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.