Shooting Physics

[Sparks333]

I realized the equation still isn't in a state that could be considered 'simple' - solvable with the Magic of Algebra, yes, but not simple.

Isolating for V_frisbee:

V_frisbee = (sqrt((4*M_wheel + 3*M_frisbee)*M_wheel) * omega_start * r_wheel)/(4*M_wheel + 3*M_frisbee)

This makes the assumption discussed above that the shooter wheel is a homogenous cylinder with mass M_wheel spinning at omega_start radians per second prior to launch. This equation is only good for single-wheel shooters - otherwise, you'll have to go back and solve and add the individual moments of inertia. As previously discussed, this assumes the frisbee doesn't skid on either the wheel or the outer edge of the shooter, that system is closed (i.e. the motor driving the shooter wheel doesn't add much energy in the time it takes for the frisbee to be shot), and that all 100% of the energy lost by the spinning wheel is transferred to the frisbee.

Luck,

Sparks