# Is Moment of Inertia Dependent on Surface Movement Away from Axis?

• Guillem_dlc
In summary: So I put that changing the angle to ##45## would leave the ##x## as the minimum.Apart from finding this principal angle, it was requested that the ##x'## axis (that is, the ##x##-axis of the principal axis) should be the minimum and the ##y##-axis the maximum. That is to say, about the principal axes, that there is one that must have a maximum moment of inertia and another one a minimum. So they told you that the principal ##x##-axis of inertia should be the minimum. So if you looked at the moment of inertia of each axis with the angle of ##-45^\circ##, then you got that the ##x## is the maximum and
Guillem_dlc
Homework Statement
Find the angle of the principal axes where ##x'## and ##y'## are the principal axes.
Relevant Equations
Last figure of the solution
Statement figure:

My attempt at a solution:
FIGURE 1 ##\rightarrow A=a^2##

CG ##\rightarrow \overline{x}=-a/2, y=-a/2##
$$\overline{Ix}_1=\dfrac{bh^3}{12}=\overline{Iy}_1=\dfrac{a^4}{12}$$
$$Ix_1=\overline{Ix}_1+\overline{y}^2A=\dfrac{a^4}{12}+\dfrac{a^4}{4}=\dfrac13 a^4\, \textrm{mm}^4=Iy_1$$
$$\overline{Ixy}_1=0\, \textrm{mm}^4\rightarrow Ixy_1=\dfrac{a^4}{4}$$

FIGURE 2 ##\rightarrow A=\dfrac{\pi a^2}{2}##

CG ##\rightarrow \overline{x}=0, \overline{y}=\dfrac{4\pi}3a##
$$Ix_2=Iy_2=\dfrac{\pi}8a^4\, \textrm{mm}^4,\,\, \overline{Ixy}=0\rightarrow Ixy=0+0\cdot \overline{y}A=0\, \textrm{mm}^4$$
$$Ix=\sum Ix_i=0,76032a^4\, \textrm{mm}^4$$
$$Iy=\sum Iy_i=0,76032a^4 \, \textrm{mm}^4$$
$$Ixy=\sum Ixy_i=\dfrac 14a^2$$
$$\boxed{\sigma =\dfrac12 \arctan \left( \dfrac{-2Ixy}{Ix-Iy}\right)=\dfrac12 \arctan (-\infty)=-45^\circ}$$

If you could tell me if it's ok you would do me a big favor. thank you very much!

The moment of inertia depended on how the surface moves away from the axis, right? Because if so maybe it does make sense that I got the two equal moments (##x## and ##y##), doesn't it?

Your work for obtaining ##I_x## and ##I_y## looks correct. However, when I calculate ##\pi/8 + 1/3##, I get 0.726 instead of 0.760. But, the important thing for this problem is that ##I_x = I_y##. Many books use the notation ##I_{xx}## and ##I_{yy}## for your ##I_x## and ##I_y##. Also, many books define ##I_{xy}## to have the opposite sign of your definition. But, I've seen both definitions used.

The question asks for the angles of orientation of the principal ##x'## and ##y'## axes. You calculated one angle equal to -45o. So, I think you should include some comments regarding how this calculated angle relates to the orientation of the ##x'## and ##y'## axes.

Guillem_dlc said:
The moment of inertia depended on how the surface moves away from the axis, right? Because if so maybe it does make sense that I got the two equal moments (##x## and ##y##), doesn't it?

Yes, if you study how the mass is distributed about the ##x## and ##y## axes for this problem, then I think you can see that ##I_{xx}## should equal ##I_{yy}## without doing any calculation.

Lnewqban
TSny said:
Your work for obtaining ##I_x## and ##I_y## looks correct. However, when I calculate ##\pi/8 + 1/3##, I get 0.726 instead of 0.760. But, the important thing for this problem is that ##I_x = I_y##. Many books use the notation ##I_{xx}## and ##I_{yy}## for your ##I_x## and ##I_y##. Also, many books define ##I_{xy}## to have the opposite sign of your definition. But, I've seen both definitions used.

The question asks for the angles of orientation of the principal ##x'## and ##y'## axes. You calculated one angle equal to -45o. So, I think you should include some comments regarding how this calculated angle relates to the orientation of the ##x'## and ##y'## axes.
Yes, if you study how the mass is distributed about the ##x## and ##y## axes for this problem, then I think you can see that ##I_{xx}## should equal ##I_{yy}## without doing any calculation.
##0,726## yes sorry I left out the ##2##.

That would be fine then, wouldn't it?

Guillem_dlc said:
##0,726## yes sorry I left out the ##2##.

That would be fine then, wouldn't it?
Yes. Hopefully, the orientations of the principal axes are clear to you from your answer for the angle ##\sigma##.

TSny said:
Yes. Hopefully, the orientations of the principal axes are clear to you from your answer for the angle ##\sigma##.
I was told that the ##x##-axis was going to be the minimum and with ##-45## I was left with the maximum.

So I put that changing the angle to ##45## would leave the ##x## as the minimum.

Guillem_dlc said:
I was told that the ##x##-axis was going to be the minimum and with ##-45## I was left with the maximum.

So I put that changing the angle to ##45## would leave the ##x## as the minimum.
I don't understand what it means to say that the ##x##-axis is a minimum or a maximum. The important thing is to be able to draw or describe the orientation of the principal axes.

TSny said:
I don't understand what it means to say that the ##x##-axis is a minimum or a maximum. The important thing is to be able to draw or describe the orientation of the principal axes.
Apart from finding this principal angle, it was requested that the ##x'## axis (that is, the ##x##-axis of the principal axis) should be the minimum and the ##y##-axis the maximum. That is to say, about the principal axes, that there is one that must have a maximum moment of inertia and another one a minimum. So they told you that the principal ##x##-axis of inertia should be the minimum. So if you looked at the moment of inertia of each axis with the angle of ##-45^\circ##, then you got that the ##x## is the maximum and the ##y## is the minimum. And then I thought about the angle of ##45^\circ##, which in the end is the same. And then the moment of inertia of ##x## was already the minimum, you know?

OK. I see now. I agree that the principal moment of inertial about the axis at +45o would be smaller than the principal moment of inertia about the axis at -45o.

TSny said:
OK. I see now. I agree that the principal moment of inertial about the axis at +45o would be smaller than the principal moment of inertia about the axis at -45o.
Okay, so it would be fine then, wouldn't it?

I believe so.

## Is the moment of inertia dependent on the distance of the mass from the axis of rotation?

Yes, the moment of inertia is highly dependent on the distance of the mass from the axis of rotation. It is calculated as the sum of the products of each mass element and the square of its distance from the axis of rotation. This means that as the distance increases, the moment of inertia increases significantly.

## How does the distribution of mass affect the moment of inertia?

The distribution of mass affects the moment of inertia because it determines how far the mass elements are from the axis of rotation. Masses that are further away from the axis contribute more to the moment of inertia than masses that are closer. Therefore, an object with mass distributed far from the axis will have a higher moment of inertia compared to an object with the same mass concentrated close to the axis.

## Can the moment of inertia change if the shape of the object changes?

Yes, the moment of inertia can change if the shape of the object changes. This is because reshaping the object can alter the distances of the mass elements from the axis of rotation. For example, stretching a mass outward from the axis will increase the moment of inertia, while compressing it inward will decrease it.

## Does the orientation of the axis of rotation affect the moment of inertia?

Yes, the orientation of the axis of rotation affects the moment of inertia. The same object can have different moments of inertia depending on the axis about which it is rotating. This is because the distances of the mass elements from the axis can vary with different orientations, leading to different moments of inertia.

## Is the moment of inertia the same for all types of motion?

No, the moment of inertia is not the same for all types of motion. It is specifically a property of rotational motion and depends on the axis of rotation. For linear motion, the concept of moment of inertia does not apply; instead, mass and velocity are the primary factors. In rotational motion, both the distribution of mass and the axis of rotation are crucial in determining the moment of inertia.

• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
29
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
908
• Introductory Physics Homework Help
Replies
5
Views
3K
• Math Proof Training and Practice
Replies
60
Views
8K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
19K
• Introductory Physics Homework Help
Replies
7
Views
18K