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Centripedal Motion and Tires

Question: How do tires stay on the wheel of a car?
Answer: This has to with how the tire is designed and mounted, as well as with its pressure.

Introduction | Experiment | Videos | Analysis
Introduction

Tire manufacturers have to consider many things when designing their products. Firstly, they must consider what their product is being used for: the conditions, requirements, and the expected longevity. This leads to many types of ties; however, almost all tire require air pressure. Air pressure helps hold the tire onto the wheel, and also ensures proper contact with the road. Wheels need proper contact to ensure even wear, solid grip, and good distribution of forces. However, considering all the forces acting on a wheel, what would happen if things were not designed properly?


Experiment

We decided to mount a non-inflated tire to a wheel, and test its grip at high RPMs. For this experiment I mounted two 1.5V electric motors onto a plastic case. I used a potentiometer to regulate the voltage (and therefore the RPM). Once this system was put together, I lifted the rig off the ground with a retort stand, and hooked it up to some batteries. This is what it looked like:

car.jpg test.jpg


Test Videos

Well, we gave it a try and this is what happened. If you can notice, at high RMPs the rubber actually separates from the rim. Try and have a look:

movie.jpg


Analysis and Conclusion

We know that the force acting on the tire was a centripetal force (one related to rotation). Therefore, we chose the appropriate equations and did some number crunching:

The wheel had a diameter of 6.35 centimeters. However, we were interested in its circumference (i.e. the distance the car travels per rotation). This came out to be roughly 20cm or 0.2 meters.

We then used a tachometer to find the maximum rotations per second (RPS) done by the motor. The motor, when over clocked (at 9V) did about 130RPS.

This means that the cars maximum speed was 0.2(130)=26ms-1 and thus, by using F=(mv2)/r we can find the centripedal force acting on the tire.

Given that the tire’s mass is 0.05kg:

F=(262*0.05)/6.35=5.32N

That means a force of just over 5 Newtons was pushing the tire off the wheel.


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