
#91
Apr2808, 11:37 AM

P: 3,966

I tend to agree with your observation. There appears to be a small mistake in DaleSpam's original equation that has the ball rotating counter clockwise while translating from left to right in the positive x direction which does not agree with the contact point being at the bottom. This is a non fatal error that is easily corrected by reversing the sign of the y coordinates. Of course I agree that events that are simultaneous in one frame are not simultaneous in the other. The point I was trying to make is that using eq7 we are still required to determine t for a given t' in eq7 by numerical aproximation using eq5. Do you agree? By reversing the signs of the y coordinates (see above comment to Bill) to get the ball rotating in the same direction as the linear motion and by using the solve function of a spreadsheet to determine t in eq7 I get the following results for t'=0 in frame S' for four particles on the rim of the hoop with coordinates expressed as (t,x',y') where t is the proper time of each particle. (Assume w = v =0.8c where v is the relative velocity of frame S and S') a' = 0.683390604 0.512542953 0.519881721 b' = 0.000000000 0.000000000 1.000000000 c' = 0.683390604 0.512542953 0.519881721 d' = 0.000000000 0.000000000 1.000000000 where in the frame at rest with axis of rotation the (x,y) coordinates for the particles are a = (1.0), b = (0,1), c = (1,0) and d = (0,1). It can be seen that the y' coordinates of particles a and c are the same and above the centre of the axis rotation by a distance of 0.519881721 so we can assume the centre of mass for this pair of particles is at (x',y') = (0,0.519881721) Now if we look at the pair of particles at the top and bottom of the ball (b and d), the particle at the contact point has a momentary velocity of zero in frame S' and the particle at the top has a velocity of 0.975609756c by velocity addition. If we assign a value of 1 to the rest mass of the particles then the particle b at the contact point has a inertial mass of 1.00000 and particle d has an inertial mass of 4.55555. The centre of mass of these two particles assuming radius (R=1) is found from 2*R*mb/(mb+md)R = 0.64R or (x',y') = (0, 0.64) in terms of coordinates. From the above it can be seen that the centre of mass when the particles are vertically alligned in S' is at y'=0.64 and moves to y'=0.519881721 after the particles on the hoop have completed a quarter rotation. Therefore the height of the centre of mass does not remain constant in time for a rolling object with constant linear and rotational velocity in the transformed frame. That is something we have to live with and it neccessary to find a compensating factor such as stress in the connecting rod or hoop joining the test masses, or a counter rotating ghost torque to sort things out (to justify no wobble observed in the rest frame of the rotation axis). P.S. DAleSpam, I am not being critical of your work (I think it very good and the most positive and technical contribution to this thread so far ;) 



#92
Apr2808, 04:47 PM

Mentor
P: 16,476

The hoop has more than two (or even four) particles. I think you need to integrate around the hoop, or at least do a numerical approximation to the integral. 



#93
Apr2808, 05:16 PM

P: 303

[tex] s=(t,\cos (\phi t \omega ),\sin (\phi t \omega ),0) [/tex] Regards, Bill 



#94
Apr2808, 05:30 PM

P: 3,966





#95
Apr2808, 06:42 PM

Mentor
P: 16,476

That's a good argument, although it really doesn't apply for the hoop geometry described by my equations. For a hoop I think the best conclusion is that the center of mass is stationary.
I would not recommend using my hoop equations to draw conclusions for the barbell geometry. 



#96
Apr2808, 08:27 PM

P: 303

Regards, Bill 



#97
Apr2808, 10:27 PM

Mentor
P: 16,476

You have to get rid of the variable phi and add a variable r.
I think it is an interesting problem, but it is a different problem than the one I worked. 



#98
Apr2908, 07:38 AM

P: 145

Forgive me I just deleted this post and have to retype it so I am bound miss something I put before.
I have been studying the Dale's attachments on post 80 and they have been bothering me for some reason. Last night I think I finally realised why. To begin with some basic information about the drawings as I understand them. 1) The rings are in the roads view. 2) The colors represent 30 degrees in the starionary axis view. 3) c=1 4) r=1 5) v = .9c I looked at the velocity of a particle in the roads view. Using the basic equation for velocity: V = [tex]\frac{D}{T}[/tex] I asked what if the wheel rolled forward pi/4. This yields the equation:.9c = [tex]\frac{pi/4}{T}[/tex] Then I looked at the distance the left most particle on the wheel would have traveled. To do this I first noticed that because of length contraction the bottem half of the wheel only contains 90 degrees. There for when the wheel turns pi/4 the left most particle will travel to the very bottem of the wheel. This is a distance of at least r (just look in the Y direction only). So using the same time as a pi/4 roll T, by particle has moved with velocity:V' = [tex]\frac{1}{T}[/tex] Dividing the V' equation by the V equation I get:V'/.9c = [tex]\frac{1}{pi/4}[/tex] Which yields the particles velocity V' to be 1.15c.This obviously can not be correct though. I assume I am missing something. Can anyone help me understand what I did wrong? 



#99
Apr2908, 09:05 AM

P: 321

In the frame S, comoving with the object, the center of mass (ycoordinate) is: [tex]y_M=\frac{\int y \rho dV}{\int \rho dV}=\frac{\int y dV}{\int dV}[/tex] The center of mass coincides with the center of rotation because of the radial symmetry. In the observer's frame S': [tex]y'_M=\frac{\int y' \rho' dV'}{\int \rho' dV'}=\frac{\int y \gamma dV}{\int \gamma dV}= y_M[/tex] 



#100
Apr2908, 10:44 AM

P: 3,966

I am not sure how you got to your conclusions because you have not provided much detail. For example what is V? Is it volume? Anyway, you seem to conclude that y coordinate of the centre of mass in the S' frame is at same height as the centre of mass in the S frame. Dalespam reached the opposite conclusion that the height of the centre of mass of the rolling ball is higher in S' than in S and his conclusion is in agreement with the fairly well known fact that the centre of mass of a rotating object is not invariant under Lorentz transformation. It seems you have reached a false conclusion. 



#101
Apr2908, 10:59 AM

Mentor
P: 16,476





#102
Apr2908, 11:02 AM

P: 3,966

There is a minor mistake in DaleSpams drawing that shows the wheel rotating counter clockwise from while moving from left to right so that after one eighth of a turn (pi/4) in frame S the left most particle will move to the top and not to the bottom. Also, you do not seem have taken time dilation into account. If it takes t = pi/(4v) seconds to complete a quarter turn in frame S then it takes t' = pi/(sqrt(1v^2)4v) seconds in frame S'. Also, it should be pointed out that the left most particle with ball moving from left to right will not rotate through 90 degrees in either frame after a rotation of pi/4 = (45 degrees) in frame S. So after taking everything into account including simultaneity and relativistic velocity addition I am sure you will not get v>c anywhere in the transformation of the rolling ball particles, using DaleSpam's equations. 



#103
Apr2908, 11:18 AM

P: 321





#104
Apr2908, 11:22 AM

P: 3,966

Hi DaleSpam, Using your initial equation: [tex]s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)[/tex] and substituting phi = 0 degrees, t=0 and w = 0.9c then the coordinates for the particle are (t,x,y,z) = (0,1,0,0). Staying with t= 0 and w =0.9 and using phi = 90 degrees the coordinates are (t,x,y,z) are (0,0,1,0) so that the convention you are using for phi is that a rotation from 0 to 90 degrees is counter clockwise in the rest frame of the axis of rotation. I think that is the cause of the confusion. If I keep phi = zero degrees then when t=0.1 seconds the coordinates of our right most particle change from (t,x,y,z) = (0,1,0,0) to (0.1, 0.99, 0.09, 0) in frame S and the coordinates change from (t',x',y',z') = (2.06, 2.29, 0, 0) to (2.28, 2.49, 0.9) in frame S'. I am sure you will agree that is a counter clockwise rotation in both frames that is easily corrected by starting with: [tex]s=(t,\cos (\phi +t \omega ), \sin (\phi +t \omega ),0)[/tex] 



#105
Apr2908, 11:27 AM

P: 3,966





#106
Apr2908, 11:36 AM

P: 321





#107
Apr2908, 11:55 AM

P: 3,966

I have done the calculations for t = pi/4/w which is one eighth of a turn in the rest frame of the hoop axis which equates to t' = 1.636246174 seconds in frame S' when w and v is 0.8c. The transformed coordinates (t, x', y') of particles a, b, c and d are: a' = 0.191000000 1.901914626 0.152206102 b' = 1.772000000 0.715519375 0.152597799 c' = 1.355638753 1.028609890 0.884074990 d' = 0.608000000 1.589744078 0.884021391 It is fairly easy (due to the symmetry) to work out that the centre of mass for the four particles at this posion is now y'=0.551 That means the centre of mass for the four particles is y' =0.58 every quarter of a turn (your calculation) and y' = 0.55 every eighth of a turn. It is reasonable to assume that as the number of particles on the rim increases the centre of mass will stabilise somewhere around y' = 0.56. This answers the question of the OP that a homogenous ball of even density will be grape shaped and move without wobbling. However, we have a new problem (perhaps needing a new thread?) that there appears to be a wobble of the centre of mass for a system of 4 or less particles with circular motion under Lorentz tranformation. 



#108
Apr2908, 01:14 PM

P: 321

The correct way is to do the following: [tex]dx'=\gamma sin(\phi+t \omega) d \phi[/tex] [tex]dy'=cos (\phi+t \omega) d \phi[/tex] The arc element is : [tex]ds'=\sqrt (dx'^2+dy'^2)[/tex] The center of mass is obtained by integrating from 0 to [tex] 2 \pi[/tex] the expression: [tex]\frac{\int y' ds'}{\int ds'}[/tex] It is very easy to show that [tex]\int y' ds'=0[/tex] 


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