1. When a bicycle goes by you at top speed, is it true that the top of the wheel is going faster than the bottom of the wheel?

www.noggintwisters.com  2.

3. Certainly. All you have to do is imagine the wheel an instant later. The point that had just been the top is now more forward than the point that had been the bottom. Clearly the top had to move faster during this small time interval.  4. Yes, and it’s more amazing than that. If you follow a point on the circumference of the wheel you will see that it actually goes backward for a moment at the bottom of the loop.

www.noggintwisters.com  5. Originally Posted by Noggintwisters
Yes, and it’s more amazing than that. If you follow a point on the circumference of the wheel you will see that it actually goes backward for a moment at the bottom of the loop.
www.noggintwisters.com
I don't think that is right.

The equation for the postion of a point on the wheel on a flat road where the wheel never slips and ignore any flattening of the wheel from the weight of the bicycle would be given by

x=r*cos(theta)+v*t, y=r*sin(theta)

where theta = v*t/r

so taking the derivative of x wrt t

dx/dt = v sin(v*t/r)+v

but now it is obvious that dx/dt is NEVER negative, which means that a point on the rim of the bicycle is never going backwards.

PS: I hope you didn't just trick us into doing a homwork or takehome test problem for you.  6. Originally Posted by Noggintwisters
Yes, and it’s more amazing than that. If you follow a point on the circumference of the wheel you will see that it actually goes backward for a moment at the bottom of the loop.
It could, if it's the rear wheel and it's slipping. Otherwise, if it's in contact with the road and not slipping, its speed is zero.  7. Originally Posted by Harold14370 Originally Posted by Noggintwisters
Yes, and it’s more amazing than that. If you follow a point on the circumference of the wheel you will see that it actually goes backward for a moment at the bottom of the loop.
It could, if it's the rear wheel and it's slipping. Otherwise, if it's in contact with the road and not slipping, its speed is zero.
I take it you mean its speed along the road exactly at the time it is in contact with the road.

to generalize the equations above to look at any point on the wheel we need to add an angle to vt/r

x=r*cos(theta)+v*t, y=r*sin(theta)

where theta = v*t/r + theta0

so taking the derivative of x wrt t

dx/dt = v sin(v*t/r+theta0)+v

now looking at x=r*cos(theta)+v*t, y=r*sin(theta)
The lowest position is given by theta = 3pi/2.
So we want theta0=3pi/2 so that at t=0 we will be looking at the lowest point on the wheel.

But now dx/dt = v sin(3pi/2) + v = -v + v = 0

Of course at any other time dx/dt will not be 0 but greater than 0

At the top of the wheel where theta = pi/2
dx/dt = v sin(pi/2) + v = v + v = 2*v
The point on the wheel will be going twice as fast as the bicycle.  8. Faster relative to what? The epicenter of the wheel, to the surroundings, or to the bicycle? To the center of the wheel, they are always at the same speed; why wouldn't they be? Unless, like stated the you are considering wheel compression. To the surroundings, the bottom would be going in the negative direction, so the top would be going faster. To the bicycle, in a relative cumulation to the surroundings, you could say that both points are not moving, but that would be stretching it.  9. Originally Posted by Cold Fusion
Faster relative to what?
With respect to the road of course. We usually don't consider the road to be moving unless specifically stating a special frame reference or consering motion on a much larger scale than a bicycle.  Bookmarks
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