Parallel Axis Theorem Not Working

[laser]

Homework Statement

See description

Relevant Equations

I=1/2mr^2 for a disk
I=Icom + md^2

Using the trivial way, you get I = M(r+L)^2. This is definitely correct.

Out of curiosity, can the parallel axis theorem be applied here? I took the axis of rotation to be into the page, then calculated the moment of inertia for the disk, which is I = 1/2Mr^2. Next, I calculated the distance from the centre of the bob to P, which is r+L. So the moment of inertia "should" be I = 1/2MR^2 + M(r+L)^2, which contradicts my previous answer.

Why is the parallel axis theorem not applicable here?

 

[PeroK]

Using the trivial way, you get I = M(r+L)^2. This is definitely correct.

Is it?

 

[PeroK]

As the disk is symmetrical we can say that all of its mass acts through its centre of mass.

The distance from P to its centre of mass is L + d.

Is this not correct?

That's not correct. The simplest proof (that it's not correct) in this case is to apply the parallel axis theorem!

 

[laser]

That's not correct. The simplest proof (that it's not correct) in this case is to apply the parallel axis theorem!

sorry, I meant to say L + r

To rephrase:

As the disk is symmetrical we can say that all of its mass acts through its centre of mass.

The distance from P to its centre of mass is L + r.

 

[PeroK]

sorry, I meant to say L + r

To rephrase:

As the disk is symmetrical we can say that all of its mass acts through its centre of mass.

The distance from P to its centre of mass is L + r.

It wasn't the "d" that was the problem. You disproved your own statement yourself in the original post by applying the parallel axis theorem.

 

[laser]

It wasn't the "d" that was the problem. You disproved your own statement yourself in the original post by applying the parallel axis theorem.

Ah okay. I thought that there was a problem in the parallel axis theorem proof, some limitation to it that I was missing.

On the other hand, why is M(L+r)^2 wrong? The definition of moment of inertia is the integral of y^2 dm. Because we can say that it's a point particle (symmetrical disk), y is constant. And the integral of dm is just M.
(y = L + r)

 

[kuruman]

On the other hand, why is M(L+r)^2 wrong?

It's not wrong, it's an approximation in the limit $L>>r.$

Because we can say that it's a point particle (symmetrical disk), y is constant.

We cannot say that. The mass element $dm$ nearest to the axis of rotation is at $y=L-r$ and the farthest at $L+r.$ You have to do the integral between these two limits.

 

[PeroK]

We cannot say that. The mass element $dm$ nearest to the axis of rotation is at $y=L-r$ and the farthest at $L+r.$ You have to do the integral between these two limits.

... or use the parallel axis theorem!

 

[Orodruin]

It's not wrong, it's an approximation in the limit L>>r.

No it is not. It is most definitely wrong. It has terms at linear order in r, which do not appear if you do it correctly. For example by applying the parallel axis theorem.

 

[kuruman]

... or use the parallel axis theorem!

Sure. Doing the integral $\int dm~y^2$ for an arbitrarily shaped object is how the derivation of the parallel axis theorem proceeds.

 

[Steve4Physics]

@laser, if you haven't yet understood your mistake, here’s a little moment of inertia (MoI) exercise that should help.

P . . . . . . A-B

Point mass A = 1kg and distance PA =10m.
Point mass B = 1kg and distance PB =12m.
A and B are connected by a rod (length 2m) of negligible mass.

1. What is A’s MoI about P?
2. What is B’s MoI about P?
3. What is the total MoI about P (of the obect consisting of A, B and rod)?
4. What is the MoI about P of a 2kg point mass a distance 11m from P?

Compare your answers to 3 and 4 and think about why they are different.

 

[kuruman]

No it is not. It is most definitely wrong. It has terms at linear order in r, which do not appear if you do it correctly. For example by applying the parallel axis theorem.

Yes, it's not. I assumed that $L$ is the length to the center of the disk.

 

[Orodruin]

No it is not. It is most definitely wrong. It has terms at linear order in r, which do not appear if you do it correctly. For example by applying the parallel axis theorem.

… or rather … if L is actually the length of the wire (meaning the distance to the disc’s com) then the y-values of the integral will vary between L and L+2r.

We cannot say that. The mass element dm nearest to the axis of rotation is at y=L−r and the farthest at L+r. You have to do the integral between these two limits.

 

[Orodruin]

Yes, it's not. I assumed that $L$ is the length to the center of the disk.

Then this is wrong instead, as I pointed out in the cross post:

The mass element dm nearest to the axis of rotation is at y=L−r and the farthest at L+r. You have to do the integral between these two limits.

 

[laser]

It's not wrong, it's an approximation in the limit $L>>r.$

We cannot say that. The mass element $dm$ nearest to the axis of rotation is at $y=L-r$ and the farthest at $L+r.$ You have to do the integral between these two limits.

Just realised I messed up my notations... I was taking L to be P to nearest, so closest is L and farthest is L+2r. And the question says L is to the centre to P...

Regardless, let me get this straight. You cannot say that the disk is a point particle with all its mass acting through its centre of mass?

Because the disk is symmetrical though, wouldn't the integral just be L^2? (L is what I previously called L+r, let's just say it's the distance from the centre of mass to P.)

To give an analogy (I hope I haven't carried this misconception with me for several years :D), when doing Newton's law of gravitation with planets, we take "r" in F=Gmm/r^2 to be the distance from its centre of mass, because even though you have to do an integral will all the dm, because Earth is "basically a sphere", we can say that all its mass acts on its centre of mass. Is this also incorrect?

 

[erobz]

Regardless, let me get this straight. You cannot say that the disk is a point particle with all its mass acting through its centre of mass?

Just forget about the pendulum for a moment, if you could collapse the disk to a point, what would be its MOI through its center of mass?

 

[kuruman]

Then this is wrong instead, as I pointed out in the cross post

It is wrong too but consistent with my original misreading of the length $L$.

 

[laser]

@laser, if you haven't yet understood your mistake, here’s a little moment of inertia (MoI) exercise that should help.

P . . . . . . A-B

Point mass A = 1kg and distance PA =10m.
Point mass B = 1kg and distance PB =12m.
A and B are connected by a rod (length 2m) of negligible mass.

1. What is A’s MoI about P?
2. What is B’s MoI about P?
3. What is the total MoI about P (of the obect consisting of A, B and rod)?
4. What is the MoI about P of a 2kg point mass a distance 11m from P?

Compare your answers to 3 and 4 and think about why they are different.

OHH, it's the square, isn't it? 10^2 and 12^2 is not the same as 2*11^2

 

[Orodruin]

Because the disk is symmetrical though, wouldn't the integral just be L^2?

No. The integral is on the form:
$$
\int_{a-r}^{a+r} f(y) y^2 dy
$$
where $f(y)$ is symmetric around $y=a$ (actually it is not, but it is to first approximation). However, $y^2$ is not linear so you cannot use the symmetry argument to just take the value at $a$.

 

[kuruman]

Just forget about the pendulum for a moment, if you could collapse the disk to a point, what would be its MOI through its center of mass?

That's a good question. Is the point of collapse at the original position of the CM, namely at distance $L+r$ from the support or at the tip of the string at distance $L$?

 

[erobz]

That's a good question. Is the point of collapse at the original position of the CM, namely at distance $L+r$ from the support or at the tip of the string at distance $L$?

I see what you are saying about where the string is connected being vague. That was not my intention though.

I was just trying to have the OP use "the disk collapsing to a point mass by symmetry" argument, to find the MOI of a disk about its own center of mass, to let them find out whether that agrees with the accepted formula?

Maybe I'm under thinking what they aren't understanding though. To me it seems like they believe that you can take a disk, convert it to a point mass and side step the integration.

 

[Orodruin]

To me it seems like they believe that you can take a disk, convert it to a point mass and side step the integration.

I agree with this. However, the mere existence of the parallel axis theorem tells us that it is not the case. The parallel axis theorem tells us that the moment of inertia is what it would be if the object mass was collapsed to the center of mass plus the moment of inertia relative to the center of mass.

 

[laser]

I see what you are saying about where the string is connected being vague. That was not my intention though.

I was just trying to have the OP use "the disk collapsing to a point mass by symmetry" argument, to find the MOI of a disk about its own center of mass, to let them find out whether that agrees with the accepted formula?

Maybe I'm under thinking what they aren't understanding though. To me it seems like they believe that you can take a disk, convert it to a point mass and side step the integration.

You are right, I thought it was trivial that one could convert it to a point mass. However, thanks to this thread, I now realise that I cannot! :)

 

[Lnewqban]

The distance from P to its centre of mass is L + r.

Carefully read the wording of the problem.
There are 50 cm separating P from the center of the disc.

How the 60 cm diameter thin disc can remain vertical and not turning around the string, I don't know.

 

[haruspex]

I am mystified by this thread. The question clearly states that L is the distance from P to the centre of the disc, yet so many seem to have read it as the distance to the circumference.

But I see no discussion of how exactly the disc is moving. If only held by a wire at its centre (a loop of wire passing through the centre, say), with no friction between string and disc, it is unlikely to rotate on its own axis. So no parallel axis theorem needed: it can be reduced to a point mass, leading to $mL^2$.
On the other hand, part (a) says the string-disc combination is rotating about point P, which suggests the combination moves as a rigid body. Maybe the string is stuck to the sides of the disc.

How the 60 cm diameter thin disc can remain vertical and not turning around the string, I don't know.

You must be seeing some issue that eludes me. Are you suggesting it would flop sideways? Not if the whole wire is 80cm long, looping through the centre of the disc and back to itself at the perimeter.

 

[Lnewqban]

You must be seeing some issue that eludes me. Are you suggesting it would flop sideways? Not if the whole wire is 80cm long, looping through the centre of the disc and back to itself at the perimeter.

Yes, to flop sideways and also to rotate about the line made by the string.
Even if it would not, we would still need the string to drag the disc to make it simultaneously oscilate and rotate about its center (like a regular plumb would).

Yours would be a good assumption.
Perhaps with 90 cm we could make a knot tight enough to force the disc to rotate on the plane of the paper.

 

[PeterO]

Homework Statement: See description
Relevant Equations: I=1/2mr^2 for a disk
I=Icom + md^2

View attachment 337373

Using the trivial way, you get I = M(r+L)^2. This is definitely correct.

Out of curiosity, can the parallel axis theorem be applied here? I took the axis of rotation to be into the page, then calculated the moment of inertia for the disk, which is I = 1/2Mr^2. Next, I calculated the distance from the centre of the bob to P, which is r+L. So the moment of inertia "should" be I = 1/2MR^2 + M(r+L)^2, which contradicts my previous answer.

Why is the parallel axis theorem not applicable here?

The length of the cable is 50cm, the radius of the disk is 30 cm. Why do the setters of such questions provide such hopelessly out of scale diagrams?

 

[haruspex]

The length of the cable is 50cm, the radius of the disk is 30 cm. Why do the setters of such questions provide such hopelessly out of scale diagrams?

So that you don't arrive at the right answer by invalidly inferring some geometric relationship from the diagram? Nah, it's because the diagram was recycled with different parameters and the setter could not be bothered drawing a new one.

 

[PeterO]

So that you don't arrive at the right answer by invalidly inferring some geometric relationship from the diagram? Nah, it's because the diagram was recycled with different parameters and the setter could not be bothered drawing a new one.

Agreed, but so many diagram are poorly drawn How many times are angles are all drawn as 45 degrees with other sizes marked on them. Of course the sensible student re-draws their own "reasonable" diagram.