Moment of inertia composite body

In summary, the problem is that you are trying to calculate the MoI for a 3D object using a 2D formula, and this is causing confusion. You should use a table of MoI to help you calculate the MoI for a 3D object.
  • #1
antonov
5
2
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?
Skärmavbild 2021-06-02 kl. 23.56.49.png

20210602-222925-5.jpg
 
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  • #2
Take it in smaller steps.
Post all your steps in getting the MoI of the 10x1 strip about O.
 
  • #3
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.
 
  • #4
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.
 
  • #5
antonov said:
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?
 
  • #6
Skärmavbild 2021-06-03 kl. 23.29.21.png

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.
 
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  • #7
antonov said:
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?
 
Last edited:
  • #8
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.
 
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  • #9
antonov said:
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)?
 
  • #10
We get that the total mass is m.
 
  • #11
antonov said:
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?
 
  • #12
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}##
 

FAQ: Moment of inertia composite body

What is a moment of inertia composite body?

A moment of inertia composite body is a measurement of an object's resistance to rotational motion. It takes into account the distribution of mass around an axis of rotation.

How is the moment of inertia of a composite body calculated?

The moment of inertia of a composite body is calculated by summing the individual moments of inertia of each component of the body. This can be done using the parallel axis theorem, which takes into account the distance of each component from the axis of rotation.

Why is the moment of inertia important in physics?

The moment of inertia is important in physics because it is a key factor in determining the rotational behavior of an object. It affects how much torque is required to make an object rotate and how quickly it will rotate.

How does the moment of inertia change for different shapes?

The moment of inertia depends on the distribution of mass in an object and the distance of that mass from the axis of rotation. Therefore, the moment of inertia will be different for different shapes, even if they have the same mass.

Can the moment of inertia of a composite body be negative?

No, the moment of inertia of a composite body cannot be negative. It is always a positive value, as it represents the object's resistance to rotational motion. However, the direction of the moment of inertia vector can be positive or negative, depending on the direction of rotation.

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