & 0 \
0 & c+a & 0 \
0 & 0 & a+b \
end{pmatrix}$$



See also the link here.



P.S.:



By symmetry, $ineq j implies I_{ij}=0$



Let $x=a X$, $y=b Y$ and $z=c Z$



begin{align*}
iiint_{x^2/a^2+y^2/b^2+z^2/c^2<1} x^{2} dV &=
abc iiint_{X^2+Y^2+Z^2<1} a^{2} X^{2} dX dY dZ \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
left( int_{-sqrt{1-X^2-Y^2}}^{sqrt{1-X^2-Y^2}} dZ right) dY
right] a^2X^2 dX \
&=
abcint_{-1}^{1}
left[
int_{-sqrt{1-X^2}}^{sqrt{1-X^2}}
2sqrt{1-X^2-Y^2} dY
right] a^2X^2 dX \
&= abcint_{-1}^{1} pi (1-X^2) a^2X^2 dX \
&= pi abc times frac{4a^2}{15} \
rho &= frac{3m}{4pi abc} \
I_{11} &= rho iiint (y^2+z^2) dV \
&= frac{m}{5} (b^2+c^2)
end{align*}



Finish the rest by symmetry.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 12 '16 at 7:36

























answered May 10 '16 at 2:41









Ng Chung TakNg Chung Tak

14.8k31334




14.8k31334












  • $begingroup$
    Don't you mean $z^2$ for the third term?
    $endgroup$
    – Jim M
    May 12 '16 at 3:19










  • $begingroup$
    @JimM Typo corrected
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 3:24










  • $begingroup$
    How do you evaluate this triple integral?
    $endgroup$
    – Chill2Macht
    May 12 '16 at 4:50












  • $begingroup$
    See PS in answer
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 5:55










  • $begingroup$
    Also, once you do the substitution that reduced the integral to a sphere, you can finish with spherical coordinates and eliminate most of the work (instead of a triple horror of square roots), or just "steal" the known answer from the sphere.
    $endgroup$
    – orion
    May 12 '16 at 6:40


















  • $begingroup$
    Don't you mean $z^2$ for the third term?
    $endgroup$
    – Jim M
    May 12 '16 at 3:19










  • $begingroup$
    @JimM Typo corrected
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 3:24










  • $begingroup$
    How do you evaluate this triple integral?
    $endgroup$
    – Chill2Macht
    May 12 '16 at 4:50












  • $begingroup$
    See PS in answer
    $endgroup$
    – Ng Chung Tak
    May 12 '16 at 5:55










  • $begingroup$
    Also, once you do the substitution that reduced the integral to a sphere, you can finish with spherical coordinates and eliminate most of the work (instead of a triple horror of square roots), or just "steal" the known answer from the sphere.
    $endgroup$
    – orion
    May 12 '16 at 6:40
















$begingroup$
Don't you mean $z^2$ for the third term?
$endgroup$
– Jim M
May 12 '16 at 3:19




$begingroup$
Don't you mean $z^2$ for the third term?
$endgroup$
– Jim M
May 12 '16 at 3:19












$begingroup$
@JimM Typo corrected
$endgroup$
– Ng Chung Tak
May 12 '16 at 3:24




$begingroup$
@JimM Typo corrected
$endgroup$
– Ng Chung Tak
May 12 '16 at 3:24












$begingroup$
How do you evaluate this triple integral?
$endgroup$
– Chill2Macht
May 12 '16 at 4:50






$begingroup$
How do you evaluate this triple integral?
$endgroup$
– Chill2Macht
May 12 '16 at 4:50














$begingroup$
See PS in answer
$endgroup$
– Ng Chung Tak
May 12 '16 at 5:55




$begingroup$
See PS in answer
$endgroup$
– Ng Chung Tak
May 12 '16 at 5:55












$begingroup$
Also, once you do the substitution that reduced the integral to a sphere, you can finish with spherical coordinates and eliminate most of the work (instead of a triple horror of square roots), or just "steal" the known answer from the sphere.
$endgroup$
– orion
May 12 '16 at 6:40




$begingroup$
Also, once you do the substitution that reduced the integral to a sphere, you can finish with spherical coordinates and eliminate most of the work (instead of a triple horror of square roots), or just "steal" the known answer from the sphere.
$endgroup$
– orion
May 12 '16 at 6:40


















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Are they similar matrix

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0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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