Cycloid help

Let's consider a cycloid. You have a piece of gum stuck to your bicycle tire. When is the gum at rest, and why?
I want to understand how this works mathematically.

Attached: sykloidi.png (820x476, 10K)

The gum is never at rest

It actually is, for an instant. It happens at the bottom of the cycloid curve, where you have that narrow peak.
Intuitively it makes no sense so I wanna understand the math behind it.

called an infinitesimal it goes up and then down so there has to be a point in between where it does nothing same thing with the slope of a line being zero at one point

if you know calc find the derivative and where the slope is 0 for velocity. circular motion is very weird to think about

Attached: 90FFA309-8ED1-4DA6-B89A-C16C9C75DEB6.gif (291x355, 1K)

so its in relation to the wheel or ground?

Relative to the ground the bike is riding on, the gum is at rest for a fraction of a second, just as it touches the ground. Relative to the wheel, it's at rest when the wheel is at rest relative to the ground, or you're doing a tokyo drift with the bike.

it's still spinning at that moment so it's never at rest *dabs*

Sorry, I was the first post and was just being dumb for a minute. Cyloids aren't my field

I mean, think of it in terms of vertical velocity and horizontal velocity.

The vertical velocity when plotted on a velocity time graph makes a sine graph (if the gum starts at the bottom) because when the gum is at the top and bottom, it has no vertical velocity since it is changing direction.

The horizontal velocity causes a graph of sin(2x) because when the gum attached to the wheel moving at x ms-1 consider that the independent motion of the gum is in the direction of -x ms-1 therefore the total velocity acting on the gum is 0 causing the total velocity to be zero as well.

Attached: do your own homework faggot.png (1875x421, 65K)

>this entire post
el atrocidad...

Search for Euler's equations.
To calculate the point you have to use the following equation (all 3D-Vectors):
V (point) = V (translatory speed of your circle) + W (rotatory speed) x r (radius of the rotatory speed)

This is a task from university, you can see that the speed of point B is 0 but point A is moving.

Attached: kugel rollt.jpg (1445x259, 71K)

>if you know calc find the derivative and where the slope is 0

The track in OP's pic is just (x,y) points, it's not time on the x axis so you can't just look at where the derivative for thecycloid is 0.

btw omega was 10s^-1, V0=5m/s and R=0.5m

Your handwriting is so horrible I don't even want to try and think what it says there

>kugel rollt
Svensk?

When the gum touches the ground, it is at rest. The lowest part of the circle has no relative velocity (considering the ground). If you want a purely mathematical explanation, you can differentiate parametric equations for the the cycloid curve: x=radius(angle-sin(angle)) and y=radius(-cos(angle)). x'=r(1-cos(angle)) and y'=radius(sin(angle)). At angle=2pi, (x',y'), the velocity of the curve, is equal to (0,0), showing that the point that touches the line has no speed. I'll post another explanation below, tell me if you need help figuring out the parametric equations of the cycloid or understanding what have I done.

excuse no line breaks
Here's 2 physical explanations:
1-intuitive: Should the lowest point have any speed in relation to the ground, the circle would skid.
2-analysis: The circle moves forward with speed V. As it does not deform, all of it's points also have speed V. However, as the circle is also rotating, there is another component in the velocity of a given point, created by rotation. Only the center of the circle does not rotate. Let T be the time the circle takes to finish a rotation. In T units of time, the circle will have moved forward V times T which is equal to 2pi times the radius, and rotated 2pi radians -> it's angular velocity is 2pi divided by T. Analysing the lowest point, let Vr be it's velocity caused by rotation: linear velocity= angular velocity times radius -> (1)Vr=2pi/T times radius. However, V times T = 2pi times radius -> (2) 2pi/T=V/radius. Substituting (2) in (1), we get Vr= V/radius times radius -> Vr=V. The vector sum is equal to zero since the Vr is in clockwise direction and V is to the right and we conclude the total velocity in that point is zero.

It is possible to find the point with zero speed without analysing just the right point. In , should you draw the parametric curve/cycloid, it is easy to see which point on the curve has no velocity. In the analysis above, you could write a general equation for velocity in any point and solve it.

All my effort, in vain.
Who else /threadkiller/?

this is my last post in this thread. should anyone seek something from me dorp some contact info

I have nothing to do. Even here I am ignored . I feel hollow

Isn't it at rest when it touches the ground because the wheel is kinda pivoting around the point that contacts the ground? That's what I'd assume anyway, since the wheel hast to rotate at the same speed the bike moves.

I lied
yes, that is a valid way to interpret it.