In between AP calculus, AP English 12 and Honors Physics, I'm working on a research project. The official title is "Optimum Mass of a Small Rocket Propelled Vehicle."
The basic premise of the research is to find the factors that determine the optimum mass of a model rocket. Does it occur when the maximum momentum is at motor burnout, or maximum kinetic energy?
The idea of optimum mass is to find the mass of a rocket that will yield the highest flight. You can't throw a bowling ball as high as a basketball, but you can throw the basketball a lot higher than a beach ball. So for any given shape, there is a mass that is ideal for a given impulse. (Of course, on the moon the optimum mass would always be zero, because there is no atmosphere to slow the rocket down, so you want it to be as light as possible).
My idea was to fly a rocket multiple times with all variables held constant save its mass. I built a small ballast compartment to fly different amounts of modelling clay. The next step wasto measure the altitude the rocket attains using two theodolites. I was hoping to see some correllation between mass and altitude, then determine whether the optimum mass was determined by maximum momentum at burnout or maximum kinetic energy at burnout.
There was a problem. I discovered that Optimum Mass can lead to Lost Rockets. Especially during September/October when the corn stalks are twelve feet tall.
I was hoping to finish this project and send it to New Jersey by October 1st for a science competition. But all that changed when I watched my little orange rocket drift over the horizon, never to be seen again.
I am still going to finish this research project, even though I missed the deadline for the competition. Maybe I'll enter it in some other competition or science fair.
P.S.
(For the scientists on my blog): I think that it might be possible to use logic to solve this hypothesis. The equation for momentum is given byp = mv
where p is the momentum, m is the mass of the object and v is the velocity of the object. The equation for kinetic energy is given by
E = 1/2(mv^2)
In the first equation, momentum rises with velocity. In the second equation, kinetic energy rises with the square of the velocity, meaning that when you double the velocity, it will take four times the energy to cancel the forward energy.
This is key: the drag equation also rises with the square of the velocity.
D = 1/2(rho*Cd*v^2*A)
where D is the drag force, rho describes the atmospheric conditions, Cd is a coefficient that sums up the complex dependencies, v is the velocity, and A is the fronal area of the rocket. Doesn't this equation look astoundingly similar to the equation for kinetic energy?
Is any of this making sense? I haven't exactly figured it all out yet, but when I do, I'll post my whole research report.
Happy Rocketeering!
