Nothing in the description indicates whether the LENGTHS or DIAMETERS or MASS of metal/wood are equal...only that balance is achieved. There is insufficient data to answer the question. and a 12 year old would say "I don't know..why is it important anyway' and that IS the correct answer!
You are correct that as the metal expands (evenly) the center of gravity (c.g.) of the all-metal (denoted M)side will push out further than the center of gravity of the wood/metal (denoted w/m) side. But consider this. The rod is now in balance. If the weight of the w/m side is less than the M side, then it's elementary. Prior to heating, the c.g. of w/m will have to be further from the fulcrum than the c.g. of the M side, in order for the moments to balance out. Upon heating, the c.g. of M moves a greater absolute distance from the fulcrum. Since the weight M is larger than (w + m) by initial stipulation, you have a greater weight times a larger distance vs a lesser weight times a smaller distance, and it's obvious that the rod tilts on side M. This is pretty trivial, as you rightly say. But if the weight of the w/m side is GREATER than that of the M side, then it's more complicated. Prior to heating, the c.g. of w/m is now closer to the fulcrum than the c.g. of M in order for the moments to balance out. When heat is applied it is true that the absolute displacement (from the fulcrum) of the c.g. of w/m will be less than that of M. But in this case, when we compare moments, we're comparing a greater weight times a smaller distance against a lesser weight times a greater distance. This is not quite so easy to see. I have to resort to simple algebra to help you see this. Let the extra displacement from the fulcrum caused by heating be d for the w/m side and D for the M side. The weight of the wood is w, that of the small metal part is m and that of the large metal part is M. d is smaller than D while (w + m) is greater than M. After heating we need to compare the following moments : On the w/m side, the moment increases by (w + m)*d ---(1) On the M side, the moment increases by M*D ---(2) I think you can see that these are difficult to compare with each other because one term in the multiplication is big, the other is small. So I needed something more rigorous. The algebra I did used symbols to represent the relative coordinates of the respective c.g.s and for the weights, lengths, etc. This way I was able to prove that no matter what the stipulations, the M side would always go down. I'm only satisfied when I've proven something rigorously and unequivocally. BTW, the problem does not even require one to assume that the wood is of lower density. We can pose the problem as a high density non-expansile material attached to lower density expansile metal, and still the cases would come out the same, with the metal part going lower after heating. **** In any case, the actual exam question was far simpler, and therefore not the best. The actual question stated that the rod was made half of wood and half of iron and was balanced in the middle. This leaves me scratching my head at the implicit assumption that the densities of wood and iron are assumed to be the same. Not a great question. A big scientist type has "weighed in" (pardon the pun ) in the local paper that the metal side would go down. Of course, given that question, it's easy to see the answer, because expansion only occurs on one side. Still, another smart aleck layman-type has brought up the question of convection currents on the heated side pushing up the metal, and confounding the issue. I considered this, but figured this wasn't what the question was going for, and assumed the thing was done in a vaccuum. But if it's done in air, he's right, convection currents do become important.
Lengths, diameters, assumptions about density do not affect the meat of the question. The answer comes out the same way. This is why this is an elegant question.
If you have two equal umbrellas balanced on a fulcrum (taped together) and you extend one side, it will go down because the weight is farther away. I forgot what this means, but it means it will be 'heavier' in terms of relating to the other umbrella and the fulcrum.
This has to do with what are known as "moments of force", alternatively known as "torque" (although I prefer the latter term for rotational scenarios). The formula for moment is simple : mass (weight) times distance (from fulcrum). When you have two masses (weights) in balance on a beam that's supported by a fulcrum (pivot), it means the moments on each side are equal. When you have two closed identical umbrellas, they balance each other in moments, because same weight times same distance equals the same. When you open one umbrella up, you cause the weight distribution in the umbrella to change. More of the umbrella's "stuff" (matter) goes far away from the support point. So although the weights on each side are the same, the distance is greater on the opened side, and it tips down. This problem (the rod) is based on a similar priciple.
That is the real answer (metal end drops down) at least in a vacuum. In air, it can go either way because of convection currents.
Since wood and iron don't have the same density, this question implies that the rod consists of two "half rods" laminated together. So the rod merely bends when heated. A complicated way to make toothpicks.
LOL, good answer. Reminds me of the guy who, when asked in a Physics exam "How do you measure the height of a building using an aneroid barometer ?", answered "I would tie string to the aneroid barometer and lower it from the top of the building to the ground and measure the length of the string." Low on knowledge, high on ingenuity.
whoever said the metal moves down is correct, however it wouldnt move that much. Since the metal can only expand outward from the free end (because one end is fixed on a pivot) its center of mass moves farther away from the pivot. The moment that results from the cg (which was previously equal and opposite to that created by the wood) of the rod being some distance from the pivot increases because the distance from the pivot increases. The metal rod will then tip down until the horizontal distance from the cg of the metal rod to the pivot creates a moment that is equal and opposite to that of the wood.
That guy was Niels Bohr. When the professor asked him about his answer he saaid you could also throw the barometer off the top of the building and time how olng it took to fall, or you could go to the superintendend and tell him you will give him this nice barometer if he tells you how high the building is. Thats what i call thinking outside the box.
No, it will tip down arbitrary low (until it the metal side hits the ground). This is a rigid rod. When the metal side begins to tip down, the moment depends on the horizontal distance between the c.g. and the fulcrum (as you said). This is actually the length of the rod on that side times the cosine of the angle of tip, and it is of course less than the moment before. But the moment on the wood side changes the same way (the moment arm becomes length times cosine of angle of tip) since this is a rigid rod and the angle of tip is equal throughout. So when comparing the moments after tipping has begun, the cosines cancel out, and the metal side's moment remains greater than the wood side's. So the rod continues tipping until one of two things happens : either a) the metal part hits a solid obstruction (the support or ground), or b) the static friction between the rod and the fulcrum is overcome and the rod slips. Even as the rod slips, the metal part becomes more dependent (lower) and the rod continues to tip and slip faster until the rod comes off the fulcrum entirely or the metal end hits the support.
I think it will tip until the moments balance, restoring the entire rod to equilibrium. Last year i did a lab which demonstrated moments and this was what we observed. If i have time I will replicate the situation (or my perception of it) on Working Model 2D (which i dont think i have on this laptop so i have to get a copy from my buddy) to see if this is what happens. We may be thinking of two different situations though.
What if the rod is "part metal and part wood" along the long axis, not (as I suspect most assumed), perpendicular to the long axis? Then the balance point is exactly mid-length and changes in moment due to thermal expansion will not affect balance.
