Re: Does wheel size affect a car's speed if you keep everything else constant?

The question: "Does wheel size affect a car's speed if you keep everything else constant? The car is going to be propelled by gravity. I am going to set a car at the top of a ramp and release it. I will time it's speed and then I will repeat the process with a different size wheels. I am going to make the car out of Legos. I will be keeping mass, height, road surface, aerodynamics constant. I have done a lot of research but I have not found anything that relates to my question. All the web-sites and books I have found discuss the effects of aerodynamics on speed, how the mass of the car affects speed, how the road surface affects speed and how the heat of the tires affects friction. The stuff I have read makes me think the wheel size won't matter, but I don't really have the physics to back up that idea."

First, look through the following previous answers in our archives, and then I will add a few comments:

one

two

three

The main thing you need to take away from the above information is that rotational inertia, called "moment of inertia", depends on the distribution of mass around the axis. You can see in the first previous answer above that a uniform disc of material will have a moment of inertia that is equal to 0.5mr², whereas a disc that has most of its mass at the rim will have a moment of inertia that approaches 1.0mr² (As long as there are spokes or other means of supporting the rim, the coefficient in front of "mr²" can not be exactly equal to 1.0).

You will find the most difference in your experiment between wheels that have more mass out towards the rim versus those wheels that have less mass out towards the rim (perhaps by tapering the thickness of the wheel to be thinner out at the rim: a "knife-edge" wheel).

The math you need to compute the difference is not too difficult. We start with the principle of the conservation of energy, and we ignore friction (A favorite trick of physicists!). At the top of a ramp of height "h" the total energy of the car can be taken to be only gravitational potential energy, with magnitude
E_i = PE_i = Mgh = (m_c + m_w)gh
where the suffix "i" denotes initial condition, "M" is the total mass of the car, m_c is the mass of the body of the car, m_w is the total mass of all the wheels, "g" is the acceleration of gravity, and "h" is the (vertical) height of the ramp. At the bottom of the ramp, where "h" is zero, there is no gravitational potential energy (in relation to the ramp! Of course it has nonzero gravitational potential energy if we measure "h" from some place other than the bottom of the ramp.), so all the energy is kinetic, composed of linear kinetic energy (KE) and rotational kinetic energy (RKE). It is the wheels which absorb the rotational part of the kinetic energy, and both the car's body and the wheels obtain linear kinetic energy. So,
E_f = KE_f + RKE_f = 0.5Mv² + 0.5Iw²
where the suffix "f" denotes final conditions and the new quantities are "v", the final linear speed of the car, "I" is the total moment of inertia of the wheels, and "w" is rotational speed of the wheels (measured in radians per second). Let's call the coefficient in front of the formula for moment of inertia "f". That is, for the wheel that has all its mass at the rim, f would equal 1.0, for instance. So then
I = fm_wr²
where m_w is the total mass of all the wheels, and "r" is the radius of the wheels. So we have
E_f = 0.5Mv² + 0.5fm_wr²w²
The linear speed and the wheels' rotational speed are related:
v = wr, or w = v/r, so
E_f = 0.5Mv² + 0.5fm_wr²(v²/r²)
or, after some algebra,
E_f = 0.5Mv² + 0.5fm_wv² = 0.5(m_c + m_w)v² + 0.5fm_wv²
Due to conservation of energy E_f = E_i so
0.5(m_c + m_w)v² + 0.5fm_wv² = (m_c + m_w)gh
and after a little more algebra we have
v = sqrt { 2(m_c + m_w)gh / ((m_c + m_w) + fm_w) }
where "sqrt" is the squareroot. This can be rearranged slightly to obtain
v = sqrt { 2gh(m_c + m_w) / ( m_c + m_w ( 1 + f ) ) }

The bottom line: For wheels whose "moment-of-inertia fraction" f is smaller, the final velocity of the car will be greater because f appears in the denominator. In fact, if you could magically produce wheels that have zero moment-of-inertia fraction the equation reduces to one you may have seen for a block sliding on a frictionless inclined plane:
v = sqrt { 2gh }

In all this we have neglected friction, but if you do a good job polishing and/or lubricating the axles then friction can actually be a small part of the answer, but if you find results that vary significantly from what you expect you can suspect that friction is at least one of the culprits.

John Link, MadSci Physicist