How to calculate the moment of inertia of a rectangular cuboid?

[Lotto]

Homework Statement

My task is to calculate the moment of inertia of a homogenous rectangular cuboid around the z-axis, so $I_{zz}$. We know, that $|x| \le a$, $|y| \le b$ and $|z| \le c$.

Relevant Equations

$I_{zz}=M({x^2}+{y^2})$ ....there should be \overline above $x^2$ and $y^2$, but it doesn't work here...

In my textbook, a hint is the formula above, which can be used when we have a homogenous body. $M$ is the body's mass, but what does the remnant mean?

 

[kuruman]

Does the z-axis pass through the CM of the cuboid? Are the principal axes perpendicular to the faces?
What do you mean by remnant?
Can you set up the triple integral $~I_{zz}=\int dm~(x^2+y^2)~$?

Please show what you tried that "doesn't work."

 

[Lotto]

Does the z-axis pass through the CM of the cuboid? Are the principal axes perpendicular to the faces?
What do you mean by remnant?
Can you set up the triple integral $~I_{zz}=\int dm~(x^2+y^2)~$?

Please show what you tried that "doesn't work."

Well, in the task there is written only that $|x| \le a$, $|y| \le b$ and $|z| \le c$. I don't understand what it means. But I suppose that all axes go through its CM and that its faces are perpendicular to x,y,z.

By the "remnant" I mean that \overline $x^2$ and \overline $y^2$. I think it should be calculated somehow via triple integrals, but I don't know why the formula above is valid.

 

[kuruman]

Well, in the task there is written only that $|x| \le a$, $|y| \le b$ and $|z| \le c$. I don't understand what it means.

It means that $-a \le x\le +a$, $-b \le y\le +b$ and $-c \le z\le +c$. This means that you have a cuboid of dimensions (2a)×(2b)×(2c) with the origin of the Cartesian axes at the center of the cuboid.

By the "remnant" I mean that \overline $x^2$ and \overline $y^2$. I think it should be calculated somehow via triple integrals, but I don't know why the formula above is valid.

The expression that you call "remnant" is the mass-averaged distance squared from the z-axis, $$\overline{(x^2+y^2)}=\frac{\int(x^2+y^2)~dm}{\int dm}=\frac{1}{M}\int(x^2+y^2)~dm.$$ Since $~I_{zz}=\int dm~(x^2+y^2)$, it follows that $$I_{zz}=M\overline{(x^2+y^2)}.$$ BTW, \overline works fine here as you can see. Use

##\overline{(x^2+y^2)}##

 

[jbriggs444]

But I suppose that all axes go through its CM and that its faces are perpendicular to x,y,z.

Yes.

By the "remnant" I mean that \overline $x^2$ and \overline $y^2$. I think it should be calculated somehow via triple integrals, but I don't know why the formula above is valid.

When you say "\overline", it seems that you mean x bar and y bar. e.g. $\bar{x}^2 + \bar{y}^2$. Here, x bar ($\bar{x}$) denotes the mean value of $x$ and similar for y bar.

Unfortunately, the mean of a set of squares is not equal to the square of the mean. Nor is the weighted average of a set of values equal to the average value multiplied by the average weight. [A moment of inertia is a sort of weighted average of a set of mass elements where the square of the radius is used as part of the weight]

We can try a quick test: The mean of 1 and 3 is 2. Squared, that is 4. The mean of $1^2$ and $3^2$ is 5. Four is not equal to five.

In addition, if we have a cuboid centered on the origin then it is clear that $\bar{x} = \bar{y} = 0$. So the idea of using those in the correct formula is a non-starter.

 

[haruspex]

By the "remnant" I mean that \overline $x^2$ and \overline $y^2$.

That’s a curious use of the word. Remnant means that which remains; leftovers. Perhaps you are thinking of some other word?