Moment of Inertia (no calculus) - What am I missing?

[Femme_physics]

Homework Statement

Calculate the moment of inertia to the Y and X axis

Diagram attached.

Homework Equations

Everything is in the diagram...there's no need to use calculus, just the formulas for geometrical shapes for moment of inertia and center of mass...

Answers are

Ix = 95.4 cm^4
Iy = 263.7 cm^4

The Attempt at a Solution

Attached. I got Ix, I'm not getting Iy from some reason. Question is on the left, my attempt on the right

 

[I like Serena]

You're not applying Steiner's theorem quite right.
It only works relative to the center of mass.

So for instance for I_y2 you start out with a half-circle with an axis that is not through the center of mass.

So first you need Steiner's theorem to find the moment of inertia through the center of mass.
That is, you need to subtract:

$$\frac 1 2 \pi a^2 (\frac {4a} {3\pi})^2$$.

And then you need to find the moment of inertia relative to the y-axis.
That is, you need to add:

$$\frac 1 2 \pi a^2 (3a - \frac {4a} {3\pi})^2$$.

 

[Femme_physics]

It only works relative to the center of mass.

Oh. See, I wasn't sure whether I need to subtract or add in Steiner's theorem, I thought you always add and then LATER when you do the Iy total, you subtract. But, guess I was wrong.

Anyway, I did what you said but the answer is still off


$$\frac 1 2 \pi a^2 (\frac {4a} {3\pi})^2$$.

That equals 45270.7393

$$\frac 1 2 \pi a^2 (3a - \frac {4a} {3\pi})^2$$.[/QUOTE]


That equals 1667217.45

-45270.7393 + 1667217.45 = 162196.711

If I now do Iy total =
The value I got for I1 = 62831.853
Minus this new value I got for I2 = 162196.711
Plus the value for I3 (square's moment of inertia) = 1645479

Iy = -I1 -I2 +I3

= 1420450.436


Still not the answer :-/

 

[I like Serena]

Well, your I_y3 is not quite right yet either.

It seems to me that you inserted the surface of the circle disk instead of the surface of a square.

The proper formula would be:

$$I_{y_3} = \frac {3a \cdot (3a)^3} {12} + (3a)^2 \cdot \left(\frac {3a} 2\right)^2$$

 

[Femme_physics]

You're right, I did accidentally used circle's area, but I've tried it with the square's area result too, like in your formula (3600), still not the answer

4320000 - 62831.853 - 1729914.273 = 2527253.874

It seems to be approaching it. Is there a mistake at the answers? It seems a lot of the other students I've asked didn't get the result

 

[I like Serena]

You're right, I did accidentally used circle's area, but I've tried it with the square's area result too, like in your formula (3600), still not the answer

4320000 - 62831.853 - 1729914.273 = 2527253.874

It seems to be approaching it. Is there a mistake at the answers? It seems a lot of the other students I've asked didn't get the result

Your result for I_y2 is still a little off.
It should be 62831.853 - 45270.739 + 1667217.450 = 1684778.564.

So I'm getting:

4320000 - 62831.853 - 1684778.564 = 2572389.583 mm⁴

This is indeed a little bit off from 263.7 cm⁴.
So I think there is a mistake in the answer as well.

 

[Femme_physics]

Agreed :) Thank you.

 

[Femme_physics]

I'm actually a bit confused where you took the 4a/3pi formula. The formula for half a circle center of mass says 4R/3pi. Since R and a just so happen to be the same value, I guess it doesn't matter. But I presume that is an error on your part?

 

[I like Serena]

I'm actually a bit confused where you took the 4a/3pi formula. The formula for half a circle center of mass says 4R/3pi. Since R and a just so happen to be the same value, I guess it doesn't matter. But I presume that is an error on your part?

I'm not sure if I understand your question.
As you say 4R/3pi is the formula for half a circle's center of mass.
Applying it to your particular case, a is substituted for R, and it becomes 4a/3pi.
Where's the error?

 

[Femme_physics]

Oh, nevermind, sorry my bad. I'm just so invested in this question that I got confused. Been working on it for a few days if not more. I keep thinking I understand how to solve moment of inertia without calculus but I keep getting some of the answers wrong, and it's just really frustrating with the exercise sheet has errors in it. I notice I'm unable to solve the following two simple moment of inertia questions...I wonder if it's the same pattern of errors, but I'm not sure if it's the paper side or my side. Just frustrating. Thanks anyway. Appreciate the help.

 

[I like Serena]

Oh, nevermind, sorry my bad. I'm just so invested in this question that I got confused. Been working on it for a few days if not more. I keep thinking I understand how to solve moment of inertia without calculus but I keep getting some of the answers wrong, and it's just really frustrating with the exercise sheet has errors in it. I notice I'm unable to solve the following two simple moment of inertia questions...I wonder if it's the same pattern of errors, but I'm not sure if it's the paper side or my side. Just frustrating.

The trick I use to find out if I have made a mistake, is to calculate the same thing using a different method (calculus in this case). If the answer comes out the same (as it did in this case), I become confident that I didn't make a mistake.

Thanks anyway. Appreciate the help.

You're welcome .

 

[Femme_physics]

The trick I use to find out if I have made a mistake, is to calculate the same thing using a different method (calculus in this case). If the answer comes out the same (as it did in this case), I become confident that I didn't make a mistake.

Ah, problem is we haven't started calculus with these sort of problems. Can't wait, though!