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zaphod
Please help with the second part only :)
@amistre64 @ParthKohli @Compassionate
we have a half solid and an arc?
if you know where that object is being hung from, the the center of gravity; and that it must be at equilbrium .... then you should be able to measure the angle
as a rough idea, if hung from A, then wouldnt we want 10N on each side? which suggests to me that the Ab line moves left and the ring is lowest
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how much of an arclength is equal to 1N? 180 = 11N, so 1N = 180/11 degrees?
prolly more involved than that, but that just what i get from my perspective
Haha, that looks good to me! :)
:) all i know is the picture hangs crooked and i got to fiddle around with it till i give up
Haha! It doesn't state which side the center of mass is, but you probably found that to find part (i). Either way, I think we can guess based on the weights and their distribution.
it 'disk' balances out at on center of the x axis; and therefore can only center out to the right of the y axis since we have 9 to the left and 11 to the right ... its already right heavy and so we schooch it over
if my arclength idea is correct, then an arclength of 1/2 a degree is needed. or pi/360 radians and it might work out to the expected value
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you will find the angle by re-drawing so that the centre of mass (CoM) of the combined body lies directly below A, whence it is suspended. you know that the line from the O to the CoM is at right angles to the line A-B because the overall shape is symmetrical. simple trig follows. you have triangle with sides A-O (you know its length), and O - CoM (ditto) and a right angle. clear?
not allowing for distances and thats a .7 gives me .3355 which is to big
@IrishBoy123 is right. tan θ = e/R The lowest point belongs to the arc. |dw:1398356952148:dw|
@amistre64 I only skimmed this post since I last replied, but it looks like there's some debate. But I realize a lapse we might have made when analyzing this too quickly before. We thought there should be \(10\rm\ N\) on both sides of the divide, but that's not true. There's the issue of torque, so its the torque that must be equal on both sides of the line that connects the center of mass to point A. And so an improper assumption gave us an improper result! Easy quick mistake. So just using the geometry, knowing that the center of mass will fall below the hanging point, is the easiest way to go. That's my analysis. Anyone correct me if I'm wrong!
so it is the arc that is at the lowest point? thanks guys i got the same answer, but the marking scheme says it is the lamina.
@zaphod of course the lowest point must belong to the lamina as it weighs more. i though that was obvious so i did not mention it. if it belonged to the lighter arc, then gravity is working the wrong way round....
No, it's not the lamina; The lamina weighs slightly more (11 vs 9), but its centre of mass is much closer to O (0.30 vs 0.45).
hmm, yeah, trying to switch back and forth between windows i seem to have switched the weights as well. I was thinking lamina was 9 and ring was 11
@amistre64 I made the same confusion with the \(11\) and \(9\). Either way, it's still believable that the center of mass of the system lies on the side of the arc. All of its mass is far off to the right. It's very dense, there. The mass of the lamina is uniform, so its extra 0.2kg (or around there if this is Earth (m = 2/g)) is spread closer to its base, not just on the edges where it would shift the lamina center of mass most leftward. I wouldn't be able to guess, given the shapes and weights, but I would consider it a toss-up until the math was done. @Vincent-Lyon.Fr found similar center of masses for each the lamina and arc as I did. Instead of integrating with those values, I just took the general formulas from my book. I'm going to use this coordinate system:|dw:1398456215345:dw|because we know that the center of mass for each, and thus the whole thing, will be on that axis \(x\) and having the \(y\)-axis where it is makes it convenient to find distance from \(O\) to the center of mass. And the \(y\)-axis there means I can use my formulas!! The lamina is in the negative \(x\) region and arc in the positive. \({\sf \large Lamina}\\\qquad x_{cm}=-\dfrac{4a}{3\pi}=-0.29708....{\rm\ m}\approx0.297{\rm\ m}\\ {\sf \large Arc}\\\qquad x_{cm}=\dfrac{2a}{\pi}=0.44563....{\rm\ m}\approx0.446{\rm\ m}\) So.....
When I find the average center of mass, though, I get a different answer. I think this is due to rounding. I get \(x_{cm}=\rm\dfrac{-0.297\ m\times 11\ kg\quad+\quad0.446\ m\times9\ kg}{20\ kg}\\\ \\\qquad\qquad=0.03735\ m\approx0.0374\ m\) Anyway, the center of mass is at a positive \(x\) and so it is on the right in the picture, on the side with the arc.|dw:1398457186914:dw|Even though just barely...
When it hangs by point \(A\), the center of mass will be below \(A\) and so the whole shape will shift accordingly.|dw:1398457375627:dw|
I think the markup might be wrong, then...