# Calculating a wheel’s moment of inertia

## As long as you move a rigid object in a straight line, it does not matter how its mass is distributed throughout it; the amount of work to move it will be the same. This is not true if you drive the object by rotating it; then how the mass is distributed does play an important role in how much energy it takes to move it. In the case of a wheel, it is probably obvious that it will take more work to accelerate it if the mass is concentrated out at its edge than at its center. But how do you quantify that?

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### The math behind the VeloNews Tech Report: Light as a Feather, Stiff as a Board

*Editor’s note: The following is an explanation of the inertia test used in the Aug. 18, 2008 issue of VeloNews (and later in the September 2010 issue). Grab a cup of coffee and your thinking cap. You’ll need them both as Lennard busts out the physics.*

If you move a rigid object in a straight line, it does not matter how its mass is distributed; the amount of work to move it will be the same. This is not true if you drive the object by rotating it; then, how the mass is distributed plays an important role in how much energy it takes to move it. In the case of a wheel, it is probably obvious that it will take more work to accelerate it if the mass is concentrated out at its edge than at its center. But how do you quantify that?

*Rotational inertia, or moment of inertia,* is the rotational equivalent of mass; this is the quantity that we want to measure to see how much energy it takes to accelerate a wheel. We can measure the mass of the wheel easily enough, but it is not necessarily the case that the lightest wheel will have the lowest moment of inertia, or vice versa.

That’s where the torsional pendulum comes in — a device that measures a wheel’s rotational inertia.

Ever since I built this torsional pendulum for bicycle wheels and tested the moment of inertia of some wheels in the June 28, 1999 issue of *VeloNews*, I’ve wanted to provide readers with a full explanation of the physics involved. So here it is.

**Using the Torsional Pendulum**

If we twist the vertically hanging rod to a relatively small angle without a wheel attached, it will twist back and forth rapidly, and we can measure how many times the horizontal rod welded to it oscillates back and forth in a given amount of time. We don’t want to twist so far that we exceed the elastic limit of the rod; we want to be well within the range where it springs back and forth in a repeatable manner.

If we affix a wheel on the end of the vertical rod and twist it to a similarly low angle, it will twist back and forth more slowly. The higher the moment of inertia of the wheel, the more slowly the torsional pendulum will twist back and forth.

And as long as we keep the deflection to low angles, it does not matter what angle we are actually twisting to; one of the characteristics of oscillating harmonic motion is that the period and frequency are independent of amplitude.

Our vertical rod behaves like a spring, and we aim to discover the torsional spring constant of the rod. From that, we can determine the moment of inertia of the entire apparatus — the wheel and the horizontal rod twisting back and forth. The moment of inertia of the horizontal rod must be subtracted from the total moment in order to be left with the moment of inertia of the wheel alone. Fortunately, the moment of inertia of a uniform solid cylinder about a central diameter is easy to find.

**Rotational Inertia or Moment of Inertia**

If we have a rigid body rotating at an angular speed Ω about a fixed axis, each particle in it will have a certain amount of kinetic energy. A particle of mass m at radius r from the axis of rotation moves in a circle of radius r with an angular speed Ω about this axis. It therefore has a linear speed of v = Ωr, and its kinetic energy K is ½ mv^{2} = ½ mΩ^{2}r^{2}. The total kinetic energy is the sum of the kinetic energies of its particles, and each and every particle in a rigid body moves with the same angular speed Ω (it moves through the same number of degrees of angle per unit time), but the radius r may be different for different particles. So the total kinetic energy is

K = ½ (m_{1}r_{1}^{2} + m_{2}r_{2}^{2} + …)Ω^{2} = ½ (∑m_{i}r_{i}^{2})Ω^{2}

The sum of the products of the masses of the particles by the squares of their distances from the axis is ∑m_{i}r_{i}^{2}. This is the moment of inertia or rotational inertia and is denoted by I.

I = ∑m_{i}r_{i}^{2}

So the kinetic energy of the entire rigid body can now be written as,

K = ½ I Ω^{2}

This is not a new kind of kinetic energy; as you saw from the derivation, it is the same translational kinetic energy K = ½ mv^{2} expressed for a rigid body rotating about a fixed axis.

Rather than adding separate terms for discrete point masses, we integrate I for a rigid body with a continuous distribution of matter, yielding

I = ∫r^{2}dm

And calculating it for a uniform solid cylinder rotating about a central diameter, like the horizontal rod in our fixture that rotates perpendicular to the vertical rod, we find:

I = MR^{2}/4 + ML^{2}/12

M is the total mass of cylindrical rod, R is the radius of the rod, and L is the length of the rod. In the case of our torsional pendulum, M = 87 grams, L = 70cm, and R = 0.225cm, so

I_{rod} = 3.5 X 10^{4} g-cm^{2}

**Equation of motion**

The equation of motion of a simple harmonic oscillator — a particle moving back and forth about an equilibrium position (think of a steel block attached to a spring and sliding on a frictionless surface, boinging back and forth as the spring extends and compresses) — can be derived from thinking about its potential energy, U. The expression for the potential energy of an “ideal” spring (one that doesn’t lose energy through friction) compressed or extended through a distance x is:

U_{(x)} = ½ k x^{2}

Where k is an arbitrary constant. First-year physics and basic calculus tells us that the force acting on the block is

F_{(x)}= – dU/dx = – d(½ k x^{2})/dx = – kx

This is called Hooke’s Law, and using Newton’s second law, F = ma, we get,

m d^{2}x/dt^{2} = – kx

or the differential equation of motion for a simple harmonic oscillator,

d^{2}x/dt^{2} + kx/m = 0

The rotational version of this differential equation for our torsional pendulum is similar, but we are now interested in angular displacement, θ, not the linear position, x, the moment of inertia, I, not the mass, m, and the torque, τ, not the force, F. Hooke’s law then becomes

τ = I d^{2}θ/dt^{2} = – kθ

or,

d^{2}θ/dt^{2} + kθ/I = 0

The constant k now depends on the properties of the vertical rod and is called the torsional constant. The solution requires that θ(t) be a function whose second derivative is the negative of the function itself, save for the constant factor k/I. The sine or cosine function has this property and is not affected by being multiplied by a constant A, so one solution to this equation is

θ = A cos(Ωt)

where Ω = √(k/I) is the frequency of angular oscillation in radians/sec, and A is the maximum angular displacement; in other words, it is the amplitude of the angular oscillation.

(A general solution to this equation is actually

θ = A cos(Ωt + σ)

where the constant σ allows for any combination of sine and cosine solutions, since

cos(Ωt + σ) = cos(Ωt)cos(σ) – sin(Ωt)sin(σ) = a cos(Ωt) + b sin(Ωt)

However, σ only affects the phase of the oscillation, so for the purposes of finding the moment of inertia, it provides no benefit, nor does including both sine and cosine functions. So we can simplify the equation to θ = A cos(Ωt), and we’ll still retain all we need to calculate the moments of inertia of the wheels.)

It can be shown that cos(Ωt) is an oscillatory function whose value repeats every time Ωt increases by an integral multiple of 2π, meaning an integral number of cycles.

θ = A cos[Ω(t + 2π/Ω)]

= A cos(Ωt + 2π)

= A cos(Ωt)

In other words, if the time t is increased by 2π/Ω, the expression repeats itself, so 2π/Ω is therefore the period of the motion T.

**Finding the torsional constant k**

We can find k two different ways.

1) We can measure the torque required to twist our torsion rod to a given angle θ, since τ = kθ. We did this by winding up the torsion rod to a given angle θ by pushing on the end of the horizontal rod of the torsional pendulum at right angles to a given angle θ with a digital scale. The scale read 82 grams at 30 degrees, or π/6 radians, and since there are 980 dynes in a gram of force (i.e., adjusting for the acceleration of gravity), the force to displace the rod end π/6 radians was

F _{(π/6)} = 82 * 980 = 8 X 10^{4} dynes

The measured forces at π/4 (45 degrees), π/3 (60 degrees) and π/2 (90 degrees) were 133gf, 199gf, and 273gf, yielding

F _{(π/4)} = 133 * 980 = 1.3 X 10^{5} dynes

F _{(π/3)} = 199 * 980 = 1.9 X 10^{5} dynes

F _{(π/2)} = 273 * 980 = 2.7 X 10^{5} dynes

Since the magnitude of torque τ is r * F, where F is the force applied perpendicular to the radius r, and r = 35cm for our horizontal rod, using τ = kθ, we get at our smallest angle

k π/6 = r * F = (35cm)( 8 X 10^{4} dynes)

k = 6(35)( 8 X 10^{4})/π dyne-cm

= 5 X 10^{6} erg

Using the same method, we get

k = 6 X 10^{6} erg

at the other three angles.

2) We can also determine k by measuring the period of oscillation of the apparatus without a wheel attached. Since Ω = √(k/I) and T = 2π/Ω, then

(2π/T)^{2} = k/I

and

k = (2π)^{2}I/T^{2}

Inserting the period of oscillation of the rod alone (0.5 sec) and the moment of inertia we calculated above for the rod alone, I_{rod} = 3.5 X 10^{4} g-cm^{2}, we get

k = (2π)^{2}(3.5 X 10^{4} g-cm^{2})/(0.5s)^{2}

= 6 X 10^{6} erg

Now that we have k, we can find the moment of inertia for any wheel we attach to our torsional pendulum! Here we go!

Two equivalent ways to solve for I

1) We saw above that the potential energy of a linear harmonic oscillator is

U(x) = ½ k x^{2}

The equivalent expression for our torsional pendulum is

U(θ) = ½ k θ^{2}

We also know that the total energy of the system is equal to the sum of the potential energy and the kinetic energy, or E = K + U. When the potential energy is at a maximum, the pendulum is stopped at the end of its motion, so K = 0. When the kinetic energy is at a maximum is at θ = 0, where there is no twist in the rod, so U = 0. Since

θ = Acos(Ωt)

the potential energy U will be at its maximum when cos(Ωt) is at its maximum, in other words, when cos(Ωt) = 1. Thus,

U(max) = ½ k [Acos(Ωt)]^{2}

= ½ k A^{2}

If we were to graph the angular velocity Ω as a function of time, we would see that the maximum angular speed is ΩA, and since T = 2π/Ω, we get

K(max) = ½ I Ω^{2}

= ½ I (2π/τ)^{2}A^{2}

We know that K_{(max)} = U_{(max)}, so

½ k A^{2} = ½ I (2π/τ)^{2}A^{2}

or,

I = k τ^{2}/4π^{2}

2) Another way to look at this is that since Ω = √(k/I) and T = 2π/Ω, then

(2π/T)^{2} = k/I

then,

I = k τ^{2}/4π^{2}

Remember, this is the moment of inertia of the entire system; we need to subtract off the moment of inertia of the horizontal rod alone to be left with just the moment of inertia of the wheel, so

I = k τ^{2}/4π^{2} – I_{rod}

**Plugging in the numbers and finding the moments of inertia for the wheels**

*Editor’s note: The following applies specifically to the climbing wheels tested for the **Aug. 18, 2008 issue.*

* *

If we pick one of the wheels that came out in the middle of the readings, namely 20 oscillations in 30 seconds, we find both the front and rear Reynolds and Shimano wheels as well as the front Easton wheel. The period of oscillation T is thus 30/20 seconds, or 1.5 seconds. Plugging this and the k = 6 X 10^{6} erg , and the I_{rod} = 3.5 X 10^{4} g-cm^{2} that we got earlier, we get

I = (6 X 10^{6})(1.5)^{2}/4π^{2} – 3.5 X 10^{4} g-cm^{2}

= 3 X 10^{5} g-cm^{2}

for the front and rear Shimano and Reynolds and front Easton wheels.

It is always interesting and informative to double check our answers. Let’s first imagine that all of the mass of a theoretical wheel is concentrated at its outer edge, so I = MR^{2}, where M is the total mass of the wheel and R is the wheel’s radius (which in the case of all of these wheels is 31.75cm, since they roll out to a circumference of 199.5cm). So, R^{2} =1008 ≈ 10^{3}. To get a wheel with all of its mass concentrated at its outside edge to have the same moment of inertia as we calculated for the front and rear Shimano and Reynolds and front Easton wheels, how heavy would it have to be?

M = I/R^{2}

= 3 X 10^{5}/10^{3} grams

= 300 grams

That’s believable, eh?

Let’s look at it another way. What if the entire mass of, say, the Shimano front wheel, were concentrated at its outer edge, what would its moment of inertia be?

I_{(max)} = MR^{2}

= 526(31.75)^{2} g-cm^{2}

I_{(max)} = 5.3 X 10^{5} g-cm^{2}

How about if only half the Shimano front wheel’s mass were out at the rim’s edge and the other half of its mass were located directly at its center at r = 0?

I_{(1/2max)} = ½ MR^{2} + ½ M (0)^{2}

= 263(31.75)^{2} g-cm^{2}

= 2.7 X 10^{5} g-cm^{2}

We are getting believable numbers in the ballpark with each other. I think we can calculate the rest of the wheels with some confidence now.

## Oscillation Frequency Front Wheel |
||||

Avg. cycles in 30s | Period (sec.) | I (g-cm^{2}) |
Weight (g) | |

Bontrager Race XXX Lite | 19 | 1.58s | 3.4 X 10^{5} |
511g |

Campagnolo Hyperon | 19.33 | 1.55s | 3.2 X 10^{5} |
541g |

Easton EC90SLX | 20 | 1.50s | 3 X 10^{5} |
514g |

Reynolds MV30T | 20 | 1.50s | 3 X 10^{5} |
503g |

Shimano WH-7850-C24-TU | 20 | 1.50s | 3 X 10^{5} |
526g |

Zipp 202 | 21 | 1.43s | 2.7 X 10^{5} |
464g |

## Oscillation Frequency Rear Wheel |
||||

Bontrager Race XXX Lite | 19 | 1.58s | 3.4 X 10^{5} |
693g |

Campagnolo Hyperon | 19 | 1.58s | 3.4 X 10^{5} |
711g |

Easton EC90SLX | 19 | 1.58s | 3.4 X 10^{5} |
714g |

Reynolds MV30T | 20 | 1.50s | 3 X 10^{5} |
662g |

Shimano WH-7850-C24-TU | 20 | 1.50s | 3 X 10^{5} |
725g |

Zipp 202 | 21 | 1.43s | 2.7 X 10^{5} |
594g |

As you can see, the Zipp has the lowest weight and the lowest moment of inertia. Its rotational inertia is the same as if it weighed 270 grams, all concentrated at its outer edge. Sounds like a fast wheel, all other things being equal (which they never are).

Interestingly, the Shimano wheel is considerably heavier than the Bontrager in both front and rear, yet it has a lower moment of inertia. And the Shimano wheels share the same rotational inertia with the Reynolds wheels, which are lighter yet. This indicates that the Shimano’s rim and the outboard ends of its spokes are lighter. The Shimano’s and Easton’s rotational inertias are the same as if they weighed 300 grams, all concentrated at the rim’s outer diameter, while the Bontrager’s rotational inertia is the same as if it weighed 340 grams, all concentrated at its outer rim diameter.