No it doesn’t. Cyclotrons accelerate charged particles. The interaction of the electric charge and the magnetic field produces lateral acceleration. This electromagnetic force does not exist for wheels or masses on a string.
No they’re not. The forces are unbalanced, so by definition, the mass must be accelerating. F=ma.
Then I’d like to see what your free-body diagram of the mass-on-a-string system is, and how it differs from a mass not on a string.
To treis,
I’me talking about the linear and angular accelreation of a givenobject for the applcation of a given force. Not the linear acceleration of one object (e.g a treadmill) and the angular acceleration or a second object (e.g a solid disc). We don’t appear to actually disagree on anything.
To David Simmons,
Very clear explanations. I’m now going to have to agree with you that if we throw a whole bunch of Theoretical Physics Land equipment on this plane and treadmill (perfectly rigid homogeneous solid discs for tires/wheels with frictionless bearings and a perfectly rigid conveyor surface with ideal no-slip condition between the two) the treadmill can do anything it wants and it won’t impart so much as a fraction of a Newton towards anything but rotating the plane’s wheels.
Whoa, spoke too quick in that last post, I think.
So you have a free mass M and a point P and a line MP connecting them. Force F is applied to M at right angles to line MP. This will push M further away from P. Line MP will get longer. The mass will move in a spiral, not a circle.
I’m not really sure what that spiral motion would analogize to in the solid disc scenario. It might simply be that the cases are not analogous… or it might be linear acceleration.
If I read this right, he’s saying that though forces e and d are of equal magnitude they are applied at different distances from the center of the circle… so the torques produced by the two forces do not cancel.
Oh come now. The magnetic force on the charged particle produces a force on the particle and that force drives the particle into a circle because it is always at right angles to the velocity vector of the particle the same as the force on the mass is in my post on angular acceleration and torque. A force is a force, of course, of course.
Rather than answering this directly I’ll just refer to to this much betterexplanation of why there isn’t any lateral force on the wheel.
No it won’t because as the radius vector (what you call MP) rotates so does the direction of the force so as to keep it at right angles to the radius. The force is always at right angles to the radius vector between M and P. This at all times moves the particle so that the distance to the center point stays constant. Think about it some more and draw some sketches. For each small increment of motion the lateral motion is the time multiplied by the linear velocity along the path times the cosine of the angle of the force. The vertical velocity is the same except the multiplier is the sine of the same angle. The way the sine and cosine work as the angle increases makes the point move in a circle. Think of a lissajous pattern on an oscilloscope with equal sine waves at time quadrature on the vertical and horizontal deflection plates. The sinusoids are intimately connected with a circle. That’s why they are called “circular functions.”
But see
the link way down at the end of this post.
Holy mackerel guys. I guess we need a short course on vector diagrams.Here it is.
Ah, er, um, cough, cough. I’m not so sure now that the cyclotron and my case are identical. Similar … maybe, but not identical.
Um, what? Since when is the force on the mass in your post on angular acceleration at right angles to the velocity vector? It’s pretty clearly parallel to the velocity. I will assume, from your later post, that you caught the same. However, note this: what is at right angles to the velocity vector is the tension in the string. The magnetic force on the particle is analogous to this tension, as I said earlier.
That doesn’t really explain anything. There’s still an unbalanced lateral force on the cross-bar mass thing. You’ve even got it labeled as F. One more time: F=ma. If you have an unbalanced force F, it must produce an acceleration.
Not so much. You can’t move vector forces around anywhere you want them to be. You can slide them along their line of action, as you do with force b. You can’t move them perpendicular to their line of action, like you’re doing with force a. Force a still has to line up with the original point of application of F.
And, finally, to make one thing crystal clear:
F=ma
If there’s a force on a body, there must be an acceleration in the direction of the force. You can’t just ignore it because it’s not through the center of mass. F=ma.
From NASA:
Sigh, chalk one up for ignorance.
Since line of action of the original force went through P there must be at least one component of force a through p. Therefore the start of vector c is at P which means both a and d vectors also start at P. So d is on the vertical centerline pointed down.
Understood. Think of it like this:
On a line passing thourhg M, oriented 90 degrees to PM, is the point M’. Basically, the force F is pushing the mass M towards M’. PM, PM’, and MM’ form the sides of a right triangle. PM’ is the hypotenuse.
I’m not finding any force that would push M back towards P. Only the force pushing it towards M’. PM’ is longer than PM.
To paraphrase an old saying: If all you have is sine and cosine, everything looks like uniform circular motion.
All the vectors that act on M point away from P. This is not conducive to keeping PM at constant length.
That’s why I said all that fun stuff about perfectly rigid homogenous solid discs with frictionless bearings in ideal no-slip contact with perfectly rigid coveyor belts. In the real world there would be all sorts of small unaccounted for sources that would allow you to hold a plane back if you could just run the conveyor fast enough.
So, what does slow down the wheel in this case?
I can’t get over how easily you just brush aside irrefutable evidence that your analysis is wrong.
Hmmm. This requires some thought. The thing I keep thinking is that at every instant the force is pointing along the path at right angles to the radius. I sort of question whether or not there actually is an M’. When the distance between successive points approaches zero what happens? You cannot pick an M’ that is so close to M that I can’t put an infinite number of points between them. The motion is continuous so are there any such things as discrete, successive points on the path … ?
My math skills aren’t up to it tonight and maybe never. My head hurts.
See this post.
My physicist coworker doesn’t think it’s wrong. Whenever there is a difference you dismiss all others as wrong.
What about it? You said the force doesn’t slow down the wheel and I asked you how. The stuff about the momentum is just what happens not how it does. We have an acceleration and only one force on the body. What else, besides the force, could cause the acceleration.
First of all, I didn’t dismiss anything as wrong. I showed that if you used a different geometry in your analysis energy is magically created. This, unequivocally, shows that your analysis is wrong. I will tell you again why it is wrong. You did not include torsional work into your analysis and a net force, no matter where its applied, results in linear motion.
You complain about all of these thought experiments but I am offering them becuase they prove the physical relationships that I am arguing. The ice experiment, for example, shows that a friction force acting on a wheel does result in an acceleration. I have no idea what else I can do to prove that F=ma really means that. All that matters for linear acceleration is the netforce, not the location of that force.
I mean you are arguing that this wheel has linear acceleration. If all of the tangential force goes into rotation then that means the force through the center of wheel is unopposed. I don’t know what else to say but that this is just wrong.
If you want rotational motion without translation you need to apply a force couple. A few relevent quotes from that site:
The essence is that if you want to induce purely rotational motion you need to have no net force on the object. In our situation there is a net force so there is rotational and linear motion.
This is completely and utterly wrong. Please review some type of statics book. You can move a force along its own line of action, which you did with force b. But the remaining component force a is not through point P, so you can’t arbitrarily move it there. Just because you can move a vector along its line of action © doesn’t mean that you can take the other component (d) along for the ride.
Look. You’re completely wrong about an entire string of things, and with every explanation you add more and more incorrect notions. You really need to review some kind of dynamics/statics textbook. Since we’re discussing these issues in the context of a rotating wheel, I’d like you to develop the equations of motion in accordance with, for instance, the development in Beer and Johnston. Thanks to Amazon, I can point you to exactly where this is discussed. You might also want to take a look at this sample problem. In fact, do that first. Please explain how, in this sample problem, acceleration is caused by only a tangent force, and yet it isn’t in a wheel.
One more time: F=ma. Always.
A physicist co-worker? Good. Then draw a circle with a force applied. Ask this person, “under what conditions will this force not result in an acceleration?” Ask this person to produce the set of equations that will demonstrate a lack of acceleration.
Interesting (and at times infuriating!) thread. However, I think the question is ultimately futile.
Here’s my take on it.
In the real world: yes, of course the aeroplane will take off. Sure the wheels will spin like the blazes, but the power of a jet engine is just going to force the plane forward and there will never be enough friction between the tyres and the conveyor belt to prevent it reaching take-off speed. I wouldn’t like to try and land the plane afterwards though, as its landing gear will not be in a pretty state.
In the theoretical sense: the problem is meaningless. http://en.wikipedia.org/wiki/Irresistible_force_paradox
The conveyor belt is effectively an irresistible force. Why? Because, by definition, it can accelerate arbitrarily to match the speed of the wheels. F=ma. If a is arbitrarily large, F is also arbitrarily large. Ergo, F is effectively infinite. That is an irresisitible force.
And equally, the aircraft is now an immovable object. Huh? But I thought you just said it would move and take off? Yes I did, but that was in the real world, remember.
The aircraft is an immovable object because it has to be to satisfy the restriction in the question. (Assuming of course that the belt is matching the rotation speed of the wheel.)
If the conveyor belt is moving backwards at velocity v, and the wheel is rotating forwards at velocity v, the plane is not moving, by definition, assuming of course that the wheel is not sliding (which it would in the real world as I said above).
So basically, all this question represents is a rehashing of the age-old irrestible force/immovable object “paradox”, which is not really a paradox at all:
an irresistible force, such as our putative conveyor belt, would represent infinite energy, and therefore, by Einstein, infinite mass.
There - you’ve just caused the universe to collapse in on itself into a singularity, just for a stupid prank with an airplane. Bet you feel kinda dumb now, huh?
I have finally seen the light. I’ve been retired for nearly 25 years so my time is really not very valuable. Despite that lack of value, it is being wasted on this particular pastime.
But I did learn one thing for sure. F certainly does equal ma.
