The new Vex game “nothing but net” might involve rotating shooter wheels. We know that if all the mass of a wheel was located on its rim then the moment of inertia about its rotating axis (I_rim) would be

**I_rim = r^2 * m ** where m is the mass of the wheel and r = radius of wheel. But we know that the wheels actually have mass that is unevenly distributed along the radius so the moment of inertia I_wheel will be less than I_rim.

**Easy experiment to determine I_wheel if we know its mass.**

We can determine I_wheel experimentally using the parallel axis theorem and the dynamics of a pendulum.

Parallel axis theorem says that any object that is rotated about an axis parallel to and a distance , d, from an axis going through the centroid of the object will add an amount = m*d^2 to the moment of inertia about its centroid. I.e.

I_parallel = I_centroid + m*d^2 .

Suppose we now swing the mass, m, about the parallel axis like a pendulum using just the torque from gravity pulling on the mass. It is easy to show that the period, T , of the pendulum is related to the distance , d, and the moment of inertial , I_parallel, by the following formula:

**T = 2*pi*sqrt(I_parallel/(d*m*g)) . g is gravitational constant and assumes swing angles smaller than say 10 degs from the lowest point on the pendulum path. **

If we measure the period of the pendulum we can rearrange the equation and find I_parallel

**I_parallel = T^2*d*m*g/(2*pi)^2 **

Once we have I_parallel, we can now use the parallel axis theorem to determine I_wheel.

**I_wheel = I_parallel – d^2*m = d^2*m * (T^2*g/d/(2*pi)^2 -1)**

(This assumes that the string has negligible mass relative to the mass of the wheel)

**Vex 5 in Wheel experiment:**

Given….r_wheel = 2.5 inches ,

wheel mass, m = 180 gm (0.180 kg)

The pendulum is created by suspending the wheel with a thread 2.75 inches from its center so

d =. 07 m (approx. 2.75 inches)

The average period **T = .668 sec **

**I_wheel = d^2*m*(T^2/d*9.8/(6.28)^2 -1) **

** = d^2*m*( .248*T^2/d -1)**

** = .07^2*.180*(.248*.668^2/.07 -1) = ****0.00051 kg m^2**

Equivalent radius with a rim only mass r_e

**r_e = sqrt( I_wheel/m) = .0533 m ( 2.1 inches) **

This means that the wheel behaves as if the mass if located at 84 % of the radius of the wheel which one could almost guess by looking at it.