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Will a hole cut into a metal disk expand or shrink when the disc is heated?

The metal disc of diameter D1 has a hole in it and the diameter of the hole is D2 (D2

Now, if I heat the metal disc, it expands. As far as I know, the metal expands along its free ends upon heating. In this case, the disc has got the scope to expand along both the inner and outer circumferences of the disc.

The question here is, in which way does the metal disc really expand and as a result, will the hole on the disc expand or contract. To put in other words, assuming the new diameter of the hole on the disc to be D3, will D3 be greater than, lesser than or equal to D2?

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    $\begingroup$ Imagine the bit cut out of the middle - a disc with diameter D2. What happens if you heat it? Now imagine it embedded inside your metal disc of diameter D1, and heat both at the same time: how will they behave? $\endgroup$
    – 410 gone
    Commented Mar 15, 2012 at 14:09
  • $\begingroup$ Merge answers? Is that possible? Or close? $\endgroup$
    – Manishearth
    Commented Mar 15, 2012 at 14:43
  • $\begingroup$ @EnergyNumbers, the two cases, a) the smaller Disc is cut out and b) embedded in to the bigger disc, are different. In the first case, there is no resistance whatsoever for the Disc D1 to expand inward and thereby shrinking the hole.... $\endgroup$
    – Mallik
    Commented Mar 15, 2012 at 17:38

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It expands. Take a normal disc, and draw a circle on it. The circle expands. Take a new disc, draw a hole, and make a score on that circle. It still expands. Repeat with deeper scoring. It still expands. Keep making it deeper. It still expands. At one point, your razor(or whatever) will start poking through the metal. It still expands. Cut out the circle, but leave the cut out piece in the hole. Now, doesn't it seem logical that the circle+hole will expand? Remove the circle, it still expands.

So, to answer your question, a hole in a material behaves just like a circle of that same material. It expands on heating.

What's actually happening is that if it tries to expand inwards (contract basically), it will have to compress itself, and increase its density. There will be resistance to it, so it tries to expand into a free region, namely the outside. And the inside expands just because it does when there is no hole.

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  • $\begingroup$ Simultaneous indeed. The questions is such a familiar one! :) $\endgroup$
    – Slaviks
    Commented Mar 15, 2012 at 14:10
  • $\begingroup$ @Slaviks Yep.. I had this same doubt years back and then I imagined myself an answer (same explanation as yours/mine). =D $\endgroup$
    – Manishearth
    Commented Mar 15, 2012 at 14:11
  • $\begingroup$ @Manishearth, Thank you for the explanation.. But I think there is a difference in the 2 cases, you mentioned... first when the metal piece is still in the disk, and the second - the metal piece is removed to make it a disk.. in the latter case, there is no resistance for expansion inwards... so, there is a possibility of hole shrinking... $\endgroup$
    – Mallik
    Commented Mar 15, 2012 at 17:33
  • $\begingroup$ @Mallik there is resistance. If a part at the hole border tries to move inside similtaneously with a part next to it, both of them will be trying to occupy nearly the same space. So there is resistance. $\endgroup$
    – Manishearth
    Commented Mar 15, 2012 at 17:39
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    $\begingroup$ @Manishearth, Consider the disc is heated uniformly across its area... the metal near the outer circumference, tries to expand outward because of resistance from its disk-side neighboring molecules. Similarly, the part that is near the inner circumference of the hole (cut-out part) tend to expand to its free side, and thereby covering up the hole. \ $\endgroup$
    – Mallik
    Commented Mar 15, 2012 at 17:43
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Imagine that the hole is filled with another disk of exactly the same diameter as the hole (initially $D_2$) and made from the same material. Once both discs heat up, they will expand as if they were a single solid object. Since the small disk expands, we conclude that the hole in the large disc must expand too, $D_3>D_2$.

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  • $\begingroup$ Simultaneous answers ;) $\endgroup$
    – Manishearth
    Commented Mar 15, 2012 at 14:09

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