Why is Icm not needed in the Parallel Axis Theorem?

In summary, the conversation is about finding the moment of inertia of a uniform bar with two small balls attached to its ends about an axis parallel to the bar and 0.500 m from it. The parallel axis theorem is discussed, with the correct equation being I = Md^2. The student also mentions the importance of considering the direction of rotation and the assumption of the rod being one-dimensional. Ultimately, the conversation serves as a reminder to think carefully about the given problem before applying any equations or theorems.
  • #1
ChloeYip
93
1

Homework Statement



Homework Equations


Parallel axis theorem: Ip = Icm + Md^2
Icm = I = ML²/12 + 2 * mr²

3. The attempt
Ip = Icm + Md^2 ==> wrong
I = Md^2 ==> right

Why don't I need to add "Icm"?
Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
ChloeYip said:

Homework Statement


A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 5.00 kg ,while the balls each have mass 0.300 kg and can be treated as point masses.
Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.

Homework Equations


Parallel axis theorem: Ip = Icm + Md^2
Icm = I = ML²/12 + 2 * mr²

3. The attempt
Ip = Icm + Md^2 ==> wrong
I = Md^2 ==> right

Why don't I need to add "Icm"?
Thanks.

Why do you think you need the parallel axis theorem?
 
  • #3
ChloeYip said:
Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.
 
  • #4

You mean because the question has the word "parallel" in it?
 
  • #5
My teacher told me that when ever the axis isn't past though center of mass, we can use it...
Isn't it?
 
  • #6
ChloeYip said:
My teacher told me that when ever the axis isn't past though center of mass, we can use it...
Isn't it?

You can use it. But, because you haven't analysed the question properly, you haven't thought about the moment of inertia of the bar in this question.
 
  • #7
"Part A
Find the moment of inertia of this combination about an axis perpendicular to the bar through its center.
Express your answer with the appropriate units.

I = 2.27 kg⋅m2

Correct"I have calculated the Icm, and it is right...
 
  • #8
ChloeYip said:
"Part A
Find the moment of inertia of this combination about an axis perpendicular to the bar through its center.
Express your answer with the appropriate units.

I = 2.27 kg⋅m2

Correct"I have calculated the Icm, and it is right...

For which direction of rotation have you calculated the moment of inertia?
 
  • #9
Do you mean the axis of rotation (dimension) is different this time?
But why don't we need to calculate Icm in the required direction and then add Md^2?
Thanks
 
  • #10
ChloeYip said:
Do you mean the axis of rotation (dimension) is different this time?
But why don't we need to calculate Icm in the required direction and then add Md^2?
Thanks

There isn't only one moment of inertia for a rigid body for rotation about its centre of mass. It depends on the direction you rotate it. Can you see how to rotate a bar so that its moment of inertia is 0? Hint: you are assuming the bar is one-dimensional in this problem.
 
  • #11
PeroK said:
moment of inertia is 0
meaning no rotation?
PeroK said:
depends on the direction you rotate it
I understand this, that means answer in part a can't apply on this question. But, why don't we need to calculate Icm in the required direction for this question? (Let Icm pass through centre of mass and parallel to the bar)
Huhh, no radius of the rod is given, do you mean it is assuming the radius of the rod is zero?
 
  • #12
ChloeYip said:
meaning no rotation?

I understand this, that means answer in part a can't apply on this question. But, why don't we need to calculate Icm in the required direction for this question? (Let Icm pass through centre of mass and parallel to the bar)
Huhh, no radius of the rod is given, do you mean it is assuming the radius of the rod is zero?

Yes, unless the rod is given a radius, you treat it as a one-dimensional body. So, you must take it to have 0 moment of inertia if it is rotated about its axis. The masses at each end are points, so they have 0 moment of inertia about any axis through them.

Perhaps it was something of a trick question. But, it did catch you out, because you applied the parallel axis theorem without thinking about the problem carefully enough. Don't let them catch you out again!
 
  • #13
thanks
 

FAQ: Why is Icm not needed in the Parallel Axis Theorem?

What is the Parallel Axis Theorem?

The Parallel Axis Theorem is a physics concept that relates to the moment of inertia of an object. It states that the moment of inertia of an object about a certain axis is equal to the moment of inertia of the object about a parallel axis through its center of mass, plus the product of the mass of the object and the square of the distance between the two axes.

Why is the Parallel Axis Theorem important?

The Parallel Axis Theorem is important because it allows us to calculate the moment of inertia of an object about any axis, not just the center of mass. This is useful in many physics and engineering applications, such as calculating the stability of rotating objects or designing structures that can withstand rotational forces.

How is the Parallel Axis Theorem derived?

The Parallel Axis Theorem can be derived using the moment of inertia of a continuous body formula, which involves integrating the mass of the object with respect to its distance from the axis of rotation. By manipulating this formula, we can arrive at the Parallel Axis Theorem.

What are some examples of the Parallel Axis Theorem in action?

The Parallel Axis Theorem can be seen in various real-world situations, such as a spinning top (where the moment of inertia changes as the axis of rotation moves away from the center) or a figure skater performing a spin (where they can change their moment of inertia by extending or contracting their arms).

Are there any limitations to the Parallel Axis Theorem?

While the Parallel Axis Theorem is a useful concept, it does have some limitations. It assumes that the object being rotated is rigid and has a constant mass distribution. It also does not take into account the effects of external forces on the object's moment of inertia.

Similar threads

Back
Top