Shooting Physics

[Ether]

Originally Posted by Sparks333

The energy of the shooter and the frisbee together after shooting is

(1/2 * I * omega_end^2) + (1/2 * M_frisbee *V_frisbee^2)

Nope. The frisbee is translating and rotating. Its kinetic energy is (3/4)MV²

[Sparks333]

Hello!

You caught me Ether - that's why I shouldn't derive physics formulas a 2AM.

For people following at home, Ether modified the equation with the assumption that the frisbee can be treated as a thin disk of homogenous density - not strictly true, but definitely close enough for this sort of equation. The more general form of the equation is

I_wheel * omega_start^2 = (I_wheel * omega_end^2) + I_frisbee * (V_frisbee / r_frisbee)^2 + (M_frisbee * V_frisbee^2)

Due to the frisbee's large diameter compared to most shooting wheels, its moment of inertia is definitely non-negligible, despite having a much lower weight (moments of inertia of disks grow with the square of the radius, and linearly with mass).

Thanks for keeping me honest,

Sparks

[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