Parallel Axis Theorem problem help

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The discussion revolves around the application of the parallel axis theorem to calculate the total area moment of inertia after cutting a circle from a rectangle. The key question is whether the equation should be Itotal=Irectangle-(Icentral-Ad^2) or Itotal=Irectangle-(Icentral+Ad^2). It is clarified that when a positive area is removed, it contributes negatively to the total moment of inertia. The correct formulation is confirmed to be Itotal=Irectangle-(Icentral+Ad^2). Understanding the treatment of negative areas in the context of the theorem is essential for accurate calculations.
LeFerret
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Homework Statement


This is moreof a conceptual question regarding the parallel axis theorem and area.
Lets say I have a solid rectangle with length L, and Width W.
I cut out a circle of radius R at the center.
When calculating the Total Area moment of inertia of the hollow region where the circle was cut out
is it Itotal=Irectangle-(Icentral-Ad2) or
Itotal=Irectangle-(Icentral+Ad2)
because I know the region cut out has a negative area, so I subtract that moment of inertia from the rectangle's moment of inertia, but I'm not sure if that negative carries over inside the parenthesis as well

Homework Equations


Iany=Icentral + Ad2

The Attempt at a Solution

 
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If you have a rectangle and then you remove a circular area from the center, the parallel axis theorem really doesn't come into play, because the centroid of the resulting figure hasn't changed from before the circular area was removed; i.e. I(net) = I(rectangle) - I(circle)
 
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I agree, but I wanted to explain what I was asking as simply as possible, the circle could be removed anywhere and it could be any series of shapes attached together. What I am confused on is if that negative area comes to play inside of the parenthesis or if it's |Ad^2|
 
LeFerret said:

Homework Statement


This is moreof a conceptual question regarding the parallel axis theorem and area.
Lets say I have a solid rectangle with length L, and Width W.
I cut out a circle of radius R at the center.
When calculating the Total Area moment of inertia of the hollow region where the circle was cut out
is it [STRIKE]Itotal=Irectangle-(Icentral-Ad2)[/STRIKE] or
Itotal=Irectangle-(Icentral+Ad2)
because I know the region cut out has a negative area,

You cut out a positive area, that makes a negative contribution to the total moment of inertia.


LeFerret said:
so I subtract that moment of inertia from the rectangle's moment of inertia, but I'm not sure if that negative carries over inside the parenthesis as well

Homework Equations


Iany=Icentral + Ad2

The Attempt at a Solution


The second equation is correct.

ehild
 
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thank you!
 
You are welcome:smile:

ehild
 
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