Centre of gravity, moment of inertia and radius of gyration

In summary: How would you do that?In summary, the conversation discusses the calculations for finding the centre of gravity, moment of inertia for a rectangle and triangle, and radius of gyration. The calculations involve finding the mass of two bodies, using the equations X=m1x1-m2x2/m1-m2, MOI rectangle=1/3ml^2, MOI triangle=1/18md^2, and Ixx=mk^2. There is also a discussion about the correct axis to use for the moment of inertia of a triangle and the use of the parallel axis theorem to find the moment of inertia about the xx axis.
  • #1
DevonZA
181
6

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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  • #2
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?
 
  • #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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
DevonZA said:
Ixx= 1/3ml^2 - 1/18md^2
The term for the triangle is wrong. What axis would be right for that expression?
 
  • #9
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.
 

1. What is the difference between centre of gravity and centre of mass?

The centre of gravity is the point at which the entire weight of an object can be considered to act. It is the point where the force of gravity is acting on the object. The centre of mass, on the other hand, is the point at which the entire mass of an object is considered to be concentrated. In most cases, the centre of gravity and the centre of mass are at the same point, but this is not always the case.

2. How is the centre of gravity of an irregularly shaped object determined?

The centre of gravity of an irregularly shaped object can be determined through experimentation or by using mathematical calculations. One method is to suspend the object from different points and see where it balances. Another method is to divide the object into smaller, regular shapes and calculate the centre of gravity for each shape, then find the overall centre of gravity by taking into account the weight and position of each smaller shape.

3. What is the significance of the moment of inertia in rotational motion?

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is influenced by the mass and distribution of mass of an object. The larger the moment of inertia, the more force is needed to accelerate or decelerate the rotation of an object. In other words, it determines how difficult it is to change an object's rotation.

4. How does the radius of gyration affect an object's stability?

The radius of gyration is a measure of how spread out an object's mass is from its axis of rotation. A larger radius of gyration means that the mass is spread out over a larger distance, making the object less stable. This is because it takes more force to change the rotation of an object with a larger radius of gyration.

5. Can the centre of gravity of an object be outside of the object itself?

Yes, the centre of gravity can be outside of the object if the object is not a solid, homogenous shape. For example, a hollow cylinder will have its centre of gravity located at the centre of the cylinder, even though there is no mass at that point. In this case, the centre of gravity is determined by the distribution of mass within the object.

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