Parallel-Axis Theorem Clarification

  • Thread starter vladittude0583
  • Start date
  • Tags
    Theorem
In summary: The Parallel-Axis Theorem is used to calculate the moment of inertia about a parallel axis. To do this, you need to calculate the moment of inertia about the object's centroidal axis, and then add that to the moment of inertia about the new parallel axis.
  • #1
vladittude0583
40
0
Hey guys, we are currently studying chapter 10 in the Fundamentals of Physics 8th. Ed. Halliday & Resnick Textbook and I am having a little problem understanding the Parallel-Axis Theorem. I understand that the Moment of Inertia (Rotational Inertia) is derived from the Kinetic Energy Formula for Rotational Motion of a rigid body. However, the Parallel-Axis Theorem allows us to formulate the Moment of Inertia of a particular rigid body about an axis parallel to the axis of the center of mass. What I do not understand is why the formula for the Parallel-Axis Theorem has Icm + MD^2? Why do you include the Icm and not just take the differential mass element at some distance "r" from the axis of rotation and calculate it that way - is this way more complicated? The way I interpreted the Icm part is that it represents the differential mass element of the rigid body right at some distance "r" from the new axis of rotation right? I totally understand the whole Moment of Inertia is I = integral r^2 dm which is the same thing as the MD^2 portion of the parallel-axis theorem right? I guess I am just trying to figure out why there is that plus in the middle of the formula and what it all means? Could someone please help clarify this for me because its just confusing to me. Thanks.
 
Physics news on Phys.org
  • #2
When a body rotates about an axis parallel to its centroidal axis, not only do you need to move that mass around the parallel axis (which can be considered a point mass at a distance r from the rotation axis), but it also needs to rotate about its own centroidal axis, which means you need to add the centroidal moment term.
 
  • #3
All you need is this two formulas:
1. Integral[alfa*f+beta*g]=alfa*Integral[f]+beta*Integral[g]
2. Integral[(x-<x>)*dm]=0 (definition of center of mass <x>)

Below I used x' for (x-<x>) and y' for (y-<y>):

Integral[r^2*dm]=Int[x^2+y^2*dm]=Int[( (x'+<x>)^2+(y'+<y>)^2 )*dm]=
Int[(x'^2+2*x'*<x>+<x>^2+y'^2+2*y'*<y>+<y>^2)*dm]=
Int[(x'^2+y'^2)*dm]+Int[(<x>^2+<y>^2)*dm]=
Int[r'^2*dm]+Int[<r>^2*dm]=
Icm+<r>^2*m
 
  • #4
Lojzek said:
2. Integral[(x-<x>)*dm]=0 (definition of center of mass <x>)

Could you clarify this a little better? Is the x the distance along the x-axis to the particle? If so, then you are subtracting out the location on the x-axis of the CM, <x>, right? Which is basically the distance along the x-axis from the CM to the particle.

Now comes my confusion, why does the integral of {(the distance from the CM to the particle)*(mass of the particle)} equal zero. Why not int(x'*dm) = (x' as constant) x'*m?

Is the definition of center of mass supposed to tip me off that we can think of it as having no mass there (x') and all being concentrated at the CM? Therefore int(x'*dm) = x'*m, but the m is zero everywhere except x'=0, which of course makes the expression zero everywhere.

This just feels a little fishy to me, at one point (when calculating the moment of inertia) the Center of Mass is not brought into the reasoning (except to find the axis to measure from). For instance when we calculate the moment of inertia as a sum of all the infinitely small masses (dm) multiplied by the square of their respective distances (r), in other symbols integral(r^2 * dm), we do not think of the mass concentrated at the CM. Of course it would be a trivial calculation if we considered all of the mass to be at the origin of the calculation, the integral would be zero.

I feel like I am missing something so I am just wanting to get the specifics of the interpretation correct. Thanks.
 
Last edited:
  • #5
The reason for usefullness of average values is easier evaluation of the sum or integral with which that average was defined.
Center of mass is defined as:

<x>=Integral[x*dm]/m (m=Integral[dm]=total mass)

This means we can use <x> to evaluate the integral:

Integral[x*dm]=<x>*m

The equation I used follows:

0=Integral[x*dm]-<x>*m=Integral[x*dm]-<x>*Integral[dm]=Integral[(x-<x>)*dm]

In general it is not correct to replace a variable in an integral by it's average value. This is possible only if the average was defined with that integral.
 
  • #6
Ah, I see it now. Once you take the integral of the entire object, by definition of the center of mass, every m*x' will inevitably have an equal but opposite -(m*x') that then cancels everything to zero. I'm with you on that now, I appreciate the help.
 

1. What is the Parallel-Axis Theorem?

The Parallel-Axis Theorem is a principle in physics that relates the moment of inertia of an object about its center of mass to the moment of inertia of the same object about a parallel axis passing through a point at a distance from the center of mass.

2. How is the Parallel-Axis Theorem derived?

The Parallel-Axis Theorem is derived using the definition of moment of inertia and the parallel axis theorem for point masses. By considering an object as a collection of infinitesimal point masses, the moment of inertia about a parallel axis can be calculated and related to the moment of inertia about the center of mass.

3. What is the practical significance of the Parallel-Axis Theorem?

The Parallel-Axis Theorem is used to calculate the moment of inertia of objects that are not symmetrical or have irregular shapes. It allows for the determination of the moment of inertia about any axis parallel to the axis of rotation, making it a valuable tool in engineering and physics.

4. Can the Parallel-Axis Theorem be applied to all objects?

The Parallel-Axis Theorem can be applied to any rigid body, regardless of its shape or size, as long as its mass is distributed evenly. It cannot be used for objects that are not rigid, such as liquids or gases.

5. Are there any limitations to the Parallel-Axis Theorem?

The Parallel-Axis Theorem assumes that the object is rigid and that its mass is evenly distributed. It also does not take into account any external forces acting on the object. Additionally, it is only valid for rotation about a single axis.

Similar threads

  • Introductory Physics Homework Help
Replies
28
Views
544
  • Mechanics
Replies
2
Views
498
  • Classical Physics
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
634
Replies
10
Views
1K
Replies
25
Views
460
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
10
Replies
335
Views
8K
Replies
12
Views
354
  • Advanced Physics Homework Help
Replies
15
Views
1K
Back
Top