Instantaneous velocity of point on a trochoid

[ktoz]

Hi
I Googled http://mathworld.wolfram.com/Trochoid.html" but wasn't able to find formulas for the instantaneous velocity of a point on trochoid curves. Does anyone know of an online reference (or know how to find the velocity)

I can get rough estimates by calculating the distance between two "ticks" on a trochoid curve but I want to get exact values if possible.

Thanks for any help

 

[HallsofIvy]

Then you learned that parametric equations for a trochoid are
$x= a\phi- bsin(\phi)$ and $y= a- bcos(\phi)$ where $\phi$ is the angle the line from the point to the center of the "rolling circle".

I assume you also understand that a "point" on a tronchoid (or any other geometric figure) doesn't have a "velocity"- that's why just look up those words didn't give a "velocity". "Velocity" is a physics concept while "tronchoid" and "points" are purely geometric. Assuming you mean "the instantaneous velocity of a point moving with given speed on a tronchoid", then you get the instantaneous velocity, as Sir Isaac Netwons said, by differentiating the position vector: the x and y components of the velocity vector are given by
$$\frac{dx}{dt}= \left[a -b cos(\phi)\right]\frac{d\phi}{dt}$$
and
$$\frac{dy}{dt}= sin(\phi)\frac{d\phi}{dt}$$

In the simple case of "constant angular speed", $\phi(t)= \omega t$, that is
$$\frac{dx}{dt}= \omega\left[a -b cos(\phi)\right]$$
and
$$\frac{dy}{dt}= \omega sin(\phi)$$

 

[ktoz]

In the simple case of "constant angular speed", $\phi(t)= \omega t$, that is
$$\frac{dx}{dt}= \omega\left[a -b cos(\phi)\right]$$
and
$$\frac{dy}{dt}= \omega sin(\phi)$$

Thanks Halls. I'm guessing you're using $$\phi$$ to signify the angle but what are you using $$\omega$$ for? I tried Googling "omega + trig" and several variations on that, but omega seems to be used many different ways depending on the branch of mathematics. Could you give me a concrete example with actual numbers plugged in rather than $$\omega$$?

Thanks in advance.

 

[HallsofIvy]

Once again, you will not find your answer in mathematics because your question is about physics. I don't know why you are "guessing" anything since I said

1) 'where $\phi$ is the angle the line from the point to the center of the "rolling circle".'

2) (as you quoted) "In the simple case of "constant angular speed", $\phi(t)= \omega t$".
$\omega= \phi/t$ is standard (physics) notation for the angular velocity.

 

[ktoz]

Once again, you will not find your answer in mathematics because your question is about physics. I don't know why you are "guessing" anything since I said

1) 'where $\phi$ is the angle the line from the point to the center of the "rolling circle".'

2) (as you quoted) "In the simple case of "constant angular speed", $\phi(t)= \omega t$".
$\omega= \phi/t$ is standard (physics) notation for the angular velocity.

Thanks Halls. Much appreciated.