Object slides in rolling cylinder

[Nathanael]

Normal force = $mg(3+\frac mM)$.
Is that the result you got?

Yessir

Your method also uses an accelerating axis of rotation. I must say I'm a bit confused about when it is invalid to use accelerating axes of rotation.
Maybe it could be a topic for one of your insights-posts.

 

[haruspex]

Your method also uses an accelerating axis of rotation. I must say I'm a bit confused about when it is invalid to use accelerating axes of rotation.
Maybe it could be a topic for one of your insights-posts.

As I wrote in post #26, I withdraw my earlier objection to using such a frame for torque. The dangers of using an accelerating reference axis for using conservation of angular momentum is covered by one of my Insights posts (Moments).

 

[Nathanael]

As I wrote in post #26, I withdraw my earlier objection to using such a frame for torque. The dangers of using an accelerating reference axis for using conservation of angular momentum is covered by one of my Insights posts (Moments).

Okay, I didn't know you already covered it, I just knew you posted a lot about misconceptions and common tricky situations.

So torque is fine but cons.of angular momentum is not; that makes more sense, thanks!

 

[TSny]

The basic rotational dynamical law is net torque equals rate of change of angular momentum $\vec{\tau}_{net} = d\vec{L}/dt$. This law is not generally valid if you take your origin to be a point that is accelerating relative to an inertial frame.

Taking torques about the point of contact, P, between the cylinder and floor can be ambiguous.

For example, you can think of this point as a fixed point of the floor that happens to coincide instantaneously with the point of contact. In this case, P is a fixed point in an inertial frame. Then, the net torque acting on the cylinder about P will equal the rate of change of angular momentum of the cylinder with respect to P.

However, if you consider point P as moving with the point of contact so that P is always underneath the center of the cylinder, then P is now a point that is generally accelerating relative to an inertial frame. Now it will not be true in general that the net torque about P will equal the rate of change of angular momentum about P. For example the rate of change of angular momentum of the cylinder relative to P would not be $2MR^2 \alpha$ (it would be $MR^2 \alpha$). Setting up $\vec{\tau}_{net} = d\vec{L}/dt$ would lead to an incorrect answer.

As a trivial example consider a point particle falling freely as shown below. Let P be a fixed point in the inertial frame of the Earth that happens to be located at the same height as the particle at the time of interest. Then you can easily check that the torque due to gravity about P equals the rate of change of angular momentum of the particle about P.

Now let P accelerate with the particle so that P is always a distance d to the right of the particle. The torque about P is the same as before. But the particle always has zero angular momentum relative to P. So, relative to the accelerating point P, $\vec{\tau}_{net} \neq d\vec{L}/dt$.

 

[haruspex]

This law is not generally valid if you take your origin to be a point that is accelerating relative to an inertial frame.

But it's ok if it's the CoM of the object in question, because the acceleration of the body does not correspond to a (virtual) torque about that point, right?

Taking torques about the point of contact, P, between the cylinder and floor can be ambiguous.

I did clarify somewhere in the thread that I meant a fixed point on the ground.

 

[Nathanael]

@TSny
First I should say, when I said "accelerating axis of rotation" that is not really what I meant. I wasn't sure how to articulate myself (so I didn't try) but this is what I meant:
For the first instant we have a stationary axis somewhere, and then at a future instant we move the axis to somewhere else, but the axis is still stationary. So it's like it's instantaneously stationary, yet it's location is changing.
I'm still not sure if that makes sense... but that is what I've been imagining.

So in your example it's like we start the axis at the level of the particle, and then the particle falls a bit, and then we move the axis down a bit, but the axis is meant to be stationary at each instant.

Anyway, I think your example of a falling particle has made me realize when it is valid to do this:
It is valid to do this if the angular momentum is the same about all locations(that we use) of the axis. That way the torque we calculate through all these different stationary-axes is the rate of change of the same quantity.

For example, in the OP I took the torque (on the cylinder) to be about a ("instantaneously-stationary") axis which is continuously through the cylinder's CoM. The reason this works is because the angular momentum (of the cylinder) is the same through all points on that line which the CoM traces out.

Am I just making up nonsense or does this seem correct?

 

[andrewkirk]

I wonder whether some brave person is going to have to go all the way back to Noether's Theorem, to work out exactly what any valid rule of Conservation of Angular Momentum would say about the axis/axes about which angular momentum is conserved.

It's certainly not obvious that it is conserved about an axis that is accelerated relative to an inertial frame, or under what special conditions it might be conserved about such an axis.

 

[TSny]

But it's ok if it's the CoM of the object in question, because the acceleration of the body does not correspond to a (virtual) torque about that point, right?

Yes, I think that's a good way to look at it.

[EDIT: You can concoct examples where the torque equation about the CoM is not valid (even if the CoM has no acceleration!). To do this you need to have a frame of reference that's rotating relative to inertial frames. For example, consider a merry-go-round with a tangential force applied to the rim. If you choose a reference frame that rotates with the merry-go-round then the angular momentum of the system about the CoM in this frame is always zero even though there is a non-zero torque.

Maybe all this is obvious enough that I didn't need to bring it up, but I thought I would mention it anyway.]

 

[TSny]

@TSny
First I should say, when I said "accelerating axis of rotation" that is not really what I meant. I wasn't sure how to articulate myself (so I didn't try) but this is what I meant:
For the first instant we have a stationary axis somewhere, and then at a future instant we move the axis to somewhere else, but the axis is still stationary. So it's like it's instantaneously stationary, yet it's location is changing.
I'm still not sure if that makes sense... but that is what I've been imagining.

So in your example it's like we start the axis at the level of the particle, and then the particle falls a bit, and then we move the axis down a bit, but the axis is meant to be stationary at each instant.

I can see how this can be confusing. At any instant of time you can pick your origin anywhere you want. As long as this origin is thought of as a fixed point of an inertial reference frame then the torque equation will be valid relative to that origin and frame of reference. All velocities, accelerations, and torques are measured relative to that particular inertial frame and that origin for that instant of time. These velocities, accelerations, and torques do not depend in any way on how you are going to pick the frame of reference and origin at the "next" instant of time. You could use a random number generator to arbitrarily pick the inertial frame and origin at each instant of time. So your "instantaneously stationary" axes could be jumping all over the place from one instant to another.

However, if at some instant of time you choose an origin that is fixed to some non-inertial frame then you would calculate your velocities, accelerations, etc. relative to this non-inertial frame at that instant of time. Then you would generally not find the torque equation to be valid.

Hope this helps some.

 

[Nathanael]

I can see how this can be confusing. At any instant of time you can pick your origin anywhere you want. As long as this origin is thought of as a fixed point of an inertial reference frame then the torque equation will be valid relative to that origin and frame of reference. All velocities, accelerations, and torques are measured relative to that particular inertial frame and that origin for that instant of time. These velocities, accelerations, and torques do not depend in any way on how you are going to pick the frame of reference and origin at the "next" instant of time. You could use a random number generator to arbitrarily pick the inertial frame and origin at each instant of time. So your "instantaneously stationary" axes could be jumping all over the place from one instant to another.

However, if at some instant of time you choose an origin that is fixed to some non-inertial frame then you would calculate your velocities, accelerations, etc. relative to this non-inertial frame at that instant of time. Then you would generally not find the torque equation to be valid.

Hope this helps some.

Right, thanks. I try to avoid non-inertial frames, but if I do use them I will keep in mind to be careful with angular momentum.

But about these "instantaneously stationary" axes: I believe this method (of switching axes at each instant) also would not work in general. It would only work if the set of points that we (or the RNG) choose from have this property: at any instant in the relevant time interval the angular momentum of our system is the same about all of these points.
I never realized this restriction before, so I'm glad to learn the limitation of using this method.

Of course, the only purpose of using this method is to simplify calculations, but in a problem like this one it really makes a difference over using an axis at a constant location (as I began to do in post #3).

 

[haruspex]

However, if at some instant of time you choose an origin that is fixed to some non-inertial frame then you would calculate your velocities, accelerations, etc. relative to this non-inertial frame at that instant of time. Then you would generally not find the torque equation to be valid.

If we take it back to first principles, for a mass element dm at $\vec r$ in a fixed reference frame O, torque $d\tau=\vec r\times \ddot{\vec r}.dm$. So $\tau=\int \vec r\times \ddot{\vec r}.dm$. If we now allow the reference point an acceleration $\vec a$, $\tau'=\int \vec r\times (\ddot{\vec r}-\vec a).dm = \tau - (\int \vec r.dm )\times\vec a$.
$\tau'=\tau$ if:

• O is the mass centre (P), or
• $\vec a=0$, or
• the acceleration is along the line OP

In the present case, the point of the cylinder making contact is valid.