1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Prove Parallel Axis theorem

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove parallel axis theorem (Steiner's theorem).

    2. Relevant equations

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

    3. The attempt at a solution

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

    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"?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 27, 2009 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    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?).

  4. Oct 27, 2009 #3
    [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 )

    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.
  5. Oct 27, 2009 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    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].

    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.
  6. Oct 27, 2009 #5
    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?
  7. Oct 27, 2009 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper


    [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]

  8. Oct 27, 2009 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

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

    V is the volume over which you do the integration.
  9. Oct 27, 2009 #8
    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]
  10. Oct 27, 2009 #9

    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.
  11. Oct 27, 2009 #10

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    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).

  12. Oct 27, 2009 #11
    I did not find such a definition about the centre of a mass in Wikipedia.
    Could you please say the exact definition, please.


    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.
    Last edited: Oct 27, 2009
  13. Oct 28, 2009 #12

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    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]

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook