The thing that always amazes me when I think about it; the thing that is hard to get my brain around, is that any part of the tread on a tyre, on a car doing 70 mph, is accelerating from zero to 140 mph and back to zero between 800 and 900 times a minute. Even more with a small wheel.
Does anyone disagree with this proposition?
I do.
Right off the bat, what you’re saying is that 800-900 times per minute ‘any part off the tread on a tyre’ comes to a complete stop. That’s just not correct.
I’m thinking that you just learned something and either don’t fully understand it or are misapplying it. If you explain it, we can probably help you out.
Nothing to add about the cycloid other than a historical note. Galileo Galilei may have been first to discover this curve, but he failed to realize that it, rather than the circle, was the tautochrone which yields a perfect pendulum. That discovery was made by Christiaan Huygens.
Sorry - the tread that is touching the road (not ANY part) MUST be stationary - therefore the tread at the top is travelling twice as fast as the car.
Found a pretty graphic - Trochoid - Wikipedia
Ummm, no.
If the part that’s touching the road was stationary, the car wouldn’t be moving. Also, that wouldn’t imply that the antipodal part is moving twice as fast (or it would catch up). Your math is wonky.
Look at the graph that septimus provided (the unit circle). When the tire touches the ground and the point on the y-axis changes direction it’s speed is zero, but the tire still has speed in the x axis. In fact, as long as the car is moving forward, the tires will always have speed in the x-axis.
In order for the tread to stop moving, there has to be no movement on both the x-axis and y-axis, not just one of them.
At 70mph the typical automobile tire experiences a constant *centripetal *acceleration of about 2500 m/s^2.
The fact that a particular point on the tire *appears *to be starting and stopping from the point of view of a stationary observer has no effect on the actual stresses the tire endures.
That’s true with reference to the road, but the tyre is rotating around the centre of the wheel and the road reference isn’t particularly relevant. With reference to the centre of the wheel any point on the tread of the tyre is always moving at 70 mph and is constantly being accelerated toward the centre of the wheel. It doesn’t seem particularly amazing from that point of view.
Its quite correct., when using the ground as the frame of reference…
But there is no work done on accelerating that piece of tyre like that.
Its rotational system, so use the axle as the frame of reference, and the acceleration is perpendicular to the movement (WRT the axle) … no work is done… the piece of tyre feels nothing much different along its path due the forces rotating it.. when its at constant rpm’s (constant vehicle speed..)
Tyres are speed limitted due to internal strain changes causing heat.. the rate of heat production increased with speed.. Its being squashed at the bottom (due to load ) and stretched at the top, strain is actual stretching, where work is done..
Of course not - since it is exactly countered by the point opposite.
If you watch the graphic Septimus used, you can easily see that when the circle reaches a half rotation, the indicated point is travelling at twice the speed of the point opposite. But that can’t be correct, since the point at the bottom is not moving… I give up.
No - The point at the bottom has speed zero, the centre has speed one and the point at the top must have speed 2 - twice the speed of the centre.
If you jacked the tire up and let it spin freely, would the bottom still not be moving?
Timed out for editing
If you have a roller, with a plank on to, the plank moves twice as fast as the roller.
So if you turned the tire upside down would the car drive twice as fast?
Now you are just being silly.
Yes - but the centre, ie the axle and the rest of the car, would not.
It sounds silly, but I’m really not being silly. I’m just trying to use you’re words and take them to their logical end. The tire doesn’t know top from bottom. You said a plank on top rolls twice as fast. Pretend the plank was the ground and the ground was the plank. Why can’t the car move forward twice as fast as it normally does?
It’s just a case of you misunderstanding something and it’s going to take someone explaining it to you in a way that makes sense for it to ‘click’.
Wait, I got that wrong, the car is connected at the middle of the tire, not at the top, so ignore that part.
But still, if the tire tread was stationary at the point where it made contact with the ground, the car wouldn’t move. Besides, the whole tread is connected, you can’t make one part stop without stopping the rest of it.
What is stopping is the y component, but since the x component keeps traveling forward, you can’t say that it’s [some specific point] stopped.
Now, a ball, thrown straight up into the air DOES stop before it comes back down. As long as it goes straight up and straight back down, it doesn’t travel on the x axis, therefore when it’s speed goes from positive to negative it passes through 0 (intermediate value theorem IIRC) and stops just for an instant.
This concept is no different than a continuous track vehicle? i.e. caterpillar tread
That helps show the point since it’s so exaggerated. What’s going on is that on top the tread is moving with the speed that it’s being driven at PLUS the speed of the vehicle, at the bottom, you have to subtract the speed of the vehicle. If you take the plank and cylinders that bob mentioned earlier and lifted the system off the ground, the plank would no longer move twice as fast.
I guess we have to ask bob++ what his frame of reference is.
Since none of our in-house mathematicians have shown up, let me try. (FWIW, I did win some math contests before most of the in-house mathematicians were born.) The equations of a cycloid (using Planet Earth as “reference”) are
x = r (t - sin t)
y = r (1 - cos t)
Angle t can be considered equivalent to time. Velocity is given by ds/dt or, in Lagrange’s notation:
x’ = r (1 - cos t)
y’ = r sin t
When t = 0 or 2pi, or 4pi, etc.
x’(0) = 0
y’(0) = 0
Q.E.D.
This is true if you measure speed relative to the ground.
But from the point of view of the tire, all parts of the tread are moving at a constant speed and are being accelerated toward the center by a constant amount. The magnitude of their velocity vector is unchanging, but its direction is.
Still, you are correct that there’s a 140 mph different in velocity between the top and the bottom of the tire, and any point on the tire changes direction twice each revolution.
