How Is the Parallel Axis Theorem Applied in Physics IA?

[soopo]

Homework Statement

Prove parallel axis theorem (Steiner's theorem).

Homework Equations

[tex] I_{parallel axis} = I_{cm} + Mr^2 [/tex]

The Attempt at a Solution

I know Wolfram's site which seems to use http://scienceworld.wolfram.com/physics/ParallelAxisTheorem.html"

I am not sure whether the hint of my friend implies to calculate similarly.
He suggests
1. present [tex] A_{mi}, A_c,[/tex] and [tex]C_{mi} [/tex] as vectocs. A is "moment of inertia", while C is "center of mass", perhaps [tex]mR^2[/tex] -term.

2. write moment of inertia's definition as [tex]A = \Epsilon m_i \rho_i ^2[/tex]

3. write [tex] \rho_i^2 = \bar{\rho_i} * \bar{\rho_i} [/tex]
4. Put the vector [tex] \bar{l} + \bar{r_i}[/tex] into the kro at (3).
5. then calculate open

This way one term is zero, because there is a vector in that term which
length is the distance of the center of the mass from itself.

I am not sure whether my friend suggests me to use tensors in the steps (1-5).

I am unsure how the steps which he outlines proves the parallel axis theorem.
It does not seem to be a robust for me.

There seems to be also different kinds of moments of inertias so I am not sure whether the steps (1-5)
form a general result.


How would you prove the theorem in the course "Physics IA"?

 

[Andrew Mason]

Start with the definition of moment of inertia:

[tex]I = \int r^2 dm[/tex] where the integral is taken over the whole surface.

If the axis is the center of mass of the object, then putting the axis at the origin, the equation for the moment of inertia is:

[tex]I = \int (x^2 + y^2) dm[/tex]

Now, do the integral about a new axis at co-ordinates x_a and y_a

The displacement of a mass element dm located at (x,y) from the new axis at (x_a,y_a) is (x_a - x, y_a - y) which has length ((x_a - x)^2 + (y_a - y)^2)^1/2

So the moment of inertia about the new axis is:

[tex]I_a = \int ((x_a - x)^2 + (y_a - y)^2)dm[/tex]

Simplify that integral.

Keep in mind that x_a and y_a are constant. x and y are the only variables.

What do [itex]\int x dm[/itex] and [itex]\int y dm[/itex] work out to? (hint: what is the definition of centre of mass?).

AM

 

[soopo]

So the moment of inertia about the new axis is:

[tex]I_a = \int ((x_a - x)^2 + (y_a - y)^2)dm[/tex]

Simplify that integral.

Keep in mind that x_a and y_a are constant. x and y are the only variables.

What do [itex]\int x dm[/itex] and [itex]\int y dm[/itex] work out to? (hint: what is the definition of centre of mass?).

AM

[tex] \int y dm = ym[/tex] and [tex] \int x dm = xm [/tex].

So we get
[tex] \int (x_a - x)^2 dm = x_a \int dm - 2x_a \int x dm + \int x^2 dm
= x_a *m - 2x_a *m + x^2 * m
= m( x_a^2 - 2x_a *x + x^2 )
[/tex]

Similarly for [tex] (y_a - y)^2 [/tex] so we get

[tex] \int (x_a - x) ^2 + (y_a - y)^2 dm = m( x_a^2 - 2x_a *x + x^2 + y_a^2 - 2y_a *y + y^2 ) [/tex]
I am not sure how this proves Parallel axis theorem.

 

[D H]

[tex] \int (x_a - x) ^2 + (y_a - y)^2 dm = m( x_a^2 - 2x_a *x + x^2 + y_a^2 - 2y_a *y + y^2 ) [/tex]
I am not sure how this proves Parallel axis theorem.

You did the math wrong. x and y are not constants that you can take out of the integral. [itex]dm = \rho(x,y,z) \,dx\, dy\, dz[/itex].

Edit
The inertia tensor is original about the center of mass. You need to take advantage of this. In particular,

[tex]\int_V \vec x dm \equiv 0[/tex]

when position [tex]\vec x[/tex] is measured with respect to the center of mass.

 

[soopo]

You did the math wrong. x and y are not constants that you can take out of the integral. [itex]dm = \rho(x,y,z) \,dx\, dy\, dz[/itex].

This suggests me

[tex] \int (x_a - x) ^2 + (y_a - y)^2 dm = m( x_a^2 - 2x_a *x + x^2 + y_a^2 - 2y_a *y + y^2 ) [/tex]
[tex] = \int ((x_a - x) ^2 + (y_a - y)^2 ) \rho(x,y,z) \,dx \,dy \,dz [/tex]

What should I do next?

The inertia tensor is original about the center of mass. You need to take advantage of this. In particular,

[tex]\int_V \vec x dm \equiv 0[/tex]

when position [tex]\vec x[/tex] is measured with respect to the center of mass.

How can you take advantage of [itex] \int_V \vec x dm = 0 [/itex]?
What is V?

 

[Andrew Mason]

This suggests me

[tex] \int (x_a - x) ^2 + (y_a - y)^2 dm = m( x_a^2 - 2x_a *x + x^2 + y_a^2 - 2y_a *y + y^2 ) [/tex]
[tex] = \int ((x_a - x) ^2 + (y_a - y)^2 ) \rho(x,y,z) \,dx \,dy \,dz [/tex]

What should I do next?

How can you take advantage of [itex] \int_V \vec x dm = 0 [/itex]?
What is V?

Hint:

[tex]I_a = \int ((x_a - x)^2 + (y_a - y)^2)dm[/tex]

[tex]I_a = \int ((x_a^2 - 2x_ax + x^2) + (y_a^2 - 2y_ay + y^2))dm[/tex]

[tex]I_a = \int (x_a^2 + y_a^2) dm + \int (x^2 + y^2)dm - 2x_a\int x dm - 2y_a\int ydm [/tex]

AM

 

[D H]

This suggests me

[tex] \int (x_a - x) ^2 + (y_a - y)^2 dm = m( x_a^2 - 2x_a *x + x^2 + y_a^2 - 2y_a *y + y^2 ) [/tex]

How does that suggest that? x and y are dummy variables of integration here.

How can you take advantage of [itex] \int_V \vec x dm = 0 [/itex]?
What is V?

V is the volume over which you do the integration.

 

[soopo]

Hint:
[tex]I_a = \int (x_a^2 + y_a^2) dm + \int (x^2 + y^2)dm - 2x_a\int x dm - 2y_a\int ydm [/tex]

Is the following right?

[tex] I_a = \int r_a^2 dm + \int r^2 dm - 2x_a \int x dm - 2y_a \int y dm [/tex]
[tex] = (1/3) * m^3 + (1/3) * m^3 - 2x_a * (1/2) m^2 - 2y_a (1/2) m^2 [/tex]
[tex] = (2/3) * m^3 + m^2 ( - x_a - y_a )[/tex]

 

[soopo]

You did the math wrong. x and y are not constants that you can take out of the integral. [itex]dm = \rho(x,y,z) \,dx\, dy\, dz[/itex].

Edit
The inertia tensor is original about the center of mass. You need to take advantage of this. In particular,

[tex]\int_V \vec x dm \equiv 0[/tex]

when position [tex]\vec x[/tex] is measured with respect to the center of mass.

We do not have the z-coordinate in the integral.
This suggests me that

[tex] I_a = 0 [/tex]

Parallel axis theorem says

[tex] I_a = I + MR^2 [/tex]

so we get

[tex] I = - MR^2 [/tex]

I am not sure how this proves the parallel axis theorem.

 

[Andrew Mason]

Is the following right?

[tex] I_a = \int r_a^2 dm + \int r^2 dm - 2x_a \int x dm - 2y_a \int y dm [/tex]
[tex] = (1/3) * m^3 + (1/3) * m^3 - 2x_a * (1/2) m^2 - 2y_a (1/2) m^2 [/tex]
[tex] = (2/3) * m^3 + m^2 ( - x_a - y_a )[/tex]

You need to work on integration skills.

[tex]\int r_a^2 dm = Mr_a^2[/tex]

[tex]\int r^2 dm = I_{cm}[/tex]

[tex] - 2x_a \int x dm = -2x_a*0 = 0[/tex]

(this is because x is measured from the centre of mass so, by definition the sum of all xdm's is 0).

AM

 

[soopo]

this is because x is measured from the centre of mass so, by definition the sum of all xdm's is 0

I did not find such a definition about the centre of a mass in Wikipedia.
Could you please say the exact definition, please.

[edit]

I know seem to understand what you mean.

[itex] dm [/itex] infininately small mass. This way we can muliply any quantity with this such that the result is zero. Hmm... This leaves out this [itex] \int 0 [/itex]. I am not sure what it means. I would say that it is zero, but I am not sure why.

 

[Andrew Mason]

I did not find such a definition about the centre of a mass in Wikipedia.
Could you please say the exact definition, please.

[edit]

I know seem to understand what you mean.

[itex] dm [/itex] infininately small mass. This way we can muliply any quantity with this such that the result is zero. Hmm... This leaves out this [itex] \int 0 [/itex]. I am not sure what it means. I would say that it is zero, but I am not sure why.

Think of a long rod of length 2L with two point masses, m1 and m2, on the ends. The ends are located at -L and +L. The centre of mass of the whole thing is the point on the rod, p, at which:

[tex]m_1\vec{r_1} + m_2\vec{r_2} = 0[/tex]

where r1 = displacement p to -L and r2 = displacement p to L.

In other words [itex]\sum m_i \vec{r_i} = 0[/itex]

In a real object in which mass is spread out, you have to break the mass into infinitessimal elements and do the same process to find the centre of mass.

[tex]\int dm \vec{r} = 0[/tex]

In two dimensions, the centre of mass has to be located at the point at which:

[tex]\int dm \vec{x} = \int dm \vec{y} = 0[/tex]

AM