Ellipsoid moment of inertia matrix
$begingroup$
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
asked May 9 '16 at 20:22
Jim MJim M
777
$endgroup$
add a comment |
$begingroup$
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
asked May 9 '16 at 20:22
Jim MJim M
777
$endgroup$
$begingroup$
farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
$endgroup$
– Paul
May 9 '16 at 20:26
add a comment |
$begingroup$
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
asked May 9 '16 at 20:22
Jim MJim M
777
$endgroup$
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
asked May 9 '16 at 20:22
Jim MJim M
777
asked May 9 '16 at 20:22
Jim MJim M
777
edited Nov 30 '16 at 0:27
user137731
asked May 9 '16 at 20:22
Jim MJim M
777
asked May 9 '16 at 20:22
Jim MJim M
777
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered May 10 '16 at 2:41
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
$begingroup$
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
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
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.
answered May 10 '16 at 2:41
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered May 10 '16 at 2:41
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered May 10 '16 at 2:41
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
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.
answered May 10 '16 at 2:41
Ng Chung TakNg Chung Tak
14.8k31334
edited May 12 '16 at 7:36
answered May 10 '16 at 2:41
Ng Chung TakNg Chung Tak
14.8k31334
answered May 10 '16 at 2:41
Ng Chung TakNg Chung Tak
14.8k31334
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html
$endgroup$
– Paul
May 9 '16 at 20:26