Physics Problem: Linear vs Angular Velocity

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OK this is one of those beer & napkin discussions. My friend swears it's true, but I disagree. (I think his wheels have come off and he's heading for the ditch!). But I digress . . . "One" is viewing a tire from the side as it travels down a level road. (Vehicle & size don't matter, though I'll say the tire diameter is 2 feet). One puts a LED on the center of the tire and another on the outer edge. So as the tire travels down the road, one sees a straight horizontal line and another of circular arcs. Assume no slippage between tire and road and constant velocity. My friend says that since there is no slippage and the ground isn't moving, that the velocity of the tire when it contacts the road (at 6 o'clock) is zero. Further, the velocity of the tire is max when it's at the top (12 o'clock). So the tire velocity increases from 0 to max and then decreases back to 0 as the LED on the tire moves from 6 o'clock to 12 and back to 6. And that this 'average' is summed up, resulting in the linear velocity of the wheel moving in a straight line (the LED on the axle). I say: No-Way-Jose. Since velocity is constant, one part of the tire can't have a velocity of 0 while 180 degrees from it, the velocity is max. V must be the same at every point along the circumference of the tire. In other words, there is no 'special' point. Furthermore, acceleration is zero and so is torque because the car is coasting at constant V on a level, smooth road. (Ignore wind resistance). He retorts that there is a special point: The one in contact with the ground and if the ground isn't moving, neither can the tire, as there is no slippage. Therefore, it's V must be zero. I retort that the vehicle/bicycle is moving forward and every point of the tire contacts the road as the wheel turns AND that angular velocity of the tire is constant since acceleration is zero. I think the principle of conservation of angular momentum applies here. He also made another point about the wheel...something...force..but I forgot as it didn't make sense either. I searched on a couple of physics forums, but just found the standard questions/problems. So what say you? Anyone heard of this physics problem before? Please weigh in.....
 
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Originally Posted By: sleddriver
He retorts that there is a special point: The one in contact with the ground and if the ground isn't moving, neither can the tire, as there is no slippage. Therefore, it's V must be zero.
This is bunk. The instantaneous velocity (linear or angular, whatever) of a "particle" at the tire's outer edge is not 0. The contact surface will move and, since there's no slippage, roll the tire on the ground. If its velocity were zero it would not be moving. It sounds like he misunderstood something somewhere, but I'm not sure where he got that "rule".
 
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The contact point is, in fact moving with zero velocity with respect to the earth, being the vehicle's horizontal velocity v minust the vector that is Wr (equivalent to v), i.e. 0, and the top is moving with velocity v+Wr. Instantaneously, the tyre could be prtraid as a long stilt with an axle in the centre of it, continuously being replaced by a new stilt the moment that it's done it's job. From the perspective of the frame of reference of the car, the bottom of the tyre is moving backwards at V, the top forwards with V, the front downwards with V, and the back upwards with velocity v.
 

JOD

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There's an easy answer to this question: if the velocity of the tire were zero at 6:00, there would be no skin friction. Obviously, that's not the case. Car and bike tires derive most of their rolling losses from hysteresis, not skin friction--but look instead at rail car wheels, where skin friction is the dominant force in Crr. They get really hot! If they had a velocity of zero, they wouldn't be generating heat...
 
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1. Velocity is relative. Let's assume you are talking about velocity with respect to the ground. 2. The velocity of bottom of the tire is close to zero. This assumes low slip. 3. The velocity of the top of the tread is about twice the velocity of the car.
 
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Originally Posted By: spasm3
if the car is moving forward and the ground is not moving, the velocity can't be zero or there would be friction.
Of course there is friction.
 
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Shannow is correct. The linear velocity of the tire relative to the earth is 0 at 6 O'Clock and 2x at 12 O'Clock, but relative to the axles of the car they are equal in magnitude but opposite in direction. It didn't violate the law of physics (Newtonian) and without using calculus you are looking at the unit from the point of the center of gravity. Any relative difference between different point of the same piece is preserved if you do the angular physics correctly. The angular velocity and acceleration, must be preserved unless you apply external force or you factor in deformation. Friction is independent of velocity. I think what you guys mean is whether there is a slip, and assuming nothing is deformed, no slip means the wheel at 6 O'Clock is at zero velocity relative to the ground. The static friction is what hold the wheel and the road together without slip. Also the structural rigidity of the wheel is what's holding the different point of the wheel together without flying off from centripetal force. If you make your wheel out of cream cheese and the car is weightless, and you spin it at 100mph, some of the cheese will fling out.
 
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Originally Posted By: JOD
Car and bike tires derive most of their rolling losses from hysteresis, not skin friction--but look instead at rail car wheels, where skin friction is the dominant force in Crr. They get really hot! If they had a velocity of zero, they wouldn't be generating heat...
I don't think that cars (or trains) go fast enough for skin friction to be much of a factor...would make radiators a little ineffective.
 
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OK imagine the wheel spinning clear of the ground, and the speed at the circumference of the tire is 60mph. All parts of the tires circumference are moving at 60mph weather at 6 o'clock, 12 o'clock or whatever. OK. Now fire a 2x4 at at the outer edge of the tire at 60mph, so that it just contacts the tires circumference forming a Tangent (the 2x4 is traveling in the same direction as the rotation of the tire) The circumference speed of the tire is still 60mph. The 2x4 is traveling at 60mph. BUT their RELATIVE speed at the point of contact is Zero.
 
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Even my 6 year old daughter is thinking of this stuff. As we were on a bike ride last night, she asked me, "daddy, why does my tire roll faster than I am?" She was talking about looking down at the top of her bike tire and seeing it spin faster than she is going, but the tire never rolls away from her. I explained to her exactly what the "friend" in the OP's post said: if you spin the tire, the top of the tire moves away from her and the bottom of the tire moves towards her. But the center of the spinning tire stays on her bike, so the tire doesn't roll away. That's a pretty advanced topic to try to explain to a 6 year old. Velocity here is relative. Relative to the earth, any given point on a tire is constantly accelerating and decelerating. The velocity of any given point on the tire relative to the earth will be zero when on the ground and will be 2x the velocity of the center of the tire (axle, or vehicle in the larger sense) when at the top of the circumference. But the tire will have a "constant velocity" around the circumference of the tire.
 
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Akin to this: When I was in High school I had an argument with a teacher, He posed a very similar problem to this, but he illustrated by taking the example of an old style steam locomotive (Piston/Con Rod connected directly to the wheel) HE claimed that at BDC (6 o'clock) the piston was at zero velocity in relation to the track. I claimed that the piston was moving forward! Why?
 

Astro14

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Shannow's correct. It's a matter of reference frame and relative motion. Further, the rotating LED spot doesn't inscribe a series of circular arcs...it's more just arcs that go up to the top of the tire and back down to the ground...
 
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Actually, the tire's RADIUS at ground contact is what matters, NOT the diameter of the tire [unless you have a theoretical perfectly round tire with no distortion - not a real case scenario] . Therefore the ground contact point is moving slower than the top of the tire. And there is no zero speed if there is rotation and the car is moving.
 
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Yeah, the radius thing in rail parlance is "creep", the difference between the diameter x revs, and the actual distance travelled, rather than trying to determine that effective radius in the herzian zone. The point of contact has zero relative displacement to the road whilever traction is being maintained...the contact point by definition is zero.
 
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I have a question about the "rotating led spot". If we were to take the path traveled by the LED and lay it flat, what would be the distance traveled by the led if the car had traveled say one mile? I am having tough time visualizing that path. For some reason, I think that the path (obvious) and its first derivative (not so obvious) both have to be contiguous. Do you have a picture showing it? As far as the discrepancy between radius and the diameter of the tire, we are ignoring that to make problem easy to understand and solve. In reality we know that the contact surface can NOT be a single point. As a matter of fact, it is trivial to compute it given the number of tires and tire pressure. But for this type of problems, we use contact point. - Vikas
 
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Oh, the horror, all that Calculus comes flooding back in... The question is always relative to what. Your friend is correct on both counts. Angular velocity is simply the degrees or radians (depending on which units you are using) per second. If the tire is rotating, it's angular velocity can be described by such a value. Angular velocity is the velocity relative to the axis of rotation. The LED at the center is this axis of rotation. So relative to that reference point, the LED at the outer edge of the tire will have a constant angular velocity. Linear velocity is relative to some fixed point in space. That point is where the tire is in contact with the road. For any slice of time, there is one point where the tire is in contact with the road. The LED is either catching up, falling back or at that point at any given point in time. It's velocity relative to that point depends on where it is in that cycle of catching up to, falling behind, or at the reference point. As it approaches the reference point, it's velocity relative to that point approaches zero. The further away from that point the greater it's velocity relative to that point. Now if I had more time to think about this, I'd ponder if we are just talking about the X component of that linear velocity. (Velocity is a vector after all, and the outside LED will have both X and Y components to it's velocity vector.) Of course, it's been about 25 years since I've done anything serious with vectors, so I reserve the right to be wrong, or to have grossly oversimplified the issue.
 
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Originally Posted By: javacontour
Linear velocity is relative to some fixed point in space. That point is where the tire is in contact with the road. For any slice of time, there is one point where the tire is in contact with the road. The LED is either catching up, falling back or at that point at any given point in time. It's velocity relative to that point depends on where it is in that cycle of catching up to, falling behind, or at the reference point. As it approaches the reference point, it's velocity relative to that point approaches zero. The further away from that point the greater it's velocity relative to that point.
Actually, these are all covered in the "statics" and "dynamics" classes in mechanical engineering. When mixing angular and linear physics on the same object, the practice is to use the center of gravity for the linear physics if you are not going to do calculus across the entire volume. That's why balancing a wheel is important. The difference between the center of gravity and the rotational axle will generate lots of vibration.
 
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