Ball rolling with varying deceleration

1. Jan 28, 2008

inflation

I'd like to see an example of a formula describing the position of a rolling ball over time. It has been given one impuls at the start, after which air resistance and friction decelerates it (the ground is flat). At high speed the ball might be sliding, until it decelerates to some treshold speed when it starts to roll. The deceleration might behave quite different in those two modes, I guess.

How should I formulate the balls position as a function of time based on those characteristics?

I intend to observe the position of a rolling (sliding) ball at different times, and I want to formulate a function to fit with those observations in order to figure out where the ball has been at any given point in time. I thought it might make more sense to estimate the parameters of a model with some basis in real physics, rather than to fit any curve.

2. Jan 29, 2008

inflation

BUMP
Let's skip the "two modes" sliding and rolling for now, and just look at air drag:

I want an expression for the position of an object over time. I know its position x0 and speed v0 at time t=0. I know that it is decelerating due to air drag and can assume that the deceleration is proportional to speed squared: a=k*v² (can I do this and assume that k has the unit "1/meter" so that the units work out?)

Now, step by step:

1) The position given constant speed is x = v0*t.
2) Given constant deceleration x = v0*t-a*t²/2.
3) With air drag changing the deceleration x = v0*t-k*v²*t²/2.

My problem is to get rid of that "v" in equation 3. How does v change over time?

If I try something like this:
x = a*t²/2
a = 2x/t²
2x/t² = k*v²
v = root(2x/(k*t²))

And insert in equation 3:
x = v0*t-k*[2x/(k*t²)]*t²/2.

Then t cancels out from everything but v0*t so that the motion is linear. So I was just going in a circle there.

Last edited: Jan 29, 2008
3. Jan 30, 2008

Staff: Mentor

Your equation #2 is only valid for constant acceleration, so it won't apply here. To see how the velocity varies with time, you must integrate the acceleration function.

4. Jan 30, 2008

PatPwnt

Yea you can assume that, but I believe at low speeds drag is proportional only to v. I think you can write a differential equation: v' = (k/m)*v then just find the solution to
v' - (k/m)*v = 0 which I got was v(t) = c1*e^(kt/m) c1 being a new constant from integrating. This is just a hunch as I am new to applying DEs and this only takes in drag as the only thing affecting the motion. You can add friction in too.