- #1

- 156

- 59

- Homework Statement
- I need to calculate the inertia tensor again around a principal axes showing the following ##I_{23}=0## and ##I_{22}=I_{33}##.

- Relevant Equations
- none

Hi,

it's about the task e)

Since the density is homogeneous, I have assumed the following for ##\rho=\frac{M}{V}##.

I then started the proof of ##I_{23}##, the integral looks like this:

$$ I_{23}=\int_{}^{} -\frac{M}{V}r'_2r'_3 d^3r$$

Now I apply the transformation

$$ I_{23}=\int_{}^{} -\frac{M}{V}\Bigl(r_2cos\theta+r_3sin\theta)\Bigr)\cdot \Bigl(-r_2sin\theta + r_3cos\theta \Bigr) \ d^3r$$

$$ I_{23}=\int_{}^{} -\frac{M}{V}\Bigl(cos\theta sin\theta(r_3^2-r_2^2)+r_2r_3cos{2\theta})\Bigr) \ d^3r$$

Now I just used the clue, so ##\frac{1}{2\pi} \int_{0}^{2\pi} I_{23} d\theta ##

$$I_{23}=\int_{}^{} \frac{1}{2\pi} \int_{0}^{2\pi} -\frac{M}{V}\Bigl(cos\theta sin\theta(r_3^2-r_2^2)+r_2r_3cos{2\theta})\Bigr) \ d\theta d^3r $$

$$ I_{23}=\int_{}^{} 0 d^3r=0$$With ##I_{22}=I_{33}## I proceeded as follows

$$ \int_{}^{} \frac{M}{V}{r'}_1^2+{r'}_3^2 d^3r $$

$$ \int_{}^{} \frac{M}{V}{r'}_1^2+{r'}_2^2 d^3r $$

Then I did the transformation,

$$ \int_{}^{} \frac{M}{V}\Bigl(r_1^2-r_2^2sin^2\theta-2r_2r_3sin\theta cos\theta + r_3^2cos^2\theta\Bigr) d^3r $$

$$ \int_{}^{} \frac{M}{V}\Bigl(r_1^2+r_2^2cos^2\theta+2r_2r_3sin\theta cos\theta + r_3^2sin^2\theta\Bigr) d^3r $$

Unfortunately, I am now stuck on how to show that the two terms in the integral are equal.

it's about the task e)

Since the density is homogeneous, I have assumed the following for ##\rho=\frac{M}{V}##.

I then started the proof of ##I_{23}##, the integral looks like this:

$$ I_{23}=\int_{}^{} -\frac{M}{V}r'_2r'_3 d^3r$$

Now I apply the transformation

$$ I_{23}=\int_{}^{} -\frac{M}{V}\Bigl(r_2cos\theta+r_3sin\theta)\Bigr)\cdot \Bigl(-r_2sin\theta + r_3cos\theta \Bigr) \ d^3r$$

$$ I_{23}=\int_{}^{} -\frac{M}{V}\Bigl(cos\theta sin\theta(r_3^2-r_2^2)+r_2r_3cos{2\theta})\Bigr) \ d^3r$$

Now I just used the clue, so ##\frac{1}{2\pi} \int_{0}^{2\pi} I_{23} d\theta ##

$$I_{23}=\int_{}^{} \frac{1}{2\pi} \int_{0}^{2\pi} -\frac{M}{V}\Bigl(cos\theta sin\theta(r_3^2-r_2^2)+r_2r_3cos{2\theta})\Bigr) \ d\theta d^3r $$

$$ I_{23}=\int_{}^{} 0 d^3r=0$$With ##I_{22}=I_{33}## I proceeded as follows

$$ \int_{}^{} \frac{M}{V}{r'}_1^2+{r'}_3^2 d^3r $$

$$ \int_{}^{} \frac{M}{V}{r'}_1^2+{r'}_2^2 d^3r $$

Then I did the transformation,

$$ \int_{}^{} \frac{M}{V}\Bigl(r_1^2-r_2^2sin^2\theta-2r_2r_3sin\theta cos\theta + r_3^2cos^2\theta\Bigr) d^3r $$

$$ \int_{}^{} \frac{M}{V}\Bigl(r_1^2+r_2^2cos^2\theta+2r_2r_3sin\theta cos\theta + r_3^2sin^2\theta\Bigr) d^3r $$

Unfortunately, I am now stuck on how to show that the two terms in the integral are equal.