When is the parallel axis theorem not appliable?

[Haorong Wu]

TL;DR

When is the parallel axis theorem not appliable?

Hello, there. A friend asked me a problem last night.

Suppose that a system consists of a rod of length $l$ and mass $m$, and a disk of radius $R$. The mass of the disk is negligible. Now the system is rotating around an axis in the center of the disk and perpendicular to the plane where the rod and the disk lie in. What is the moment of inertia of the rod?

It is easy to have, by the definition,
$I=\int_R^{R+l} r^2 dm=\frac {m} {l}\int_R^{R+l} r^2 dr=\frac {m} {3l} [ \left (R+l \right ) ^3 -R^3 ]$.

Meanwhile, from the parallel axis theorem, I have
$I^{'}=\frac {ml^2} {12}+m (R+l/2)^2$.

Obviously, they do not match each other, since $I$ contains a term with $R^3$, while $I^{'}$ does not.

I read the parallel axis theorem over and over again, but I could not see where I misunderstand.

Suppose a body of mass $m$ is rotated about an axis $z$ passing through the body's centre of mass. The body has a moment of inertia $I_{cm}$ with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis $z^′$ which is parallel to the first axis and displaced from it by a distance $d$, then the moment of inertia $I$ with respect to the new axis is related to $I_{cm}$ by
$I=I_{cm}+md^2$.
Explicitly, $d$ is the perpendicular distance between the axes $z$ and $z′$.

 

[Leo Liu]

Can you post a picture of the apparatus? How is the rod attached to the disk?

 

[Haorong Wu]

Can you post a picture of the apparatus? How is the rod attached to the disk?

Hi, @Leo Liu . I update it. The rod just attach to the disk on its perimeter.

 

[kuruman]

It is easy to have, by the definition,
$I=\int_R^{R+l} r^2 dm=\frac {m} {l}\int_R^{R+l} r^2 dr=\frac {m} {3l} [ \left (R+l \right ) ^3 -R^3 ]$.

Meanwhile, from the parallel axis theorem, I have
$I^{'}=\frac {ml^2} {12}+m (R+l/2)^2$.

Obviously, they do not match each other, since $I$ contains a term with $R^3$, while $I^{'}$ does not.

It's not that obvious to me. Expand each expression and take the difference $I-I'$. Show your work here to convince me that this difference is not equal to zero.

 

[hutchphd]

Yes work it out. The $R^3$ terms clearly cancel.

 

[Haorong Wu]

Awkward. Thanks, @kuruman, and @hutchphd. I was misled by the answer along with the problem.

It is clearly wrong. The plus sign should be a minus sign before $R^3$.

I first tried it with the parallel axis theorem, and I could not find the $R^3$ term, so I got lost.

 

[sophiecentaur]

I first tried it with the parallel axis theorem, and I could not find the R3 term

Perhaps if you follow the basic derivation of the PAT you will find a similar cancellation of R³ terms.