<p>@Grisam-*she :)</p>
<ul>
<li>embarrassed*</li>
</ul>
<p>Guess 1: two pi</p>
<p>guess 2: exactly zero</p>
<p>
</p>
<p>= HSL :rolleyes:</p>
<p>Heaven: 2 pi is a number, not a thickness. 2 pi feet? 2 pi light years?</p>
<p>I think none of you has ever owned and maintained a car.</p>
<p>
</p>
<p>=[ aww I thought it was a trick question. Well, let’s see, you’ve given us the following parameters:
- We’re definitely earthbound
- We’re driving a vehicle normally, assuming that the car has wheels</p>
<p>By driving normally, I mean that we don’t just jam the gas pedal, but rather accelerate until we hit a certain speed and try to keep the speed constant, therefore when driving normally, we’ll assume that we’re driving at an equilibrium speed of S mph without accounting for the initial interval where we must accelerate.</p>
<p>Let’s say that car as a whole, plus the driver, weighs W kg, and that we designate the gravitational force at that point as g. The tire has a radius of R.</p>
<p>The frictional force = Wg/R * k where k is an unknown function that represents the relationship between the Normal/R and the frictional force. We know that the amount of rubber lost have a direct correlation to the amount of work done by the tire that is converted to heat energy (the percentage of which we’ll call H), therefore H(Wg/R<em>k)(2</em>pi*R) will give us the amount of heat energy given off after one revolution of the wheel.</p>
<p>However, since the loss of the rubber is not discrete, we’ll have to find the heat output as a time dependent function:
let D(t) = the distance traveled from an arbitrary starting point, or in this case int0->t(S), or just S<em>t,
Therefore, the HO(t) function can be re-expressed as H(Wg/R</em>k)D(t)</p>
<p>If the rubber loss is z<em>Heat, then the rubber lost during any given interval T is int(0->T)(H</em>W<em>g/(R</em>k)<em>S</em>z), where T is 2<em>pi</em>R/(S) (Which will give us the same answer as the discrete approach, given that only the work output was time dependent.</p>
<p>
</p>
<p>this .</p>
<p>I would say approximately ten micrograms.</p>
<p>I’d estimate that a car tire travels 1 meter every revolution, and every 1000 kilometers it would lose about 10 grams of mass. So it should lose about 10^-5 grams every revolution.</p>
<p>10g per thousand kilometers is kind of an overestimate. 1g might be a good estimate as well, in which case the answer would be approximately “one microgram”.</p>
<p>If these aren’t right, please guide us towards the real answer.</p>
<p>the tire is a subset of R^3
the surface of the tire is a subset of R^2
R^2 = 0% of R^3
thus the surface gone is 0% of the tire</p>
<p>:rolleyes:</p>
<p>
</p>
<p>You don’t need a degree from MIT to make that conclusion about a bunch of high school kids :rolleyes:.</p>
<p>Still working on the problem…</p>
<p>AH</p>
<p>“How much rubber comes off a typical car tire each time the tire turns while driving normally?”</p>
<p>When it says “turns”, it means as in turning through an intersection, not one revolution.</p>
<p>^ That’s what I initially thought, but he rephrased the question to indicate revolution.</p>
<p>This is not a trick question. But the question as posed in 1974 in introductory physics (8.01) at MIT to the freshman class I think was phrased as how “THICK” is the layer.</p>
<p>No calculus is required. A calculator is not required, just pencil and paper. However, some thought, knowledge, experience is.</p>
<p>I must say that when I heard the answer and had it explained the following week in class, I felt awfully dumb. That’s why after all these years I still remember the problem.</p>
<p>The first phsics class problem set was not geared at physics, but at getting an understanding of measurements, like orders of magnitude and what they mean in real life, so that we could gauge whether an answer that we dilligently obtained on a calculator made any real world sense.</p>
<p>So this is a dimensional analysis problem that doesn’t require the analysis of forces, work/energy, etc.? I’m getting the impression that the answer is calculated through a very simple and straightforward method, which I’m not getting.</p>
<p>Answer unknown?</p>
<p>Hint: XRCatD was on the right track initially in post #27.</p>
<p>Only one equation is needed. One that you learned in 8th grade. The rest is just math. (you really don’t even need that equation). And you don’t need to know physics and its equations.</p>
<p>Since this thread started on the MIT page, you guys think you have what it takes to get into and graduate from MIT. That not only takes hard work, but out of the box thinking.</p>
<p>In my first computer programming course at MIT, we had to write a program that would do some sort of mathematical calculation, I don’t remember exactly what it was right now. Writing “a” program was easy – everyone in the class could do it, even me. Our programs had 30 to 40 lines of code. However, there was one person who did it in 7 or so line of code. He thought about it and came up with the elegant solution. He’s probably now running some computer company somewhere.</p>
<p>Amount lost per x miles, divided by number of revolutions per mile?</p>
<p>If it requires knowledge then I guess I’m out.</p>
<p>So if we knew the average wear in terms of mass lost per distance of an average tire tread, we’d know the answer using a simple geometry equation? That’s highly anticlimactic lol. </p>
<p>Elegance in computer programming doesn’t translate into the sort of straightforwardness this question seems to implicate. In the former, there are multiple ways of achieving the same degree of accuracy, whereas the latter seems to be asking for an answer in terms of averages, instead of an accurate formula that takes into account the different aspects of physics and chemistry that could be applied to a question like this. It may teach the concept of useful approximations that may be “good enough,” but it’s not as satisfying as the more complicated way. I guess it’s good enough when considering real-world applications, though.</p>
<p>Pi (R1^2-R2^2)*Thickness of tire
Where R1 is the original radius, R2 is final radius</p>
<p>It works!!!1!! I’m Einstein!!! :p</p>
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