Moment of inertia composite body

[antonov]

Homework Statement

Marginal of 0,5% is OK

Relevant Equations

see picture

I have this moment of inertia problem and is a little confused on the semicircle part and if the rest is really right? I get over 10 if I calculate it in crew CAD but by hand I get 7,568032142. What is right and what am I doing wrong?

 

[haruspex]

Take it in smaller steps.
Post all your steps in getting the MoI of the 10x1 strip about O.

 

[Orodruin]

And more importantly, post all your steps for getting the MoI for the semi-circle about O as that is what you were specifically asking about.

 

[antonov]

I don't really know how to break it down, this is the way we learned it. And then looking up the formulas for MOI in a formula collection.

 

[haruspex]

I don't really know how to break it down, this is the way we learned it. And then looking up the formulas for MOI in a formula collection.

For example, how do you arrive at $(\frac 12-\frac{16}{9\pi^2})2^2$? That is clearly the result of subtracting one term from another. What do those individual terms represent, and what principle are you using to say you should do that subtraction?

 

[antonov]

I got it from this table of MOI we got from our teacher. I was torn because this is 3D and my model is 2D but yeh.

 

[haruspex]

View attachment 283928
I got it from this table of MOI we got from our teacher. I was torn because this is 3D and my model is 2D but yeh.

Ok. Would have helped if you had posted that table originally.
You have $d=(3L)^2+(-2+\frac 4{3\pi})^2m_2$.
I think you mean $d^2=(3L)^2+(-2+\frac 4{3\pi})^2L^2$, but that is not right either. In the formula you are using there is an 'r'.

Some questions:
Is A2 supposed to be the area of the semicircle?
If so, how do you get that value for it?
Why are you adding A3, the area of the removed circle?
Why are you dividing each area by the total area?

 

[antonov]

About A2 yes it's supposed to be the semicircle.
Area of semicircle A=(pi*r^2)/2 The radius is 2L.
A3 is a mistake from my part I saw yesterday total area should be 13.1415926
I divide it so we get the mass och the plate. And then multiply the mass from each area to the I of the same area. Or at least our teacher did it that way so I don't really now why.

The thing is that I tried to look on YouTube but the fact that we learned it another way confuses me to learn the other way.

 

[haruspex]

I divide it so we get the mass och the plate.

It depends what you are given. Are you given the total mass of the lamina or its density (mass per unit area)?

 

[antonov]

We get that the total mass is m.

 

[haruspex]

We get that the total mass is m.

So when you write that you get the answer 7,568032142, you mean you get the MoI is 7,568032142 mL²?

 

[haruspex]

Just spotted another error.
$d^2=(3L)^2+(-2+\frac {4r}{3\pi})^2L^2$ isn't right either.
The displacement from the horizontal line through O to the mass centre of the semicircle is $\frac {4r}{3\pi}$, not $-2+\frac {4r}{3\pi}$