# Centre of gravity, moment of inertia and radius of gyration

1. May 30, 2017

### DevonZA

1. The problem statement, all variables and given/known data

2. Relevant equations

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

MOI rectangle: 1/3ml^2

MOI triangle: 1/18md^2

3. 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=1.2kg

1.1

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

2. May 30, 2017

### haruspex

How do you get 400, why -400, and how do you get 1000/3?

3. May 30, 2017

### DevonZA

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. May 30, 2017

### haruspex

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

5. May 30, 2017

### DevonZA

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. May 30, 2017

### haruspex

Yes.

7. May 30, 2017

### DevonZA

Okay so then

X=m1x1-m2x2/m1-m2

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

= 482.35mm

Is this correct?

8. May 30, 2017

### haruspex

The term for the triangle is wrong. What axis would be right for that expression?

9. May 30, 2017

### DevonZA

The formulae given for a rectangular triangle:

I=1/18md^2
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. May 30, 2017

### DevonZA

11. May 30, 2017

### haruspex

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.