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Topic review (newest first)
Excellent to get the same answer!
ya for that curve CB.......
The expression in the square bracket is negative. When you sub in the upper and lower limits you would normally do (value of expression at upper) minus (value at lower). But because of the minus in the expression I subbed in 0 first
and then subtracted the value with x = 100
Well spotted. You can. I had already slogged through the first calculation before I realised that .... so I left it in two parts.
Yes, but you have it already. The centroid will have an x coordinate and a y coordinate. Because the shape is symmetrical you can say the y coordinate must be 100 without any calculations.
'm wondering why can't one directly find the moment of curve CB.........
and how's that equal to (200^3)/3-((200^2-100^2)^3/2)/3......???
I have put the diagrams again to save you having to keep going back to them.
These have to be multiplied by the distance of the strip from the y axis to get the moment.
These are then summed up using integration (because there is an infinite number of infinitesimal strips).
The integration limits are from x = 0 (at C) to x = 100 (at D)
That gives the whole moment down to the x axis (1st area on my second diagram) , so it is necessary to remove the moment for the square.(see second diagram 3rd area) and of the quarter circle (2nd)
The DB section has the same formula but changed limits: 100 to 100√3
That gives the green area from which a rectangle must be removed.
actually the problem's with CD and DB..............
This comes from a physics topic called moments.
and the total moment is found by adding these up.
Now suppose I want to replace the four weights with a single weight of 8 Kg (the total weight). Where should I hang it so that it has the same turning effect?
That is where the centroid of those weights is. It is the point where, from a mechanical point of view, you may consider all the weight to lie.
In calculations involving Newtonian mechanics, you may consider the weight of 8 Kg to be at that distance in all subsequent calculations.
For example, if I want to push up at B with sufficient force to hold the rod in a horizontal position then the force (for simplicity I'll stick to Kg for this) would be
I am making a couple of assumptions here so I'll say what they are:
(i) I'm using weights in Kg whereas a true weight should be measured in Newtons. But as the conversion factor (9.81) would cancel out throughout the problem, I can leave this out to keep things simple. In your centroid problem, I went one step further and used areas rather than weights, as I assumed the boomerang would have constant density across the surface.
(ii) I haven't allowed for the rod having any weight. We can say it is a 'weightless' rod. Of course that cannot be true in reality, but trying to allow for the centroid of the rod as well would have just introduced an unnecessary extra calculation and made the process harder to follow.
For the boomerang, the weight is spread out across a continuous surface. So it is necessary to use integration to summ up all the little bits of area and also to sum up all the moments of the little bits. Then you can use the formula
Hello Bob ..........
See post 26 for diagram.
Put y = 100
So rectangle is 100 by 100(√3-1).
sorry but i have another doubt.....
Oh sorry. I misunderstood. Let's see. What I did was
Looking back, I did miss out a few steps didn't I?