Centre of gravity, moment of inertia and radius of gyration

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SUMMARY

The discussion focuses on calculating the center of gravity, moment of inertia (MOI), and radius of gyration for composite shapes, specifically a rectangle and a triangle. Key formulas include the center of gravity equation X = (m1x1 - m2x2) / (m1 - m2), the moment of inertia for a rectangle Ixx = (1/3)ml², and for a triangle Ixx = (1/18)md². The radius of gyration is derived from Ixx = mk². The participants clarify the correct application of these formulas, particularly the placement of axes and the use of the parallel axis theorem.

PREREQUISITES
  • Understanding of basic mechanics principles, specifically center of gravity and moment of inertia.
  • Familiarity with composite shapes in physics and engineering.
  • Knowledge of the parallel axis theorem for calculating moment of inertia.
  • Ability to perform calculations involving mass, distance, and geometry.
NEXT STEPS
  • Study the application of the parallel axis theorem in moment of inertia calculations.
  • Learn about the derivation and application of the center of gravity for composite shapes.
  • Explore advanced topics in mechanics, such as dynamic stability and rotational motion.
  • Investigate the differences between calculating moment of inertia about different axes for various shapes.
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Students in mechanical engineering, physics enthusiasts, and professionals involved in structural analysis or design who require a solid understanding of center of gravity, moment of inertia, and radius of gyration calculations.

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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