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Physics question about expansion caused by heat

HalberdBlueHalberdBlue Registered User regular
edited March 2007 in Help / Advice Forum
Today in my AP Physics class we had a problem. The teacher made it seem pretty simple but it seemed wrong to me so I gave it some more thought and tried to explain what I thought to the teacher and she either didn't understand what I was trying to say or didn't care. So I'm turning to the anonymous internets for help!

The exact numbers in the problem don't matter, I'm going to change them just to make the math simpler (because I didn't bring my calculator home).

Basically you have a copper ring. The hole in its center is a circle of 9.98 meters squared. You have to fit a steel rod that has a cross-sectional area of 10 meters squared through it (it is also round). The temperature is 40 degrees celsius; to what temperature must the copper ring be heated to in order to fit the steel rod through?

The equation for thermal expansion for area is Change in Area = 2 times the expansion constant times the change in temperature times the original area.

equationpa5.jpg

The way the teacher shows us to solve it is that we know the change in area is 0.02 meters squared and the original area and the expansion constant so we just solve for change in temperature and its case closed.

Well, I didn't buy that. In my view what we solved was there is a circular copper disk of size 9.98 meters squared and how much would the temperature have to change to get it to 10 meters squared.

In order to solve the actual problem we would need to find out how much the radius of the hole in the copper ring would need to increase to get the larger area, and then know the width of the copper ring to find out the area of the copper ring. Then knowing the old area of the copper ring and the new area you can once again solve for the change in temperature. Am I wrong or is the teacher?

Using simple easy numbers (but maintaining the numbers given in the problem): the hole in the ring is 9.98 meters squared and you are trying to get it to 10 meters square. The ring is made out of a material that has an expansion constant of 0.5 (so we can ignore it in the equation). The ring also has a width of 1 meter (I just made up this number to use as an example, in the problem we were not given a width for the ring).

ringqs4.jpg

Since we just learned this concept today, I am not sure if the radius of the copper ring itself would expand by the same percentage as the hole in the middle. Its important but I made the assumption for this post and you all are welcome to correct me if I'm wrong :)

So I got that the original area of the copper ring is 14.34 meters^2 and the new area of the copper ring is 14.3865 meters^2, giving a difference of 0.0465 meters^2, not the 0.02 meters^2 that we used. And this figure changes depending on the radius of the copper ring that I chose!

So am I full of crap and missing something or is it a bad problem and my teacher was wrong?

HalberdBlue on

Posts

  • SmasherSmasher Starting to get dizzy Registered User regular
    edited March 2007
    Imagine for a moment that you have a copper disk with an area greater than 9.98 m^2, it doesn't matter how much. Picture an imaginary disk with area = 9.98 m^2 centered in the middle of the copper disk. Some number of the copper atoms are inside the imaginary disk, and the rest are outside of it.

    Now heat up the copper by some amount, and let the imaginary barrier expand so that the same atoms are inside and outside of it as before you heated the disk. The copper atoms on each side of the barrier will expand in the same manner no matter what is on the other side.

    Because of that, you can just get rid of the center of the disk entirely. When you look at it that way, you'll see that the rate of expansion of the hole is the same as the rate of expansion of the copper that it would take to fill the hole. So, treating the hole as a copper disk will get you the correct answer.

    Smasher on
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