Are drag coefficients negative?

[amolv06]

Homework Statement

Find an equation describing a body in free-fall in air. This is not really homework. I'm in a differential equations class write now, and I have fun finding real world applications for such things.

Homework Equations

m\frac{d^{2}y}{dt^{2}} = B\frac{dy}{dt}-mg

The Attempt at a Solution

Rewriting the above:

\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = -g

The corresponding homogeneous equation is:

\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = 0

Two solutions to the homogeneous equations are:

y_{1} = C_{1} and y_{2} = C_{2}e^{\frac{b}{m}t}

And as a particular solution:

y_{p} = \frac{gm}{b}t

Therefore by the superposition principle, we have a general equation as follows:

y(t) = y_{1}(t) + y_{2}(t) + y_{p}(t) = C_{1} + C_{2}e^{\frac{b}{m}t} + \frac{gm}{b}t

So here's where my question arises. I believe my assumption (the initial differential equation) should be a reasonable approximation of the real world. And I plugged my solution into the original differential equation. I seem to have answered it correctly. If both of these assumptions are true, then this equation doesn't make sense. It asserts that we are falling up! The only we I can think that this equation can still hold is if B were negative. I think after that, everything should make sense. Is this a correct assumption, or did I just make a mistake in the mathematics above?

Thanks in advance for your time and any help.

 

[kdv]

Homework Statement

Find an equation describing a body in free-fall in air. This is not really homework. I'm in a differential equations class write now, and I have fun finding real world applications for such things.

Homework Equations

m\frac{d^{2}y}{dt^{2}} = B\frac{dy}{dt}-mg

The drag force is - B\frac{dy}{dt} since it is always opposite to th emotion (where B is assumed to be a positive constant).

The Attempt at a Solution

Rewriting the above:

\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = -g

The corresponding homogeneous equation is:

\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = 0

Two solutions to the homogeneous equations are:

y_{1} = C_{1} and y_{2} = C_{2}e^{\frac{b}{m}t}

And as a particular solution:

y_{p} = \frac{gm}{b}t

Therefore by the superposition principle, we have a general equation as follows:

y(t) = y_{1}(t) + y_{2}(t) + y_{p}(t) = C_{1} + C_{2}e^{\frac{b}{m}t} + \frac{gm}{b}t

So here's where my question arises. I believe my assumption (the initial differential equation) should be a reasonable approximation of the real world. And I plugged my solution into the original differential equation. I seem to have answered it correctly. If both of these assumptions are true, then this equation doesn't make sense. It asserts that we are falling up! The only we I can think that this equation can still hold is if B were negative. I think after that, everything should make sense. Is this a correct assumption, or did I just make a mistake in the mathematics above?

Thanks in advance for your time and any help.

Your particular solution is incorrect, it should be proportional to t^2
(and of course, if you set B=0 you should recover the usual free fall equation)

 

[amolv06]

Sorry for the confusion, I'm using g as 9.8 rather than -9.8. I should have clarified. But I believe that should clear up the confusion with the drag force sign. Basically I set the drag force to be positive whereas the gravitational force is negative.

I'm not sure why my particular solution does not work.

\frac{d^{2}y_{p}}{dt^{2}} = 0

\frac{dy_{p}}{dt} = \frac{gm}{b}

Plugging this into the original differential equation I have

-\frac{b}{m}\frac{gm}{b} = -g which seems to be what I need. I must be missing something.

 

[kdv]

Sorry for the confusion, I'm using g as 9.8 rather than -9.8. I should have clarified. But I believe that should clear up the confusion with the drag force sign. Basically I set the drag force to be positive whereas the gravitational force is negative.

I'm not sure why my particular solution does not work.

\frac{d^{2}y_{p}}{dt^{2}} = 0

\frac{dy_{p}}{dt} = \frac{gm}{b}

Plugging this into the original differential equation I have

-\frac{b}{m}\frac{gm}{b} = -g which seems to be what I need. I must be missing something.

But it does not matter whether you use +9.8 or -9.8 (i.e. whether your y-axis points upward or downward) because the drag force is always opposite to the motion so no matter what your choice is for your direction of the y axis, the force will be -B dy/dt

 

[amolv06]

Ahh, yes! How could I have missed that.

Thanks a lot. That clears up a lot.