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24th August 2018, 04:23 PM  #321 
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24th August 2018, 04:54 PM  #322 
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Dave,
thanks, great post, I see your point. I need help to reconcile it with this textbook: It says: ... the resultant ΣF of the external forces acting on the body equals the mass m of the body times the acceleration a of its mass center G. So when we come up with the ΣF = Σ m_i a_i then we will get an average A. F = m A gives us bigger force with bigger A and the mass m does not change. Now A > r \omega^2 though. How this A is supposed to be used? Thanks, SDG 
26th August 2018, 12:10 PM  #323 
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26th August 2018, 12:20 PM  #324 
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26th August 2018, 12:26 PM  #325 
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26th August 2018, 12:34 PM  #326 
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27th August 2018, 02:13 AM  #327 
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This is from some way back, but: The rotational energy of the balls depends on their rotation rate (and mass and shape, of course), and their rotation rate is the same as the wheel (assuming we are still contemplating the model where they are tethered by inelastic strings). In other words, it is the same in ANY reference frame.
Hans 
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27th August 2018, 04:46 AM  #328 
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Well, the critical "post 253" question, point #5 in post 307, has been resolved.
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Yes, because the tension on the string is the vector sum of the m*a (as a is a vector) of all atoms. As the radius gets smaller relative to the ball, more of the atoms "pull" harder, but less of their pull is aligned with the string. Those two effects exactly cancel out when calculating the pull on the string. (The magnitude of the individual atom's centripetal acceleration is increased by a factor of 1/cosθ, but the resultant acceleration vector parallel to the string is decreased by a factor of cosθ, where θ is the angle between the string and the atom's own individual radius line.) So, what's left to address? 
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27th August 2018, 06:12 AM  #329  
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As tusenfem noted upthread a number of times, linear acceleration is not rotational acceleration. The two can be related but don't have to be. In the diagram on the scanned page some of the force vector components that contribute to the rotational aspect cancel each other in the linear aspect. So how the vector components of the applied forces sum can be different for linear then for rotational considerations.
Similarly as I tried to explain to you upthread linear momentum is not rotational momentum. In the case of the rolling wheel their directions are orthogonal to each other. To drive the train car you rotate the wheel however some of that energy and momentum applied to the wheel goes into pushing the car forward (by friction with the track) reducing the total final angular velocity obtained by the wheel for that applied energy and momentum. The difference is what goes into the linear energy and momentum of the moving. Back in the day there were SSP racers by Kenner. You pulled the zip line that spun a wheel with a large mass to increase it's moment of inertia. This kept the wheel spinning and gave it the angular momentum to drive the car forward at a decent speed when placed on the ground. Angular momentum stored in the wheel was transferred to linear momentum of the car. The direction of the angular momentum being perpendicular to the direction of movement of the car also acted as a gyroscope to tend to keep the car running strait. A lot of applied physics going on in just a kids toy. In the frame of the car the rotation of the wheel just slows down as friction with the ground acts as a brake. In the frame of the ground the rotation of the wheel slows down as the friction with the ground drives the car forward. Rotational deceleration results in linear acceleration. Even in a directly driven wheel as in an automobile what goes into linear acceleration of the car comes at the cost of less rotational acceleration of the wheel for the same input of torque. The friction of the ground and mass of the car act as brake on the wheel not as additional effective mass or moment inertia of the wheel. As the addition of linear momentum comes at the cost of rotational momentum. As noted by others this is all just basic Newtonian mechanics and Galilean relativity.


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27th August 2018, 08:56 AM  #330 
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Yes, I agree, the rotational energy of the balls is the same in both reference frames at steady state.
Having said that I have questions if the delta of the rotational energy (some additional work) is going to be transferred (essentially added) to the rotational kinetic energy of the balls at the same rate at all points along the cycloid. Here are the factors to consider: Different radius at the top, at the bottom and at the inflection points of cycloid. I have chosen the top and the bottom of the cycloid because there is no tangential acceleration at those points, only centripetal acceleration. The inflection points do not have any centripetal acceleration only tangential. If we have different radii on the cycloid then could the image bellow be linked to the issue? The additional question. The cycloid trajectory curves to the right at the top part. The ball rotation is in the same direction. The cycloid trajectory changes curvature to the left at the inflection point. The ball rotation is in the opposite direction to the cycloid rotation. Is this something that should be considered? These are the questions and I would like to find some answers to these questions. SDG 
27th August 2018, 09:24 AM  #331 
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Well, the acceleration has to be calculated this way:
If we choose point B as the center of the ball then the result will be exactly what Dave is saying. The smaller radius will have bigger acceleration because \omega is different for the \omega x (\omega x r) component. SDG 
27th August 2018, 09:42 AM  #332 
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If there would be a different prediction between two reference frames then it appears to me the locally only one can be true.
If we predicted different forces acting on the balls at different points on the cycloid on the contrary to the circle that predicts all the forces to be the same on all balls then only one prediction could be true and the other false. SDG 
27th August 2018, 10:00 AM  #333 
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We have a problem here.
Different radius, different rotational kinetic energy, different average acceleration even though the center of the ball has the same acceleration. My take on this issue is E_ball = m_ball c^2. If the E_ball is bigger because it has bigger rotational kinetic energy then m_ball has to be bigger, right? We take two photons, make them spin and we have mass particles, so what's the big deal. https://en.wikipedia.org/wiki/Pair_production ... but I see why this idea does not fly by many. Dave's take on the issue is to keep the mass the same and we have different average acceleration. Sure enough it is like that as well. Then we have an issue A > r \omega. We 'break' the calculus and the textbooks. Either way we look at it, ... THIS IS A BIG PROBLEM! SDG 
27th August 2018, 10:31 AM  #334 
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Only problem I see is I see no problem.

27th August 2018, 10:56 AM  #335 
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27th August 2018, 10:58 AM  #336 
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27th August 2018, 11:04 AM  #337 
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27th August 2018, 11:14 AM  #339 
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No, again relative motion is frame dependent. Again that is just basic Galilean relativity and Newtonian mechanics. Two frames and two predictions of the relative motion observable in each, both are true for their respective frames. The very assertion of the train frame being comoving laterally with the wheel explicitly precludes that very lateral relative motion of the wheel in that frame. This ain't a "prediction" it is an explicit requirement of your experimental conditions. Any "prediction" you might glean from whatever that asserts otherwise simply means you are not adhering to your experimental condition of observations only in the moving train frame.

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27th August 2018, 11:49 AM  #340 
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SDG. I can't see the issue. And I don't see why you have to obfuscate the problem with math.
If a ball is swinging at a constant speed around on a string attached to a fixed point, the centripetal force (the mass of the ball times the acceleration) will be the same all the time. If the fixed point is moving a constant speed (for example, riding in a train). The centripetal force will be the same. The nonaccelerated motion of the train has no effect. Now, instead of a fixed point, we attach the point to a rotating disc. So, now as the ball swings around and the point rotates around, there are times in the cycle when the force is more and times when it is less. But, the average force is the same. If the whole assembly is placed in a railway car moving at a constant speed, nothing changes as far as the centripetal force. If the contraption with the fixed point wasn't affected by the motion of the train, the more complicated contraption isn't going to be affected either. If your math isn't giving the same result, I would assume it is your math that is wrong. There is no relevant radius that is different from the viewpoint of someone in the train as it is stopped at a station and someone in the train as it travels in a straight line at 70 mph. 
27th August 2018, 12:42 PM  #341 
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27th August 2018, 12:43 PM  #342 
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27th August 2018, 12:44 PM  #343 
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27th August 2018, 12:46 PM  #344 
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27th August 2018, 12:47 PM  #345 
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27th August 2018, 12:49 PM  #346 
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27th August 2018, 01:59 PM  #347 
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A "We have a problem here" fantasy when it is only you that has a problem
28 August 2018 SDG: A "We have a problem here" fantasy when it is only you that has a problem in this thread.
E =mc^2 is the massenergy equivalence equation and is irrelevant to the classical problem of a pendulum with a spherical bob. Ignorance about photons (they have intrinsic spin) and pair production (a photon near a nucleus becoming a matter/antimatter pair). Ignorance of not known that changing mass will change rotational kinetic energy, Mass is a bit hidden. I (the moment of inertia) is the "ratio of the net angular momentum L of a system to its angular velocity ω". Angular momentum involves mass. A dumb "We 'break' the calculus and the textbooks" response to the suggestion to keep the mass constant. Have you never heard of density ? 
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27th August 2018, 02:07 PM  #348 
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Show that F_cp is different for different radii

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28th August 2018, 02:53 AM  #349 
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28th August 2018, 03:46 AM  #350 
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28th August 2018, 01:26 PM  #351 
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This post is about the side claim as stated in the post #253.
Here is how the centripetal forces will be different for the same ball that is being accelerated on different radii. I picked two points for the demonstration on the edge of the ball. The size of the arrows is arbitrary, to demonstrate the calculation. The same points contribute differently to the total sum of the forces because the \omega of the ball rotation is different. The arrows represent \omega x (\omega x r) portion of the acceleration. The green arrows represent forces for the r=100m and \omega=0.1rad/s. The orange arrows represent forces for the r=1 and \omega=1rad/s. \omega x (\omega x r) is different in magnitude but has the same direction. Fore example if ball radius r_b=0.1m then the green would give us 0.1*0.1^2=0.001m/s^2 and the orange 0.1*1^2=0.1m/s^2. The arrows bellow the ball is just moving forces to the center of the ball. The centripetal forces will be in the opposite direction of the end result middle arrows, to the center of the rotation. SDG 
28th August 2018, 01:51 PM  #352 
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An irrelevant post about centripetal forces and 2 points on the edge of a ball
29 August 2018 SDG: An irrelevant post about centripetal forces and 2 points on the edge of a ball.
Physics is not cartoons and guesses. The centripetal force is the sum of the centripetal forces on every point in the ball. What you need to do is that sum of forces and see if that sum depends on the radius of the ball. 28 August 2018 SDG: Show that F_cp is different for different ball radii (what you have not done since 20 August 2018) 
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29th August 2018, 07:50 AM  #353 
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29th August 2018, 08:03 AM  #354 
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Ahh, perhaps I get it now: When the diameter of the ball is a considerable fraction of the circumference of the orbit, some of the forces will cancel out internally: The ball is tethered to its centre of gravity, but outer parts of it will then transfer force at a different angle, through the structure of the ball.
The extreme example of this is if you, instead of a ball, contemplate a ring like the rim of a wheel; here, all centripetal force is redistributed within the ring, and none is transferred via the spokes. So yes, you are correct. But where is the problem? Hans 
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29th August 2018, 08:30 PM  #355 
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30th August 2018, 06:49 AM  #356 
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These questions are about the side claim.
The first questions  if we place accelerometers on the marked spots then what acceleration would they report? If we consider 1m radius rotation and \omega=1rad/s, the points are outside of the radius 1m geometrically. Are they going to report smaller acceleration than 1m/s^2 in spite of the geometrical position? The second question  how the angular momentum delta of the ball at different radii will affect the system? The third question  can we exploit the bigger tension, as per your correct observation, by some modified setup? SDG 
30th August 2018, 07:45 AM  #357 
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30th August 2018, 09:35 AM  #358 
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30th August 2018, 12:19 PM  #359 
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Actually, I happen to feel somewhat charitable_
Since the ball is moving as one piece, the accelerometers will indicate the centripetal acceleration that fits the radius and speed where they are. They cannot "know" that they are placed somewhere on a ball.
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Quote:
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4th September 2018, 06:46 AM  #360 
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Right, so they will report the accelerations at that point.
The accelerations have different magnitude and different direction for the r=1m and r=100m. This leads to different tensions in the ball. Each atom has a different tension at different radius.
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If you had one of these in your hands then you know how real is the angular momentum and how it affects the systems.
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This is the basis for my claim that the forces are not equal at different radii. We could create 'a ball' from smaller balls connected by springs and observe the different responses. SDG 
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