Question

What is the speed of a point on a circle that's rolling at v, with a radius of r? Basically I want to know the formula.
@steve816

Answer 1

@steve816

Answer 2

$$v = r~ \omega$$

Answer 3

That's cute, but in a rolling motion:

Answer 4

That is the graph of a cycloid.

Answer 5

yes, and the radius is constant, the angular velocity is constant, but the speed of the point on the circle changes

Answer 6

I want to know how to find the speed of that point on the edge of the circle.

Answer 7

we need the parametric equations for the cycloid first

Answer 8

okay......that's nice to know.

Answer 9

x = r ( t - sin t )
y = r( 1- cos t)

then dx/dy = (dx/dt) / ( dy/dt)

Answer 10

the circle is moving at v, and has radius r

Answer 11

Hold it....there should be a much simpler way without having to use derivatives or stuff.

Check this out: They used pythag
https://figures.boundless-cdn.com/9918/large/figure-17-06-03a.jpeg

Answer 12

update
https://www.desmos.com/calculator/fpfpo3ohzk

Answer 13

Your question was:
"What is the speed of a point on a circle that's rolling at v, with a radius of r? Basically I want to know the formula."

We can define the cycloid parametrically $$ x = r ( \theta -\sin \theta )
\\ y = r ( 1 - \cos \theta ) $$ Then by the chain rule it follows$$
\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt} = r ( 1 -\cos \theta ) \frac{d\theta}{dt}
\\ \, \\
\\ \frac{dy}{dt} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dt} = r \sin \theta ~\frac{d\theta}{dt}

$$
Since \( \large v = r \omega \) and \( \large\omega = \frac{d\theta}{dt}\), we have \( \large v = r \frac{d\theta}{dt} \) We can substitute
$$\frac{dx}{dt} = r ( 1 -\cos \theta ) ~\omega
\\ \, \\
\\ \frac{dy}{dt} = r \sin \theta ~\omega$$
We can use the parametrically defined speed formula:
$$ \frac{ds}{dt}= \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 }

$$ Plugging in we get$$\frac{ds}{dt} =\sqrt {{r}^{2} \left( 1-\cos \left( \theta \right) \right) ^{2}{{\it
\omega}}^{2}+{r}^{2} \left( \sin \left( \theta \right) \right) ^{2}{{
\it \omega}}^{2}}


$$ Which simplifies to
$$\frac{ds}{dt}= r \omega \sqrt{ 2- 2 \cos \theta } $$ or $$ \frac{ds}{dt}= v \sqrt{ 2- 2 \cos \theta }$$ where \( v \) is the linear velocity. To think of \( v\) more concretely you can imagine the circle being a front wheel of the car. The car is moving to the right at speed \( v \).

Answer 14

"Hold it....there should be a much simpler way without having to use derivatives or stuff. "
Why do you think that? Not everything is simple in math. There might be a simpler solution but there may not be. I am going to think about this further and see if there is a simpler solution.

Answer 15

Yes you definitely don't need any calculus to do this. And I'd be very careful about pulling formulae off the internet if you're not really that up-to-speed on this stuff......see below.

A drawing:


Everything you need is in that drawing. Especially look at \(\theta \) as it comes out via similar triangles, so you can create a velocity vector in x-y. The translational left -> right v, plus the circular motion. That's all.

The velocity vector for the point on the **ring** at \(t > 0\) (that's the blob in the drawing) is:

\(\mathbf v = \left(\begin{matrix}v - \omega r \cos \theta \\ \omega r \sin \theta\end{matrix}\right)\)

....and because \(v = \omega r\), that becomes:

\(\mathbf v = v \left(\begin{matrix} 1 - \cos \theta \\ \sin \theta\end{matrix}\right)\)

And because \(\theta = \omega t\), that becomes:

\(\mathbf v = v \left(\begin{matrix} 1 - \cos \omega t \\ \sin \omega t\end{matrix}\right)\)


Now go back and check that thing that was pinched off the internet and you will see that it has the assumption that \(\omega = 1\) 😳

you can generate the position vector for the blob in the same way. it requires some trig and some common sense :)

the cycloid is legendary mainly IMHO because it's the solution to the brachistochrone and the arc length is a pig to solve. but you can do all of this by being canny and using trig and basic kinematics :)