# Velocity, mass, and drag

1. Nov 2, 2011

### wvengineer

Was having a breakroom discussion today and I figured I would get an outside opinion. Anywas I have to projectiles/particles flying through the air and are unaffected by gravity. Both have the same drg coefficient, cross-sectional area, air densityetc. The only differences are particle A has less mass and it's initial velocity is faster than particle B with more mass, but less velocity. They are both decelerating due to the force of drag. There are no other forces involved. Will the faster object decelerate quicker than the slower object due to a higher force of drag? Will they eventually reach the same velocity? If they reach the same velocity will the particle we less mass continue to decelerate faster? I'm wondering if the inertia of the particle with more mass would affect this?

Thanks,
Lance

Last edited: Nov 2, 2011
2. Nov 2, 2011

### JeffKoch

Drag force is typically approximated as proportional to a power of velocity, say V3, independent of mass, etc. and depending just on shape. So you can work out how the deceleration scales with mass and velocity.

3. Nov 2, 2011

### wvengineer

Well if I give no consideration to the mass of the particles then given enough time the two particles will eventually be going the same velocity. After that they will decelerate at the same rate. Is that a true statement? I understand drag is independant of the particle's mass. I just keep thinking of deceleration, and the interia of the particle (B) with more mass would cause it to decelerate at a slower rate at that point.

I'm trying to have it both ways I guess.

4. Nov 2, 2011

### nasu

They may reach the same velocity at some point but they will not move with the same velocity after that.
The more massive body will have smaller acceleration for equal velocities.
If you plot the velocity versus time you'll see that the curves may cross in one point but they don't overlap. Unless you consider the asymptotic decrease to zero. Once they reach zero velocity they will move with the same velocity forever after.

5. Nov 2, 2011

### wvengineer

I think you're agreeing with me. My second post wasn't what I though, I was just wanting to make sure I understood Jeff's response.

Here's my side of the discussion. As a disclaimer I understand this is not a practical scenario. I separated this into two things. First is with all thing being equal between two particles (including mass) other than velocity. The faster particle's velocity will approach the velocity of the slower. Now I realize it could take forever that's why I'm saying "approach". If you go long enough they'll be traveling at the same velocity at some point because the force of drag will be equal and that's the only thing decelerating the two particles. Now if the mass is different then whatever force is acting on the particle will also have to over come a difference in inertia with respect to a difference in deceleration. So again two particles: particle A is moving faster, but has less mass than particle B. Particle A is decelerating at a faster rate then particle B due to a higher force of drag which is due to a higher velocity. Eventually they will be traveling at the same velocity (or until the force of drag is the same). At this point the heavier particle B would decelerate slower than the particle with less mass due to a difference in inertia. Now I'm wanting to make sure I'm correct, and not talking out of my rear end so to speak.

6. Nov 2, 2011

### nasu

If they have the same mass but different velocities they will never travel with same (non-zero) velocity at the same time. Both velocity versus time curves will go asymptotically to zero but they will not cross.

7. Nov 2, 2011

### wvengineer

The faster one will "approach" the the velocity of the slower one given enough time (at t=∞). That was just for simplification to separate the difference of deceleration due to the different magnitudes of the force of drag from the difference in deceleration due to the different masses of the two particles. I've been giving a disclaimer of there will be a asymptote when the mass is the same. As the difference in velocity gets smaller and smaller so does the difference in the force of drag.

8. Nov 2, 2011

### wvengineer

Now I would like to show it mathematically. The problem I'm having is we don't have a constant deceleration. I can get the force of drag for each particle. I can get the change in deceleration for each particle with respect to a certain velocity. I would now like to get this with respect to time and see how long it would take for the faster particle to go slower than the slower heavier particle with respect to a changing force of drag due to the change in velocity. Clear as mud right?

9. Nov 2, 2011

### wvengineer

My problem is that I have a change of force with respect to velocity, and a change of deceleration with respect to the change in force. I'm not sure how to set this equal to each other for the two particles and calculate at what time the two will be traveling at the same velocity. This is with two different masses by the way. Otherwise I could just assume that it happens as t approaches infinity.

10. Nov 2, 2011

### nasu

At t=∞ they will be both approaching zero velocity. Is that what you mean? That they will be at rest together?

As for the actual solution, you need to know how does the force depends on velocity. Once you have this, is quite straightforward.
For example, if F=-bv where b is the drag coefficient, you just solve the equation
m dv/dt=-bv
with initial condition v(0)=vo.

The solution is v=voExp[(-b/m)t]
Try to plot this for various values of the parameters. I hope it will clarify the things a little.
Equal drag means same b.

11. Nov 2, 2011

### wvengineer

I see what your saying about as t approaches infinity then the particles velocity will at best approach zero. As far as the solution can you elaborate on that for me. The drag coefficient is constant (and the same for both particles), but how are you showing the change in the force of drag with respect to velocity? Then again I have a changing deceleration with respect to the force of drag. Again the two particles start out a two velocities. The particle with less mass is initially traveling faster. Due to this the faster particle feels a higher force of drag. At the same time the the faster particle has less inertia than the slower particle with more mass. So there's two things affecting the deceleration of the two particles at the same time. Changing force of drag with respect to the velocity, and the changing deceleration due to the changing force of drag between the two masses. Maybe I'm not stating this correctly, but I think it's more complicated than what you want me to plot. I'll continue to look at your response and see if it clicks, I am rather rusty on my diff-e-q's.

12. Nov 2, 2011

### nasu

The equation is Newton's second law and it contains everything that you mentioned.
In general is
F=ma where F is the force and a is the acceleration, a=dv/dt.
The force for your case depends on velocity, let say F=-bv (the simplest case).
So the equations becomes
-bv=mdv/dt

or dv/dt=-(b/m)v
This shows that the acceleration (dv/dt)
- is not constant, but depends on velocity
- it will be smaller for larger m
- it will be larger for larger drag (larger b).

The solution of the equation is the one I wrote in my previous post.
The velocity decreases exponentially from the initial value (I called it vo).
The exponent (-b/m)t tells that it decreases faster if b increases or/and m decreases.

13. Nov 3, 2011

### wvengineer

I believe I see it Nasu. I appreciate the help. I'll have to try it later. It's funny how this started as a break-room discussion, and then how to prove it got stuck in my head.

14. Nov 3, 2011

### cjl

Nasu's analysis does show the correct dependence on drag coefficient and mass, but unfortunately, the velocity dependence is not quite correct. Drag force (for a subsonic, low-viscosity flow) is basically proportional to v2. This unfortunately makes the equations a bit harder to solve, but if all you care about is the dependence on mass and drag coefficient, it won't change any of the factors above.

15. Nov 3, 2011

### wvengineer

From what Nasu wrote I took it to mean b = 1/2*Cd*A*p and v=v^2

16. Nov 3, 2011

### nasu

No, I did not mean that. As I said in the post, I just picked this linear dependence as an example to illustrate the procedure. It is the simplest case and is not completely imaginary. For slowly moving objects in viscous fluids, the drag force may be proportional to the speed.

For a ball falling through air is more realistic to take the force proportional to the square of the speed, as observed by cjl.
The solution will have a different form. You cannot just replace v by v^2.
You can either solve the equation yourself or look it up. It's not much more complicated (if you neglect any other forces).

17. Nov 23, 2011

### wvengineer

Sorry it's been a while. Just had time to thnk about this. Here's what I was doing. Force of drag = k*(v^2), k=(1/2)*(rho)*(Cd)*(Aref)
m*a=k*(v^2)
m*(dv/dt)=k*(v^2)
dv/(v^2)=(k/m)*dt
-(1/v)=[(k*t)/m]+C
v=-m/[(k*t)+(C*m)]

Now I havne't gone any further because it seems to me this isn't right. At t=0 v= -(1/C). I would assume C was my initial velocity, but not in this form. Plus I can't seem to figure out how the units would cancel to m/s this way so I definitely don't have something right. So what do I have wrong here?

18. Nov 23, 2011

### nasu

It seems right to have 1/C =-1/vo.
The solution would be
$$v=\frac{v_0}{1+\frac{k}{m}t}$$