# Insights Explaining Rolling Motion - Comments

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1. Oct 7, 2016

### kuruman

There is something to be said in favor of picking the point of contact as the axis of rotation as far as simplification is concerned. One can always write the total kinetic energy as $K = \frac{1}{2} I_P \omega^2$ where $I_P$ is the moment of inertia about the point of contact or axis of rotation. One can then use Steiner's (a.k.a. parallel axes) theorem to relate $I_P$ to the moment of inertia about the center of mass. I adopted this point of view when I found out that students had difficulties "wrapping their head around" the dual concept of kinetic energy of the center of mass and kinetic energy about the center of mass. My method sorts these terms out automatically.
Although I do not dispute what you are saying in general, I differ with you on this point. It is a mistake within the context of what I am trying to accomplish with this series of demonstrations. Before writing equations of motion for rolling objects, it is necessary for students to have a clear and fundamental understanding of what an axis of rotation is and how to find it. The purpose of the demonstrations is to do just that. Having this understanding is a precursor to describing the motion, whether it be in the lab frame or the CM frame. But OK, perhaps "wrong answer" in the original essay was a harsh label. "Misinformed answer" would have been more appropriate.

2. Oct 7, 2016

### vanhees71

I disagree. You can choose any body-fixed reference point (it can even be a ficitious point outside of the body, it must only be rigid wrt. the body; of course, I assume an idealized rigid body here), and $\vec{\omega}$ is then related to this point. Of course, $\vec{\omega}$ changes when you change the body-fixed reference point.

3. Oct 7, 2016

### kuruman

True but not helpful when teaching freshman physics. Of the infinity of points that one is free choose to describe the rotation, the set of points that are at rest with respect to the observer makes the description easier. The geocentric system is not wrong, it just complicates the description of planetary motion. Instead, we use the heliocentric system that assumes an observer at rest with respect to the Sun.

4. Oct 7, 2016

### vanhees71

True! The choice of the right coordinates is the first step to the solution of a mechanical problem. I don't say that it's wrong to use the momentary point of contact. It's maybe even a clever choice for such problems, but I had a very hard time to understand this when I was a freshmen, while the treatment in the other frame was a piece of cake, but that's of course subjective.

5. Oct 7, 2016

### FallenApple

Hmm Interesting. So if the axis of rotation is the instantaeous point of contact, then why is the regular moment of Inertia used? Shouldn't it be Icm+mR2.?

6. Oct 7, 2016

### kuruman

I am not sure what you mean by "regular" moment of inertia. As I mention in post#26, one ought to use the moment of inertia about the point of contact P when writing the total kinetic energy as one term. Of course the moment of inertia about P is related to the moment of inertia about the cm through the parallel axes theorem that you quoted, IP = Icm + md2, where d is the distance between the point of contact P and the cm. For the rolling wheel, d = R.

7. Oct 18, 2016

### ObjectivelyRational

Hmm. Assume a self propelled wheel (bicycle) and a big sphere. The sphere being the size of the earth with its COG at its center. Due to conservation of momentum the little wheel spins around its COG while the big sphere VERY slowly spins around its own axis of rotation.

The contact point is very important because it is the fulcrum of applied torque(s)... the interaction point between the two bodies that transfers momentum. Still trying to get a sense why there is "one correct" way to look at the system.

8. Nov 11, 2016

### ThingsCanMove

Hello all,

First post. This is really confusing to me. Lets say I had a solid wheel on the ground being driven by a specific torque without slipping. So with the center of rotation being through the middle of the wheel the Moment of Inertia would be (1/2)*mr2. Yet with the center of rotation being the ground the moment of inertia changes to 3/2mr2. So if angular acceleration=Torque/I why is the angular acceleration different just based on my reference frame when the wheel is driven by the same torque, it's just two ways of describing the same problem.

9. Nov 12, 2016

### vanhees71

If you change the body-fixed reference point to describe the motion of the rigid body not only the tensor of inertia changes (according to Steiner's rule) but also the angular momentum and torque. Since the description of the motion of the body is independent on the choice of the body-fixed reference frame (i.e., the body-fixed point and orientation of the body-fixed coordinate basis) all equations must transform consistently. That's why I said above that it is your arbitrary choice how you describe the motion.

10. Nov 12, 2016

### kuruman

To augment what @vanhees71 wrote, an arbitrary choice of describing the motion will do the job but it can get messy. Of all the infinite choices, two reference points stand out as giving simple results: the point of contact P and the center of the wheel O. Here is why.

Assume the wheel accelerates to the right.
1. Choose P as reference. With respect to P, the center of the wheel O has instantaneous speed $v_O=\omega_PR$ from which $$\omega_P=\frac{v_O}{R} \rightarrow \alpha_P=\frac{1}{R}\frac{dv_O}{dt}$$
2. Choose O as reference. With respect to O, the point of contact of the wheel P has instantaneous speed $v_P=\omega_OR$ from which $$\omega_O=\frac{v_P}{R} \rightarrow \alpha_O=\frac{1}{R}\frac{dv_P}{dt}$$
In the general case, an observer on the ground sees point O moving to the right with speed $v_O$. An observer on the wheel (at rest relative to point O) sees point P moving to the left with speed $v_P$ and the ground moving also to the left with speed $v_O$. In the specific case of rolling without slipping, the ground is instantaneously at rest relative to point P at all times. Therefore, the instantaneous speeds are equal, $v_P=v_O$, and hence their time derivatives are equal which makes $\alpha_P=\alpha_O$.

Choosing either one of these frames of reference has the advantage that the acceleration of the center of mass is related to the angular acceleration by $a_{cm}=\alpha R$. Personally, I prefer to use point P as reference because then I don't have to find the force of static friction in order to calculate the torque generated by it.

11. Nov 28, 2016

The COG CAN be considered as an axis of rotation. What else could one possibly consider as an axis of rotation? The mathematics would be least complicated when the axis of rotation passes through COG and perpendicular to the plane of the surface of the coin.

12. Nov 28, 2016

By considering the centre of rotation as the instantaneous point of contact, it is actually harder to visualize and do problems.Usually, where I study, we consider the axis of rotation as any line of symmetry and make necessary adjustments. This way, complications will be avoided.

13. Nov 11, 2017

### greypilgrim

"I begin by defining the axis of rotation of a rigid body as the set of all points that are (instantaneously) at rest while all the other points rotate about it with angular speed ω."

Isn't that an unusual, coordinate-dependent definition of rotation? Why do you demand the axis to be at rest? If you drop this condition, the centre immediately becomes an option too.

I don't even think the argument is consistent. If I understand your definition of the IPOC correctly, each of them only exists at a point in time, so they don't even have a trajectory and the property "at rest" is meaningless.

Good point. If the IPOC really is the axis of a (non-accelerated) rotation, why aren't the points on the wheel accelerated towards it?

Further abstracting OP's argument, one could describe the motion of a point along $\begin{pmatrix}t\\1\end{pmatrix}$ by an infinitesimal rotation around the axis at $\begin{pmatrix}t\\0\end{pmatrix}$ at each point in time $t$, which would be quite silly.

14. Nov 11, 2017

### kuruman

Perhaps I should have said "... the set of all points that are (instantaneously) at rest with respect to the chosen frame of reference while ... ". All motion is relative and I pick the axis of rotation to be at rest relative to whatever frame of reference one chooses to express velocities.
I don't think so. It is based on what I said above. It indeed is coordinate-dependent, but it is not unusual because it defines the axis of rotation in the whatever reference frame has already been chosen.
They are. The instantaneous velocity vector of the point of interest on the rolling wheel is perpendicular to the position vector from the point of contact to the point of interest. In what direction is the centripetal acceleration relative to the point of contact?
I don't understand the abstraction. Can you explain it, especially the part that is silly?

Last edited: Nov 11, 2017
15. Nov 11, 2017

### greypilgrim

Sorry, I've misunderstood. I thought the IPOC was on the table, while it's actually on the wheel. But this means you're working in an upwards accelerated frame. Did you really intend to treat this in a non-inertial system?

Last edited: Nov 11, 2017
16. Nov 11, 2017

### kuruman

No, the IPOC is on the inertial frame of the table. Let P be the point of contact on the wheel and Q be the point of contact on the surface. If the wheel rolls without slipping, points P and Q are at rest relative to each another. However, point P has an upward acceleration in the inertial frame of the surface whereas point Q has zero acceleration in that same frame. The inertial frame to work in is the table surface with Q as the origin. Consider an arbitrary point S on the wheel defined by position vector $\vec{r}_S$ from point Q to S. Then the velocity of point S (relative to point Q), $\vec{V}_S$, is perpendicular to $\vec{r}_S$. If the center of the wheel is moving with velocity $\vec{V}_C$, then the angular velocity is $\omega=V_C/R$. This makes the speed $V_S=\omega~r_S=V_C(r_S/R)$.

17. Nov 12, 2017

### greypilgrim

Well you can't have both
and
While
is true for both P and Q, the centripetal acceleration vectors of points on the wheel are only directed to P in its rest frame (which is not inertial, but accelerated upwards), not to Q in its rest frame (the inertial frame of the table). If a wheel rotates with constant $\omega$, the centripetal acceleration of all points is directed radially inwards, adding a constant velocity doesn't change that.

How can you claim that Q is at rest but the center of the wheel is not? They are exactly above each other at each point in time, and the center of the wheel is obviously movint to the right.

18. Nov 12, 2017

### kuruman

I don't believe I made (or implied) such a claim.

In the inertial frame of the table,
1. Q is at rest because it is a point on the table.
2. P (on the rim of the wheel) is at rest with respect to the surface because the wheel is rolling without slipping.
3. The center of the wheel, C, is moving to the right (say) with speed VC.
Therefore the center of the wheel is moving to the right with speed VC. The translational motion of the center can also be construed as rotational motion about Q with the angular speed ω = VC / R. The centripetal direction is not straight up as you assert, but straight down.

Here is a formal derivation. Consider the center C at some point in time when it is not directly above Q and is moving parallel to the surface to the right (see figure below). At that point, we have
$$x = R \tan \theta$$
$$V_C=\frac{dx}{dt}=R(\tan^2 \theta+1)\frac{d\theta}{dt}=R(\tan^2 \theta+1)\omega$$
$$\omega = \frac{V_C}{R(\tan^2 \theta+1)}$$
Point C is rotating about Q with angular speed $\omega$ as expressed above. If point C is rotating about Q, what is the centripetal direction?

When C is directly above Q, $\theta = 0$ and $\omega=V_C/ R$. What is the centripetal direction now?

19. Nov 12, 2017

### greypilgrim

I wasn't talking about C but P.

And since $V_C$ is constant, it follows that $a_C=\frac{d^2x}{dt^2}=0$, hence C is not accelerated, as is obvious for a linear motion with constant velocity.

Of course you can define straight down as the "centripetal direction", but this doesn't change the fact that C is not accelerated towards that direction as it should be in a rotation. In fact the only point on the wheel accelerated towards Q is the uppermost point of the wheel, but the magnitude of the acceleration is wrong for a rotation around Q with $\omega$. Point P is even accelerated away from Q.

20. Nov 12, 2017

### kuruman

Please reread my post #43. My arguments apply to the inertial frame of reference. "Straight down" is not defined as the "centripetal direction". The centripetal direction is defined as the direction from the point of interest to the center of rotation. As seen in the figure, if point Q is the chosen center of rotation, point C rotates about Q with angular speed as derived.
Not so. It looks like you are limiting your thinking to one-dimensional motion. Examine the figure. Point C undergoes linear motion at constant velocity, true enough. Yet, relative to point Q, its velocity has a radial component (the radius being the line joining Q and C) and a tangential component. In two dimensions, point C moves radially away from Q while at the same time rotates clockwise about Q. One can always replace cartesian coordinates with polar coordinates.

21. Nov 12, 2017

### greypilgrim

I'm not disputing that you can choose whatever coordinate system you like to describe the situation. I'm just saying that it's very unnatural to call it a "rotation", for the following reasons:
1. What's "at rest" is coordinate-dependent.
2. The points on the wheel are not accelerated towards Q.
3. The velocity of the points on the wheel have radial components, which is weird for a rotation of a rigid body.
If we choose C as the axis (dropping the restriction that an axis must be "at rest"), as most people intuitively do, none of those problems occur.

Also, you still haven't explained how you're dealing with the fact that the IPOC Q is at a different place at every point in time. How is that compatible with it being at rest?

If a point travels along $\begin{pmatrix}t\\1\end{pmatrix}$ in cartesian coordinates, you can of course express this with polar coordinates with a changing radius and a changing angle. But this is not enough to make this motion a rotation.

22. Nov 12, 2017

### kuruman

I agree. However, I am examining the situation from the inertial frame of the table not the non-inertial frame of the wheel.
Please examine the drawing in post #43. Point Q is not moving relative to the surface. If Q were the IPOC, it would be drawn directly underneath point C.
That's a matter of opinion. Consider the simpler case of a block sliding on a frictionless surface. C in the figure marks the center of mass and point Q is fixed on the surface. The block has constant angular momentum $L=mV_cR$ relative to Q. Now consider force $F$ acting on C. Clearly the angular momentum will change because the velocity will change. Changing angular momentum implies the action of a non-zero torque. Non-zero torque implies non-zero angular acceleration. Non-zero angular acceleration implies changing angular velocity. Angular velocity implies rotation about a center, in this case Q because that's the point about which the angular momentum is expressed. To me this chain of reasoning is sufficient to justify rotation even when there is no rolling motion.

23. Dec 2, 2017

### Robert Morphis

You started with a seemingly arbitrary assertion about direction of rotation and application of force, then you "prove" your point by applying force to the center of the spool while pretending to apply it elsewhere using the pvc pipe.

The starting assertion should (IMO) be, applying force on one side of the axis creates rotation in one direction, applying it to the other creates rotations in the other, and applying force at the axis causes no rotation.

Yes, one certainly can analyze the problem using the IPOC, but it is not the only way to do it, and I don't really see that you have proved anything.

Last edited by a moderator: Dec 2, 2017
24. Dec 2, 2017

### kuruman

It looks like you missed the point that when the force is applied at a point that is vertically even with the IPOC, there is no torque generated by that force. This level marks the threshold below which the torque is in one direction and above which it is in the opposite direction. Given that information, at what level would you say the axis of rotation is?

25. Dec 2, 2017

### Robert Morphis

If I apply a force at the IPOC I will either slow down or speed up the rotation, unless friction between the surface and the wheel prevents that force from actually acting on the wheel.

Use a cylinder rolling on two rails to demonstrate.

The starting assertion should (IMO) be, applying force on one side of the axis creates rotation in one direction, applying it to the other creates rotations in the other, and applying force at the axis causes no rotation.