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- - **Trying to determine initial thrust on an object from angular velocity?**
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Trying to determine initial thrust on an object from angular velocity?Hi all,
I built an electric paper airplane launcher and I'm trying to figure out how much force or thrust is being applied to my paper airplane when it's launched. The setup looks like this: http://i46.tinypic.com/wbf1x5.png The discs are 124mm in diameter, are spinning at approximately 5800 RPM each, and are about 1mm apart (far enough not to touch but close enough to grab the airplane. The airplane is 7 inches long. I'm trying to figure out how much force is transferred to the airplane. My initial guess was to convert the angular velocity of the discs into linear velocity, and multiply that by mass of the plane.. but I don't think that's accurate. First, I'm not sure that two discs spinning in opposite directions at 5800 RPM each equals an angular velocity of 11600 RPM. I'm not sure that I can combine them that way. Second, assuming that I figure out the combined linear velocity of both discs, I am not sure how that is transferred to the airplane. The discs are applying force to the airplane for the total distance of its length, but I'm not sure how that factors into the airplane's initial acceleration. I'm sure there is a lot I'm missing here.. just looking for a point in the right direction. Thank you! |

Re: Trying to determine initial thrust on an object from angular velocHi, unfortunately you can't add the angular velocities of the wheels like that. Just like your car has four wheels going 60 mph each, but the car doesn't go 240 mph, in your launcher, the second wheel just allows there to be good grip and provide more power with a second motor.
You are correct that you take the angular velocity of the discs and convert into linear velocity of the plane. For a circular thing, linear velocity is given by [itex]v = \omega r = 5800rpm \cdot 62 mm=5900mm/s \approx 6m/s[/itex]. 6m/s will be the maximum speed of your plane. Thrust is a concept that applies to jets and rockets, so here we're just considering force. The trick is that the plane isn't actually accelerating during the whole time it passes through your launcher. I (of course) haven't seen your launcher, but I'm guessing it is like a baseball shooter? You put the plane in, and pretty much instantly it reaches full speed and shoots itself out. Force is proportional to acceleration.. so if something isn't accelerating, there isn't a net force being applied. So in your case, the plane very quickly reaches its maximum speed, then just keeps passing out from the launcher. So to get the force, you need to have an idea of how long it takes the plane to reach its maximum speed. This is a topic called impulse momentum theory, which you can read more about on wikipedia. The idea though, is that if the acceleration takes place in a sufficiently small amount of time, then you can approximate newtons law. Usually you will see it as F=ma, but this is the same as saying F=dp/dt, where p is momentum. But in our small-time approximation, we can say that [itex]F=Δp/Δt[/itex], where Δp is the change in momentum during the time interval, and Δt is the change in time.. For your plane, (if I assume the mass is ~250 mg, which is about standard for a sheet of printer paper), the change in momentum is [itex]p=m Δv=250mg * 6m/s = 1500 mg m/s[/itex]. Δt is something you would have to measure or estimate.. perhaps, as an idea, if you had a camera, film the launch with a ruler behind, look at it frame by frame, and estimate how long it takes the plane to accelerate. But say, as an example, it took 50ms to accelerate.. Then the force is Δp/Δt = 1500/50 g m/s^2 = 30 g m/s^2 = 0.03 kg m/s^2 = 0.03 Newtons = 0.00674 pounds force. A good time estimate is crucial, and this is still an estimation, but this will at least give you an idea. Hope this helps. |

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