Finding the moment of inertia of a Vex wheel using parallel axis theorem

May 17, 2015

vex 5 in wheel

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.

2014 FRC Choo-Choo wheel reset geometry calculator

January 30, 2014

This year’s FRC robot has a catapult arm that requires a reset wheel mechanism that will pull the arm down and release it automatically by just rotating a single “choo-choo” wheel.    The exact linkage lengths and relative pivot point locations are determined by the attached excel program.

2014 FRC arm pivot geometry calculation

Note: Minimum Motor Torque for Turning

November 23, 2011

I wanted to do a quick calculation for a student Vex robot that has 5 in wheels and is having one rail stalling during a turn.    The stalling could have been predicted with some basic torque calculations.

Using static moment equations you can easily derive the minimum torque to just turn  for a symmetric robot with four direct drive  motors on wheels with radius r .  Legacy three-wire 6.5 inlb motors are assumed.

Define the half spacing between axles as “l” and the half width between wheels as “w”   , the robot weight as W_lbs and the coefficient of friction for a tire side force as u.  Also assume that the weight is evenly distributed on the four wheels and the center of turn is at the cg.  Also w > l so turning can occur.


Input turning moment per wheel = w* torque/r and this must be greater than the moment resisting the turn caused by the side friction forces on the wheel.

Resistance moment = l*W_lb*u/4  .    The factor of 4 converts the W_lbs to normal force on the wheel.

So torque > r*(l/w)*W_lb*u/4

Lets plug in some typical numbers.  r = 2.5 in, l=3.5, w = 7

This gives a required motor torque of  2.5*3.5/7/4*W_lbs*u

or torque >( 2.5/8) * W_lbs* u

or W_lbs <  torque*8/2.5/u

Without omni wheels, u is typically  around 1 or higher and if the robot W_lb is 12 lbs  then the minimum torque

torque_min > 3.75 in lbs

This represents about 60% of the old three wire motors 6.5 in lb  spec torque .       Clearly there is not much margin for any torque losses in the drive train or extra side friction on the wheels or extra weight from a heaver robot or from game objects.  Also any contact forces resisting the turn will easily stall the turn.

Using a 4 in omni wheel in place of the 5 in traction wheels increases the input torque by  5/4 and reduces the u by about 1/3   so we get a combined improvement of  15/4 or 3.75 factor on the required torque.

min_torque_ 4 in omni =  3.75 / 3.75 = 1 in lb or about 15% of the available torque of an old motor.   This is where you want to be in your design.