# Parallel axis theorem, cube (Confirm)

• BoiledPotato
In summary, the parallel axis theorem states that the inertia of a body about any axis is equal to the inertia of the body about a parallel axis through its center of mass, plus the product of the mass and the distance between the two axes squared. In the conversation, the calculation of the inertia tensors for two different boxes is discussed, with one having a smaller mass than the other. The mistake in the initial calculation was pointed out and corrected, leading to the conclusion that the density should be the same for both boxes in order to accurately calculate the inertia.
BoiledPotato
Parallel axis theorem, cube! (Confirm)

// Idisp = Icenter + mass[ (RdotR)*I - RcrossR ]

So to test it out, I create a long box at the origin, and then a smaller
box, half its width, so I can offset it along the x axis, and times it by
2, so it should equal the inertia tensor of the long box.

Is this correct?... why would it be wrong?

Thanx,

Jon

// ref: http://en.wikipedia.org/wiki/Parallel_axis_theorem

// Understanding?
float mass = 1.0f;
Matrix inertiabBox = BoxMatrix(mass/*mass*/, 3/*width*/, 10/*height*/, 50/*depth*/);
Matrix inertiaLongBox = BoxMatrix(mass/*mass*/, 6/*width*/, 10/*height*/, 50/*depth*/);

Vector3 R = new Vector3(3.0f,0,0); // Offset it by R
Matrix mA = inertiabBox + mass*( Vector3.Dot(R,R)*Matrix.Identity - OuterProduct(R,R) );

Matrix inertiaSameAsLongBox = 2 * mA; ? but why?

/* REF */
Matrix BoxMatrix(float mass, float w, float h, float d)
{
float ss = (1/12.0f) * mass;
Matrix bb = Matrix( ss*(h*h+d*d), 0, 0, 0,
0, ss*(w*w+d*d), 0, 0,
0, 0, ss*(w*w+h*h), 0,
0, 0, 0, 1 );
return bb;
}

Matrix OuterProduct( Vector3 a, Vector3 b)
{
return Matrix( a.X*b.X, a.X*b.Y, a.X*b.Z, 0.0f,
a.Y*b.X, a.Y*b.Y, a.Y*b.Z, 0.0f,
a.Z*b.X, a.Z*b.Y, a.Z*b.Z, 0.0f,
0, 0, 0, 1.0f );
}

Problem sorted, my mistake, in my calculations I was making the assumption that mass was the same for both the smaller and larger box...should have been setting the density the same and recalculating the mass.

float mass0 = 0.5f;
float mass1 = 1.0f;
Matrix inertiabBox = BoxMatrix(mass0 /*mass*/, 3/*width*/, 10/*height*/, 50/*depth*/);
Matrix inertiaLongBox = BoxMatrix(mass1 /*mass*/, 6/*width*/, 10/*height*/, 50/*depth*/);

Vector3 R = new Vector3(1.5f,0,0); // Offset it by R
Matrix mA = inertiabBox + mass0*( Vector3.Dot(R,R)*Matrix.Identity - OuterProduct(R,R) );

Matrix inertiaSameAsLongBox = 2 * mA;

Thanx

Jon

## What is the parallel axis theorem?

The parallel axis theorem is a mathematical principle that states that the moment of inertia of a rigid body is equal to the sum of its moment of inertia about its center of mass and the product of its mass and the square of the distance between the center of mass and the parallel axis.

## How does the parallel axis theorem apply to a cube?

In the case of a cube, the parallel axis theorem can be used to calculate the moment of inertia of the cube about any axis parallel to one of its faces. This is useful in understanding the cube's rotational motion and stability.

## What is the formula for calculating the moment of inertia of a cube using the parallel axis theorem?

The formula for calculating the moment of inertia of a cube using the parallel axis theorem is I = (1/6) * m * (a^2 + b^2), where m is the mass of the cube and a and b are the lengths of two adjacent sides of the cube.

## Can the parallel axis theorem be applied to non-uniform objects?

Yes, the parallel axis theorem can be applied to any rigid body, regardless of its shape or mass distribution. It is a general principle that applies to all objects.

## How is the parallel axis theorem useful in real-world applications?

The parallel axis theorem is useful in engineering and physics, particularly in understanding the behavior of rotating objects such as gears, wheels, and flywheels. It is also used in calculating the stability of structures and in designing efficient structures for buildings and bridges.

### Similar threads

• Classical Physics
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
960
• General Engineering
Replies
3
Views
2K
• Linear and Abstract Algebra
Replies
31
Views
2K
• Classical Physics
Replies
1
Views
7K
• Mechanics
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
15
Views
5K
• Introductory Physics Homework Help
Replies
5
Views
4K
• General Math
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
3K