# Inertia tensor around principal Axes part 2

• Lambda96
In summary, the conversation discussed the task of finding the integral ##I_{23}## and the transformation used to simplify it. The conversation also touched on the task of finding ##I_{22}## and ##I_{33}##, and the application of trigonometric identities to show that they are equal.

#### Lambda96

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,

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.

Use $$\begin{split} \cos^2 \theta &= \tfrac12(1 + \cos 2\theta) \\ \sin^2 \theta &= \tfrac12(1 - \cos 2 \theta) \\ \cos \theta \sin \theta &= \tfrac12 \sin 2\theta \\ \end{split}$$ and recall that the average of sin or cos over a period is zero.

Lambda96 and vanhees71
Thanks pasmith for your help , I was now able to show that ##I_{22}=I_{33}##.

PhDeezNutz

## 1. What is the significance of the inertia tensor around principal axes?

The inertia tensor around principal axes is a mathematical representation of the distribution of mass and shape of a rigid body. It is used to calculate the rotational inertia of an object and is essential in understanding its rotational motion.

## 2. How is the inertia tensor around principal axes calculated?

The inertia tensor around principal axes is calculated by finding the moments of inertia about three mutually perpendicular axes passing through the center of mass of the object. These axes are known as the principal axes and the moments of inertia are represented by the diagonal elements of the inertia tensor matrix.

## 3. Can the inertia tensor around principal axes change?

Yes, the inertia tensor around principal axes can change if the mass distribution or shape of the object changes. For example, if an object undergoes a change in shape or size, its inertia tensor will also change accordingly.

## 4. How does the inertia tensor around principal axes affect the motion of an object?

The inertia tensor around principal axes determines the rotational motion of an object. It affects the object's ability to resist changes in its rotational motion and determines the direction and magnitude of its angular acceleration when a torque is applied.

## 5. What is the relationship between the inertia tensor around principal axes and the moment of inertia?

The inertia tensor around principal axes and the moment of inertia are closely related. The moment of inertia is a scalar value that represents the object's resistance to changes in its rotational motion. The inertia tensor is a mathematical representation of this scalar value for objects with complex shapes and mass distributions.