anyone know how to solve this? :h:
(a circle of radius 3cm is cut out from the circle of radius 6cm)
(that object is a flat circular plane [size=1]with a hole, of course[/size])
Forget it, I made a mistake :x
hmm and how’d you get that?
what he said
ehh but…how did you get it.
All right, I think I found it:
I replaced both discs by their center, to which I assigned a weigth depending on their area:
Let’s say the radius of the little circle is R => Area = Pi R²
Big circle: Area = 4 Pi R²
(A, -Pi R²)
|
|
|
(B, 4 Pi R²)
The weight of the first point is negative because it’s a hole. The distance between A and B is R.
The center of gravity is so that w1 GA + w2 GB = 0 (vectors, w1 is the weight of A, w2 the weight of B)
=> BG = w1 / (w1 + w2) AB
=> BG = 1/3 AB = 1/3 R = 1 cm
So the center of gravity is 1 cm under B.
But it might be completely wrong :beam:
wait…you lost me
here’s the main part i dont get…
what does A and B represent, and where are they located? (actually, I dont understand this diagram :P)
edit:
if i get it correctly…then what you’re trying to say that
A is the CG of the cutout, located in the centre of that cutout;
B is the CG of the original circle without the cutout.
then…shouldn’t B be at the centre of the diagram? ie from where the green arrows are originating. [edit 3! duh, that’s what you’re saying. my bad]
Edit Edit:
ok…although i dont quite understand your working but I learnt another method:
mass of smaller circle x radius of smaller circle = mass of shape x distance between its CG and the center of the large circle
yep and u get 1cm.
Still, i’m interested in your working, could you explain the part i dont understand?
Edit x3! :
Well. No need for that explanation, I got it!:party: Thanks!
Actually, your Edit Edit part is exactly what I’m doing The only difference is: I searched for the centre of gravity of the centre of gravity of the discs, and not directly the discs. It’s just a way to prove what you are saying:
mass of smaller circle x radius of smaller circle = mass of shape x distance between its CG and the center of the large circle
How did you find that?
yep, after awhile I realised that hey, that is what he’s trying to say (the edit edit part)
here’s the explanation for that equation…it’s almost the same as yours, just abit long, though
think of the smaller circle attached to the larger shape so the whole system can balance on the pin that is right at the center of the larger circle
its like a balance, one half consisting of the smaller circle, the other half consisting of the larger shape.
the weight of the smaller circle has a moment about the pin and this should be the same as the moment of the weight of the larger shape about the pin
now you know the weights of both sections
and you also know the the point through which the weight of the small circle is exerted
so you know the perpendicular distance between the cg of the small circle and the pin
*note, the pin is in fact where your B,w2 is, thus the whole idea is about the same
therefore, you get that equation
in fact, the equation is simply your equation put in words:
your equation: w1 GA + w2 GB = 0 *here i think you made a mistake, instead of w2 it should be w2-w1 (most likely a typo, cos you corrected it in the 2nd step)
=> [COLOR=Red]-w1[/COLOR] * [COLOR=Blue]GA[/COLOR] = COLOR=Green[/COLOR] * [COLOR=RoyalBlue]GB[/COLOR]
[COLOR=Red]mass of smaller circle[/COLOR] x [COLOR=Blue]radius of smaller circle [/COLOR] = [COLOR=Green]mass of shape[/COLOR] x [COLOR=RoyalBlue]distance between its CG and the center of the large circle [/COLOR]
Yep, pretty much the same thing except my English sucks and I didn’t use the ‘pin’ metaphor
Btw, where are you studying in Sg?
yep, i’m studying in Singapore. preparing for my o levels and came across this question.
edit: oh…you wrote ‘where’. misread that i’m currently studying in Fairfield Methodist Secondary School.
heh, actually I too dont use the pin metaphor, that’s the explanation given to me, in my case i’d rather look at it as
’moments about point B (where B is the c.g. of the big circle)’ [size=1]that’s based on your method[/size]