Centre of gravity, moment of inertia and radius of gyration

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Homework Help Overview

The discussion revolves around the concepts of center of gravity, moment of inertia (MOI), and radius of gyration in the context of a composite shape involving a rectangle and a triangle. Participants are analyzing the calculations related to these concepts and questioning the assumptions made in the setup.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to calculate the center of gravity and moment of inertia for a composite shape, discussing the mass and distances involved. There are questions about specific values used in the calculations, particularly regarding the placement of axes and the signs of distances. Some participants are clarifying the reasoning behind their calculations and questioning the correctness of the formulas applied.

Discussion Status

The discussion is ongoing, with participants providing insights and corrections to each other's calculations. Some have offered clarifications regarding the placement of axes and the application of the parallel axis theorem, while others are still exploring the implications of their assumptions and calculations.

Contextual Notes

There are indications of confusion regarding the placement of the axes and the corresponding distances used in the calculations. Participants are also reflecting on the implications of positive and negative values in relation to the axes.

DevonZA
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Homework Statement


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Homework Equations



Centre of gravity: X=m1x1-m2x2/m1-m2

MOI rectangle: 1/3ml^2

MOI triangle: 1/18md^2

Radius of gyration: Ixx=mk^2

The Attempt at a Solution



Mass of body 1: b*l*p = 0.8*1*10=8kg

Mass of body 2: 1/2b*h*p = 1/2(0.4)*0.6*10=1.2kg1.1

X=m1x1-m2x2/m1-m2

= 8*(-400)-1.2*(-1000/3)/8-1.2

= -411.7mm
1.2

Ixx= 1/3ml^2 - 1/18md^2

= 1/3*8*1^2 - 1/18*1.2*0.2^2

=2.664m^4
1.3

Ixx=mk^2

2.664=(8-1.2)*k^2

k=625.91mm
 
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DevonZA said:
= 8*(-400)-1.2*(-1000/3)/8-1.2
How do you get 400, why -400, and how do you get 1000/3?
 
haruspex said:
How do you get 400, why -400, and how do you get 1000/3?

That should be 500, from the centre of the rectangle to the XX axis is 500mm.
I believe the centre of the triangle should be 2/3*600 = 400mm
Then added to the 200mm from the start of the triangle to the XX axis should give 400+200=600mm.
I put a negative in front of the 400 and 1000/3 as they are "away" from the XX axis.
I don't know why I put 1000/3, I think I was dividing the length of the rectangle in 3 because of the triangle for some reason.

X=m1x1-m2x2/m1-m2

= 8*(-500)-1.2*(-600)/8-1.2

= -482.35mm
 
DevonZA said:
That should be 500, from the centre of the rectangle to the XX axis is 500mm.
I believe the centre of the triangle should be 2/3*600 = 400mm
Then added to the 200mm from the start of the triangle to the XX axis should give 400+200=600mm.
Right.
DevonZA said:
I put a negative in front of the 400 and 1000/3 as they are "away" from the XX axis.
But they are "away" from that axis in the positive y direction.
 
haruspex said:
Right.

But they are "away" from that axis in the positive y direction.

Meaning they should be positive?
If they were below the X axis then they would be negative? In other words if the XX axis were at the top of the rectangle.
 
DevonZA said:
Meaning they should be positive?
If they were below the X axis then they would be negative? In other words if the XX axis were at the top of the rectangle.
Yes.
 
Okay so then

X=m1x1-m2x2/m1-m2

= 8*(500)-1.2*(600)/8-1.2

= 482.35mm

Is this correct?

What about my answers for 1.2 and 1.3?
 
DevonZA said:
Ixx= 1/3ml^2 - 1/18md^2
The term for the triangle is wrong. What axis would be right for that expression?
 
haruspex said:
The term for the triangle is wrong. What axis would be right for that expression?

The formulae given for a rectangular triangle:

about aa axis
I=1/18md^2
about bb axis (through G)
I=1/18mh^2

because we want the inertia about xx axis I used =1/18md^2 where d=distance to xx axis=200mm
 
  • #10
IMG_0236.JPG
 
  • #11
DevonZA said:
about aa axis
I=1/18md^2
I don't know what you mean by aa axis here, but that expression is for the moment of inertia about an axis through the centroid and parallel to one side. d is the width orthogonal to the axis. So here, d=0.6 for an axis through the centroid and parallel to the x axis.
To get from that axis to the xx axis itself you need to use the parallel axis theorem.
 

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