Expansion of an object

1. Sep 18, 2006

castaway

Well my question is that we have a disc which has negligible thickness , now we cut a small circular disc at the center of the original disc. well now we heat the whole newly obtained body. what will happen to the radius of the disc we had cut?i mean to say the hole , what will happen to the radius of the hole? will it increase or decrease or remain same and why?

2. Sep 18, 2006

mathman

The hole (assuming the material is ordianry, ie. expands on heating) will get bigger. a heuristic expanation is that the atoms (or molecules) will be moving faster and push each other apart. In particular the particles at the edge of the hole need more room , so the circle has to get bigger.

3. Sep 18, 2006

Epicurus

Does the whole get bigger in the same proportion as the disk that was cut-out? Now that is an interesting question.

4. Sep 18, 2006

castaway

well i reckon it will be in same proportion as it is the same material and the coefficient of arieal expansion will be same.though thank you for your answer.

5. Sep 19, 2006

HallsofIvy

Of course, the disk will not stay flat!

6. Sep 19, 2006

FunkyDwarf

yes shape (in 2d) will remain the same as every part of that material expands at the same rate as you said. as for getting fatter, it will probably bulge a bit.

so yeh the whole thing expands in proportion. its only when you have materials with different thermal coefficients that you get bending and warping happening (which is often useful)

7. Sep 19, 2006

Staff: Mentor

If the material is isotropic thermal expansion will affect all linear dimension--including holes--in exactly the same way. No reason to think the disk would not stay flat (assuming you heat it uniformly).

8. Sep 19, 2006

FunkyDwarf

yeh what he said

Last edited: Sep 19, 2006
9. Sep 20, 2006

Astronuc

Staff Emeritus
A practical application of this phenomenon is the heating of a turbine disc and the cooling of a shaft in order to allow the disc to be easily positioned on the shaft. When the shaft heats and the disc cools, the disc shrinks onto the shaft.

10. Sep 20, 2006

Gokul43201

Staff Emeritus
The inside diameter expands with essentially the same temperature coefficient as the outside diameter (assuming that, if necessary, the disk is stiff enough to avoid buckling).

Plausibility argument: Assume the annular ring is heated through a unit raise in temperature. If the inner (r) and outer (R) radii increase by the same factor (1 + a), the important linear dimensions of the annulus (width = R-r, and the inner and outer circumferences) are easily shown to increase by the same factor, and the area increases by exactly the square of this factor (since $$\pi*[R(1+a)]^2 - \pi*[r(1+a)]^2 = \pi*(R^2-r^2)*(1+a)^2$$), which is what you want to see.

An alternative argument would be to roughly model the annulus as composed of a series of infinitesimally thin non-interacting rings. Being 1-dimensional creatures, each ring will increase in dimension (radius or circumference) by the same factor (1+a).