Why is Icm not needed in the Parallel Axis Theorem?

[ChloeYip]

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.

 

[PeroK]

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?

 

[ChloeYip]

Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.

 

[PeroK]

You mean because the question has the word "parallel" in it?

 

[ChloeYip]

My teacher told me that when ever the axis isn't past though center of mass, we can use it...
Isn't it?

 

[PeroK]

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.

 

[ChloeYip]

"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...

 

[PeroK]

"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?

 

[ChloeYip]

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

 

[PeroK]

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.

 

[ChloeYip]

moment of inertia is 0

meaning no rotation?

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?

 

[PeroK]

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!

 

[ChloeYip]

thanks