Does a relativistic rolling ball wobble?

[1effect]

It raises the question about where exactly is the centre of gravity of a rolling ball (or wheel) is in the relativistic context and does it stay it stay parallel to the direction of motion (no wobble)?

This question has already been answered. The distance from the center of gravity to the surface is constant.
In the frame comoving with the sphere there is no length contraction.
In any moving frame there is no Lorentz contraction for directions perpendicular to the direction of motion of the sphere. :-)

 

[JesseM]

As I mentioned to brw6 I agree with most of what you say except for the fact that lorentz transformation of a rolling body is lot more complicated than it might first appear to be. The motion involves transcendental functions that can not be solved empirically and require iteration techniques to come to an approximate solution.

Are you saying these transcendental functions would appear in the rest frame of the rolling body's center, or only when you try to transform into a frame where the center is moving? What was the # of the post where you mentioned this to brw6?

A geometrical solution of the motion of one particle on the rim of the rolling body can be arrived at as follows:

Observe that in the rest frame of the rolling body's rotation axis the particle follows a cycloid path. (The equation for a cycloid only has a parametric solution). In the moving frame the cycloid path of the rim particle is stretched by a factor of gamma. If we superimpose a moving ellipse (to represent the length contracted disk) on the cycloid path then the intersection of the elliptical locus and the stretched cycloid locus is the location of the rim particle at any instant in the transformed frame. The trouble is that the mathematical determination of the intersection is very difficult. Some of the difficulties stem from the fact that there is no exact equation for the perimeter length of an ellipse, there is no non parametric solution to a cycloid and the inverse of transcendental functions can not be found without iterative techniques.

If you have parametrized the path of any point on the rim in terms of x(t), y(t) and z(t) in the rest frame of the center of the rolling object, then presumably you would then just use the Lorentz transformation on these equations, although the resulting equations might indeed be hard to solve exactly for x' in terms of t', y' in terms of t' and z' in terms of t' (but you could at least get an arbitrarily good numerical approximation of the path just by picking events on the path in the first coordinate system and then finding their coordinates in the second system).

 

[yuiop]

This question has already been answered. The distance from the center of gravity to the surface is constant.
In the frame comoving with the sphere there is no length contraction.
In any moving frame there is no Lorentz contraction for directions perpendicular to the direction of motion of the sphere. :-)

Your solution just says a rolling ball still looks like a sphere due to the optical illusion of Terrell rotation and due to light taking longer to reach the observer from the parts of the ball furthest from the observer. None of your example include rolling rotation (only rotation around the tranverse axis).

I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.

 

[yuiop]

Are you saying these transcendental functions would appear in the rest frame of the rolling body's center, or only when you try to transform into a frame where the center is moving?

The transcendental functions enter into the picture as soon as wheel starts moving at even at non relatavistic velocities and before any transformations are considered. It is do with there being no non parametric solution for the cycloid path of a particle on the rim of a wheel in terms of x and y without knowing the angle theta of the particle with respect to the wheel axis.

What was the # of the post where you mentioned this to brw6?

It was in post #27 when I just reiterating some things you mentioned earlier:

I think as JesseM points out, if you want to understand the transverse motion of a rolling ball you should start with something simpler like a rolling disk or a rolling wheel and understand that first. JesseM is also right that we can simulate the rolling effect by taking a spinning ball or wheel and superimposing linear motion. The only stipulation is that the rim velocity of the ball or wheel in the rest frame of its axis is equal to the superimposed linear motion for rolling without slipping. ...

If you have parametrized the path of any point on the rim in terms of x(t), y(t) and z(t) in the rest frame of the center of the rolling object, then presumably you would then just use the Lorentz transformation on these equations, although the resulting equations might indeed be hard to solve exactly for x' in terms of t', y' in terms of t' and z' in terms of t' (but you could at least get an arbitrarily good numerical approximation of the path just by picking events on the path in the first coordinate system and then finding their coordinates in the second system).

The difficulty is that simultaneity of the points on the rim is continuously changing. I did a simultion of the motion in geometry software a while ago, When I have more time I will try and post some pics. As far as I know there is no simple non iterative equation for the transformed rolling motion, but I would like to be proved wrong ;)

 

[1effect]

Your solution just says a rolling ball still looks like a sphere due to the optical illusion of Terrell rotation and due to light taking longer to reach the observer from the parts of the ball furthest from the observer.

You are misquoting me, see the correct quote here.

"There are two effects at work , one is the well-known relativistic length contraction, the other one is the difference in arrival of light rays. "

As an aside, "light taking longer to reach the observer from the parts of the ball furthest from the observer" and "the Terrell effect" are one and the same thing :-)

None of your example include rolling rotation (only rotation around the tranverse axis).

Yes, this is true, I pointed that out in one of my posts. I also mentioned that it is not clear why this is relevant.

I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.

Are you saying that the axis of rotation describes a cycloid? I would be very interested in the proof.

 

[JesseM]

The difficulty is that simultaneity of the points on the rim is continuously changing.

How is that a difficulty? If you know the movement of the point in the frame of the center, the Lorentz transformation will tell you the time of any event on that point's worldline in another frame. It's really just a problem of the equations being difficult or impossible to solve exactly. For example, if the point on the rim is described by x'(t') = cos(t') and y'(t') = sin(t') in the rest frame of the center, applying the Lorentz transformation would give the equations:

gamma*(x - vt) = cos(gamma*(t - vx/c^2))
y = sin(gamma*(t - vx/c^2)

The difficulty is just in solving the first equation for x (if you could do that, you could plug the answer into the second equation to get y(t)). There may not be an exact solution, I don't know. But if you're just interested in a visual depiction of the path, it's an easy enough matter to just pick a bunch of different t' coordinates, find the corresponding x' and y' coordinates for each one, then convert each (x', y', t') to an (x, y, t) using the Lorentz transform.

 

[1effect]

How is that a difficulty? If you know the movement of the point in the frame of the center, the Lorentz transformation will tell you the time of any event on that point's worldline in another frame. It's really just a problem of the equations being difficult or impossible to solve exactly. For example, if the point on the rim is described by x'(t') = cos(t') and y'(t') = sin(t') in the rest frame of the center, applying the Lorentz transformation would give the equations:

gamma*(x - vt) = cos(gamma*(t - vx/c^2))
y = sin(gamma*(t - vx/c^2)

The difficulty is just in solving the first equation for x (if you could do that, you could plug the answer into the second equation to get y(t)). There may not be an exact solution, I don't know. But if you're just interested in a visual depiction of the path, it's an easy enough matter to just pick a bunch of different t' coordinates, find the corresponding x' and y' coordinates for each one, then convert each (x', y', t') to an (x, y, t) using the Lorentz transform.

Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations. Here is the trick:

x'^2+y'^2=1

\gamma^2 (x-vt)^2+y^2=1

y=sin(t)
\gamma (x-vt)=cos(t)
so:
x=\gamma^{-1} cos(t)+vt

 

[JesseM]

Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations. Here is the trick:

x'^2+y'^2=1

\gamma^2 (x-vt)^2+y^2=1

y=sin(t)

How did you get y=sin(t)? Wasn't the equation in the primed frame y'=sin(t'), which means it should be y=sin(gamma*(t - vx/c^2))

 

[1effect]

How did you get y=sin(t)? Wasn't the equation in the primed frame y'=sin(t'), which means it should be y=sin(gamma*(t - vx/c^2))

I parametrized :

\gamma^2 (x-vt)^2+y^2=1

There is an indefinite number of such parametrizations, they all produce the same curve.

For example, I could have chosen \frac{1-t^2}{1+t^2} and \frac{2t}{1+t^2}

 

[JesseM]

I re-parametrized

\gamma^2 (x-vt)^2+y^2=1

I don't understand what you mean by "re-parametrized" here. Can you fill in the missing steps that allow you to go from that equation to y=sin(t)? Unless you mean that y=sin(t) is meant to be the definition of a new parameter t, but then you should have given the parameter a different name than the time coordinate that appears in (x-vt).

 

[JesseM]

I parametrized :

\gamma^2 (x-vt)^2+y^2=1

There is an indefinite number of such parametrizations, they all produce the same curve.

For example, I could have chosen \frac{1-t^2}{1+t^2} and \frac{2t}{1+t^2}

But if you're just parametrizing y with a newly-defined parameter, you can't use the t-coordinate as the parameter! The t-coordinate already has a specific meaning--if you have some y' and t' which lie on the worldline of the point in the primed frame, you can't substitute in y=sin(t) unless it's true that when you do the Lorentz transform on that y' and t' on the point's worldline, the resulting y and t actually satisfy y=sin(t). If you want to define a new parameter you should give it a different name, like y=sin(p).

 

[1effect]

I don't understand what you mean by "re-parametrized" here. Can you fill in the missing steps that allow you to go from that equation to y=sin(t)? Unless you mean that y=sin(t) is meant to be the definition of a new parameter t, but then you should have given the parameter a different name than the time coordinate that appears in (x-vt).

I am not sure about that , I will need to think about it.
This would result into :

y=sin(\tau)

x=\gamma^{-1} cos(\tau)+vt

 

[JesseM]

I am not sure about that , I will need to think about it.
This would result into :

y=sin(\tau)

x=\gamma^{-1} cos(\tau)+vt

But that isn't very helpful, because you don't know what value of t corresponds to a given value of \tau.

 

[1effect]

But that isn't very helpful, because you don't know what value of t corresponds to a given value of \tau.

Yes, this wouldn't work. Too bad, I guess we are stuck with the unsolvable system of transcendental equations. :-(

 

[Antenna Guy]

Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations.

When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

Consider that the sphere traveling along x at relativistic speed with respect to an inertial observer will be described by:

\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1

[note: sorry about leaving out the velocity term - I'd have probably screwed that up anyway ]

Where r is the radius of the sphere in its' rest frame.

If the inertial observer is not at infinity, the "shape" of the object will not be a projection on a plane - and that "shape" will vary with distance/aspect angle.

If you get that far, you can let the surface rotate about its' volume...

Regards,

Bill

 

[JesseM]

When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

At infinity? I wasn't trying to calculate what would be seen visually, just the coordinate path of a point on the rim in a frame where the rolling object was moving.

Consider that the sphere traveling along x at relativistic speed with respect to an inertial observer will be described by:

\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1

I guess this would be the equation for the coordinates of the surface in the frame where the sphere is moving along the x' axis, at the moment the sphere was centered at the origin. But this is different from the shape the sphere would appear to be as seen by an observer at rest in this frame.

If the inertial observer is not at infinity, the "shape" of the object will not be a projection on a plane - and that "shape" will vary with distance/aspect angle.

What do you mean by "projection on a plane"?

 

[Phrak]

Did I miss it, or hasn't it been pointed out? It's not at all unambiguous as to what meant by a 'ball'?

First, is it a problem in dynamics, or simply kinematics? If only kinematics, the spinning ball has a circumference to diameter ratio of gamma*pi in the plane of angular momentum. So in this case, it's fair to ask 'was it a ball before spinning? (If it was a ball before spinning then its a ball of constant radius, otherwise constant circumference.)

 

[Antenna Guy]

Did I miss it, or hasn't it been pointed out? It's not at all unambiguous as to what meant by a 'ball'?

Are you aware that this relativistic 'ball' does not necessarily have an instantaneous circumference (per se) in the frame of the observer?

First, is it a problem in dynamics, or simply kinematics? If only kinematics, the spinning ball has a circumference to diameter ratio of gamma*pi in the plane of angular momentum. So in this case, it's fair to ask 'was it a ball before spinning? (If it was a ball before spinning then its a ball of constant radius, otherwise constant circumference.)

Let's say that the observed circumference of this ball spans a range of time that approaches \frac{2r}{c}.. What would you integrate over to verify that the rest length of the observed circumference is still 2\pi r?

Regards,

Bill

 

[1effect]

When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

Consider that the sphere traveling along x at relativistic speed with respect to an inertial observer will be described by:

\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1

[note: sorry about leaving out the velocity term - I'd have probably screwed that up anyway ]

The above is not the equation of the moving sphere. You need to apply to Lorentz transform correctly.

 

[JesseM]

The above is not the equation of the moving sphere. You need to apply to Lorentz transform correctly.

I think it is the equation for the sphere's surface at a given instant when it is centered at the origin and moving on the x'-axis. The equation describes an ellipsoid with radius r along the y' and z' axis, but Lorentz-contracted radius r/\gamma along the x'-axis...is this not correct?

 

[1effect]

I think it is the equation for the sphere's surface at a given instant when it is centered at the origin and moving on the x'-axis. The equation describes an ellipsoid with radius r along the y' and z' axis, but Lorentz-contracted radius r/\gamma along the x'-axis...is this not correct?

"moving" was key in my answer.
His equation is the snapshot for t=0, so we agree.

The correct equation is:

\frac{\gamma^2 (x'-vt')^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1

 

[JesseM]

Ah, OK. But one point that occurred to me is that if we're just looking at the total set of points occupied by the surface at any given instant, rather than trying to track an individual point on its surface, then in the rest frame of the sphere's center the coordinates of the surface won't be changing with time, so it will always occupy exactly the same coordinates as a certain nonrotating sphere centered at the same position. And if the surface of the nonrotating sphere occupies the same set of coordinates in this frame, its surface must occupy the same set of coordinates as the rotating sphere in every frame. So, this is one way of seeing that the rolling ball won't "wobble". The difficulty is just in figuring out how a given point on the rotating sphere will move around this ellipsoid surface in a frame where it's in motion.

 

[1effect]

Ah, OK. But one point that occurred to me is that if we're just looking at the total set of points occupied by the surface at any given instant, rather than trying to track an individual point on its surface, then in the rest frame of the sphere's center the coordinates of the surface won't be changing with time, so it will always occupy exactly the same coordinates as a certain nonrotating sphere centered at the same position. And if the surface of the nonrotating sphere occupies the same set of coordinates in this frame, its surface must occupy the same set of coordinates as the rotating sphere in every frame. So, this is one way of seeing that the rolling ball won't "wobble". The difficulty is just in figuring out how a given point on the rotating sphere will move around this ellipsoid surface in a frame where it's in motion.

I think that we can do the "tracking" by converting the problem to cylindrical coordinates and by describing the angle \phi=\phi(t) as a function of time.

 

[1effect]

I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.

I think you need to have a look at http://www.spacetimetravel.org/filme/radv_t_0.93/radv_t_0.93-xe-320x240.mpg , there is no wobble. You can see more of the same.

 

[Phrak]

Are you aware that this relativistic 'ball' does not necessarily have an instantaneous circumference (per se) in the frame of the observer?

I'm afraid I haven't been clear. The ball rolls without slipping. Therefore it spins was well as translates.

In the inertial frame of the ball's center of mass the circumference is contracted from it's length at rest. The various radii are not contracted, because their extent is perpendicular to the instantaneous velocity develped by rotation.

Knowing this, the ball cannot be considered a rigid object. In fact relativity and rigid objects are contradictory in principle. This is why most answers given to pushing a long, relativistic stick down a short hole are flawed.

Until anyone trying to answer the original question addresses the shape the ball in a given, non-translating inerital frame, and what parts of the ball remain invariant under rotation, they're blowin' smoke.

 

[yuiop]

I think you need to have a look at http://www.spacetimetravel.org/filme/radv_t_0.93/radv_t_0.93-xe-320x240.mpg , there is no wobble. You can see more of the same.

First of all, when discussing relativity we normally discount purely visual effects to get objective measurements. The links you provided have animations that include the visual effect of Terrell rotatation that obscure the physics as far a relativity is concerned unless of course we were specifically investigating Terrell rotation.

Having said that, I have taken the liberty of modifying an image from one of the pages you linked to, to illustrate the point I was making in the post you responded to. The white blobs on the rims of the attached image are equal masses placed at the end of each spoke. (Assume the spokes, hub and perimeter have insignificant mass compared to the white test masses). The red wheel is not rotating and joining opposite pairs of masses shows the centre of mass is at the physical axle of the wheel. Joining opposite pairs of test masses on the green wheel, which is moving at relativistic speed to the observer, shows that the centre of mass (the light blue dot) is now much higher than the physical axle of the wheel. I think you will agree that this vertical shift of the centre of mass is not significantly affected by the Terrell rotation that causes an optical horizontal shift of parts of the object.

...
I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.

If you look at the bold part of my statement above my guarantee was that the centre of mass is not in the geometrical centre of the ball or wheel. I did not guarantee that the centre of mass wobbles. However, I will try and show later that is some wobble of the centre of mass, using geometrical software. I suspect this wobble is corrected by a counter rotating torque, generated by torque reactions that are result of force not always being parallel to the acceleration, in a transformed reference frame.

 

[Dale]

OK, I am not working today so I decided to start this. Here is an equation for the worldline of a particle on a rotating hoop in the reference frame where the axis of rotation is stationary.
s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)
in units where c=1 and r=1
Which can be differentiated to obtain the four-velocity
u=\left(\frac{1}{\sqrt{1-\omega ^2}},-\frac{\omega \sin (\phi +t<br /> \omega )}{\sqrt{1-\omega ^2}},\frac{\omega \cos (\phi +t \omega<br /> )}{\sqrt{1-\omega ^2}},0\right)
Which can be differentiated to obtain the four-acceleration
a=\left(0,-\frac{\omega ^2 \cos (\phi +t \omega )}{1-\omega<br /> ^2},-\frac{\omega ^2 \sin (\phi +t \omega )}{1-\omega ^2},0\right)
A quick check shows u.u=1 and u.a=0

I then transformed the hoop into the frame where it rolls without friction
s&#039;=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega<br /> ^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega<br /> ^2}},\sin (\phi +t \omega ),0\right)
and then differentiated to obtain u' and a'. I verified that u'=L.u and a'=L.a and that u'.u'=1 and u'.a'=0. So, kinematically everything works out just fine. I didn't actually work out the four-forces nor the tension on a differential element.

So then, to answer the question about the shape of the hoop in the primed frame I need to look at all of the points for a single value of t'.
t&#039;=\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega<br /> ^2}}
I could not solve this equation for t as described above, but I could solve it for phi to obtain
\phi =-t \omega \pm \cos ^{-1}\left(\frac{t&#039; \sqrt{-(\omega -1)<br /> (\omega +1)}-t}{\omega }\right)

Substituting this back into the expression for s' gives an expression for the shape of the hoop at a given t' in the frame where it is rolling

\left(t&#039;,\frac{t&#039;-t \sqrt{1-\omega ^2}}{\omega<br /> },\pm \sqrt{1-\frac{\left(t-t&#039; \sqrt{1-\omega<br /> ^2}\right)^2}{\omega ^2}},0\right)

I plotted this for a couple of different values of t' and it appears to be a "grape" that does not wobble.

 

[yuiop]

I'm afraid I haven't been clear. The ball rolls without slipping. Therefore it spins was well as translates.

In the inertial frame of the ball's center of mass the circumference is contracted from it's length at rest. The various radii are not contracted, because their extent is perpendicular to the instantaneous velocity develped by rotation...

The simple assumption that the spokes of rotating wheel without linear translation are not length contracted is in fact an over simplification because the principle of equivalence tells us that the acceleration due to rotation is similar to being in a gravitational field and there will be length contraction of all the spokes, with the greatest radial length contraction of the spokes occurring near the rim because that is where the acceleration is greatest. That is drifting into GR teritory and we probably don't need to go to that extent at this point.

We should be clear that the radii parallel to the linear translation of the rotating ball will be length contracted due to the linear motion superimposed on the rotation.

 

[1effect]

OK, I am not working today so I decided to start this. Here is an equation for the worldline of a particle on a rotating hoop in the reference frame where the axis of rotation is stationary.
s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)
in units where c=1 and r=1
Which can be differentiated to obtain the four-velocity
u=\left(\frac{1}{\sqrt{1-\omega ^2}},-\frac{\omega \sin (\phi +t<br /> \omega )}{\sqrt{1-\omega ^2}},\frac{\omega \cos (\phi +t \omega<br /> )}{\sqrt{1-\omega ^2}},0\right)
Which can be differentiated to obtain the four-acceleration
a=\left(0,-\frac{\omega ^2 \cos (\phi +t \omega )}{1-\omega<br /> ^2},-\frac{\omega ^2 \sin (\phi +t \omega )}{1-\omega ^2},0\right)
A quick check shows u.u=1 and u.a=0

I then transformed the hoop into the frame where it rolls without friction
s&#039;=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega<br /> ^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega<br /> ^2}},\sin (\phi +t \omega ),0\right)
and then differentiated to obtain u' and a'. I verified that u'=L.u and a'=L.a and that u'.u'=1 and u'.a'=0. So, kinematically everything works out just fine. I didn't actually work out the four-forces nor the tension on a differential element.

So then, to answer the question about the shape of the hoop in the primed frame I need to look at all of the points for a single value of t'.
t&#039;=\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega<br /> ^2}}
I could not solve this equation for t as described above, but I could solve it for phi to obtain
\phi =-t \omega \pm \cos ^{-1}\left(\frac{t&#039; \sqrt{-(\omega -1)<br /> (\omega +1)}-t}{\omega }\right)

Substituting this back into the expression for s' gives an expression for the shape of the hoop at a given t' in the frame where it is rolling

\left(t&#039;,\frac{t&#039;-t \sqrt{1-\omega ^2}}{\omega<br /> },\pm \sqrt{1-\frac{\left(t-t&#039; \sqrt{1-\omega<br /> ^2}\right)^2}{\omega ^2}},0\right)

I plotted this for a couple of different values of t' and it appears to be a "grape" that does not wobble.

Very nice, thank you !

 

[yuiop]

..

I plotted this for a couple of different values of t' and it appears to be a "grape" that does not wobble.

The shape of the Lorentz transformed hoop (rotating or not) is simply and formally a geometrical ellipse. (I can show you that the theory of relativity is falsified if that is not the case). I think we are agreed that the transformed outline of the hoop or ball does not change with time. The question, I think is more about the location of the centre of mass. If it does not remain constant over time, then you would expect the hoop or ball to "bounce" or wobble with an oscillating centre of mass rather than move in a straight line. The simplest way to analyse this, is to take just two test "point masses "on opposite sides of the hoop and find their combined centre of mass at any instant in the transformed frame.