# How does Location of Mass affect Moment of Inertia

This question is applied to automotive wheels/tires.

I have two wheel and tire configurations, obviously these constitute a complex cylindrical shape.

Cylinder A is 74cm in diameter and has a mass of 31.3kg. At a radius of 43cm half of the mass is before that distance and the other half is in the remaining 31cm.

Cylinder B is 74cm in diameter and has a mass of 31.3kg. At a radius of 53cm half of the mass is before that distance and the other half is in the remaining 21cm.

Would Cylinder B have a higher MOI? Put another way, would Cyl B require more engery to accelerate then Cyl A?

I thought that the answer would be no, that the MOI is based upon the mass of the entire cylinder and the diameter across the axis, but I have a conversation that indicates otherwise that if the mass is centered further from the axis the MOI will be higher...

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#### Ranger Mike

Gold Member
every pound of rotational weight you remove for m a race car is like taking off 10 pounds static weight of the car..with regard to acceleration. we run magnesium wheels and i have found the Avon tires we run are 6 to 8 pounds lighter than Hoosiers that everyone else runs....so the car thinks it is 80 pounds lighter when its racing.

also the stock clutch that the car came with from Ford Cosworth had a single 7 1/4 inch clutch ..now we run a 5 1/2 inch twin clutch pack. dramatic reduction in rotational inertia and reduced stress on chankshaft during gear changes...( had to run twin disc pack to handlel 500 lbs torque)

the old drag racing trick was to take off the 11" dia. steel flywheel that was 30 lbs and put on a 10 inch aluminum 10 pound flywheel

but
there are some appliactions where heavy is better. we raced a mini stock Ford Pinto on 1/2 mile asphault track and kept the heavier flywheel and heavier tires on the car..momentum...once up to speed, you needed the momentum to carry you thru the turns at higher speed...miniscule 2 liter 4 cylinder did not have torque enuff to overpower the other cars..also we ran flat tappet cam when the hot dogs were running hi lift roller cam
all appliaction dependant
hope this helps

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#### timmay

Cylinder A is 74cm in diameter and has a mass of 31.3kg. At a radius of 43cm half of the mass is before that distance and the other half is in the remaining 31cm.
I trust you mean a radius of 74cm!

Your conversation is pointing you towards the right answer. The moment of inertia of an object is dependent on the mass distribution related to the axis of rotation. Lets treat the case of two 'complex cylinders' in your problem:

http://g.imagehost.org/view/0147/diagram1 [Broken]

Wheel 1 and Wheel 2 both have the same inner and outer radii. Their total mass is the same, but the distributions of mass are different. Wheel 1 has 50% of its mass located between radius r=a and r=b, and 50% between r=b and r=d. Wheel 2 has 50% of its mass located between radius r=a and r=c, and 50% between r=c and r=d.

For a composite body, we can sum the moments of inertia about the axis of rotation. We know that for a cylinder, the moment of inertia can be described as:

$$I_{z}=\frac{1}{2}m(r^{2}_{i}+r^{2}_{o})$$

where r_i and r_o are inner and outer radiii respectively.

For a composite body of two cylinders as drawn in our diagrams, we can add the moments of inertia for the two parts, knowing that the mass is the same in both and that the outer radius of the first cylinder is the inner radius of the second cylinder.

$$I_{z}=\frac{1}{2}m(a^{2}+b^{2}+b^{2}+d^{2})=\frac{1}{2}m(a^{2}+2b^{2}+d^{2})$$ for wheel 1

$$I_{z}=\frac{1}{2}m(a^{2}+c^{2}+c^{2}+d^{2})=\frac{1}{2}m(a^{2}+2c^{2}+d^{2})$$ for wheel 2

c (for wheel 2) is greater than b (for wheel 1), therefore wheel 2 has the greater moment of inertia.

Ranger Mike gives you a practical demonstration of how this can be seen. By changing a large diameter steel flywheel and replacing it with a smaller, lighter (presumably!) aluminium flywheel, he reduces the moment of inertia of the flywheel and thus can accelerate it more quickly to his desired speed. The trade off is that as soon as driving power is removed from the flywheel, its lower inertia will cause it to lose speed more quickly.

Try finding a textbook or googling flywheel design or moments of inertia for more detail.

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#### Bob S

This question is applied to automotive wheels/tires.
I have two wheel and tire configurations, obviously these constitute a complex cylindrical shape./QUOTE]

The race car has a static weight, based on the mass of the entire car including wheels. It also has a dynamic mass, which is the total kinetic energy of the car plus the inertial (rotational) energy of the 4 wheels, all divided by 0.5 v^2. For conventional automobiles, the dynamic mass is about 5% higher than the static mass. For a race car it could easily add 10% to the static mass. In the kinetic energy formula, (1/2)mv^2, m = dynamic mass. This extra "mass" slows down acceleration.

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