Pitching Machine Physics

[waialua359]

Originally Posted by Matt H.

I've seen several attempts to describe the physics governing a pitching machine. All of these attempts have been hard to follow and seem to miss the key point. Here I present a physical model of a pitching machine which might help teams make decisions.

Assumptions:
1) Two wheel pitching machine
2) Both wheels have the same angular speed in opposite directions.
3) Contact between the wheels and the ball occurs without slipping. This represents a good design as energy is not wasted ripping the poof ball to shreds and from my experience with 2006 is a reasonably assumption. Compressing the poof ball moderately makes this assumption hold on a real life robot.
4) No energy is added to the system during the shot. This is justified if the shot occurs very quickly or the motors spinning the wheels are running at low efficiency.

Variables:
w_i--Initial angular velocity of the wheels (radians/sec)
w_f--Final angular velocity of the wheels (radians/sec)
m--Mass of the ball (kilograms)
v--The final velocity of the ball (meters/sec)<--What we want to know
I--Moment of Inertial of a single wheel
r--radius of a wheel (meters)

Useful Equations:
Energy of a spinning wheel--1/2 * I * w^2
Energy of a moving object--1/2 * m * v^2

Calculations:

Using conservation of energy
I(w_i)^2=I(w_f)^2+1/2 * m * v^2

From the rolling without slipping assumption (3)
v=(w_f) * r
or
w_f=v/r

So combining these equations
I(w_i)^2=1/2 * m * v^2 + I(v/r)^2
v^2=(I*(w_i)^2)/(0.5*m+I/(r^2))

v=(w_i)*sqrt(I/(0.5*m+I/(r^2)))


Physical Description:

Because we assume no energy enters the system (4), energy is conserved. As the ball passes between the wheels, energy is transferred from the wheels to the ball. As such the ball's velocity increases, and the angular velocity of the wheels decreases. The rolling without slipping condition (3) tells us by how much this occurs.

Useful Units:
The above equation assumes SI unit which most teams don't use. Additionally, most teams don't know the moment of inertia of their wheels off hand. Here is an equation which assumes all of a wheel's mass is in the rim and uses the below units:


m--Mass of the ball (pounds)
w_i--Initial angular velocity of the wheels (rpm)
v--The final velocity of the ball (feet/sec)<--What we want to know
M--mass of a wheel (pounds)
r--radius of a wheel (inches)

v=0.0087*(w_i)*sqrt((M*r^2)/(0.5*m+M))

Example:
4" wheels which weight 1 lb each being spun at 5000 RPM (approximately the no load speed of a CIM

v= 140 ft/sec or roughly 40 m/s

Note that your performance will likely be worse due to wheel mass not being concentrated in the rim, energy losses compressing the ball, and because 40 m/s balls will be very quickly slowed by air resistance.

If there's interest, I can also model systems with differential RPM to produce spin (same general concept, but now conservation of angular momentum is important).

Have you considered using just 1 wheel?
When a basketball player shoots, a backspin occurs as a result of just 1 hand.
Or 2 wheels but different speeds.

Just saying, as the decision for teams may be other configurations other than just a pitching machine.