centroids

Bob · 2012-12-21 00:46:40

hi

See picture. Hope the colors help.

Bob

Bob · 2012-12-23 01:50:01

I've now checked my working. Cannot find any mistakes.

Diagram shows centroid with a black dot.

Bob

Maiya · 2012-12-26 03:05:15

hello bob;
I was actually wondering what happens to the square root in post #15...........?

Last edited by Maiya (2012-12-26 03:17:25)

Bob · 2012-12-26 03:42:21

hi Maiya

Did you mean this:

Power one half is another way to write a square root.

and

Bob

Maiya · 2012-12-26 04:00:42

yes i know that
but 'm referring to the part after substitution of x=200cos u....

Bob · 2012-12-26 04:12:57

Oh sorry. I misunderstood. Let's see. What I did was

Looking back, I did miss out a few steps didn't I?

Bob

Maiya · 2012-12-26 23:02:43

Hello bob;
sorry for the last post ......
i worked out a bit on it and i figured the step myself so.....

Maiya · 2012-12-26 23:39:16

sorry but i have another doubt.....
how is the required area obtained of that rectangle considered...........??

Bob · 2012-12-27 01:42:13

See post 26 for diagram.

The 5th part is 100 tall. To calculate the width, get the coordinate of B, and subtract 100.

B is on

Put y = 100

So rectangle is 100 by 100(√3-1).

Bob

Maiya · 2012-12-28 00:00:45

Hello Bob ..........
That's not what i actually meant...........
but anyway can explain that post #24......
in what method is the centroid found there........

Bob · 2012-12-28 00:40:49

hi Maiya

post #24......
in what method is the centroid found there........

This comes from a physics topic called moments.

I'll show with a simple example first.

Suppose we have a 10 cm rod, attached to a pin at A, free to rotate about A, and with weights attached at various distances from A. (see diagram)

Each weight will have a turning effect. The further a weight is from A the more turning effect it will have. For example, the 2 Kg weight that is at B will exsert more turning effect than the 2 Kg weight that is only 3 cm from A

The 'moment' of a weight's turning effect is found from this formula:

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

Bob

Maiya · 2012-12-28 01:28:28

Hello Bob;
i know the method of moments........
but 'm asking why[and how ] is the integration done for the components present there.............???

Maiya · 2012-12-28 01:33:45

actually the problem's with CD and DB..............
I could not understand how you obtained the moments for components CD and DB........

Last edited by Maiya (2012-12-28 01:36:22)

Bob · 2012-12-28 02:07:43

I have put the diagrams again to save you having to keep going back to them.

The yellow strips represent a 'thin' strip of the surface. Each one is y tall (on the diagram the strip stops at the line of symmetry but integration theory requires they extend to the x axis) and dx wide. So one little area bit is

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.

Bob

Maiya · 2012-12-29 03:36:09

and how's that equal to (200^3)/3-((200^2-100^2)^3/2)/3......???
i mean what happens to the -ve sign with the term (200^3)/3..........??

Maiya · 2012-12-29 03:41:02

'm wondering why can't one directly find the moment of curve CB.........
finally we need the centroid of the whole of boomerang ..........
bout here centroid of only half of the boomerang is found ......
so how come.....??

Bob · 2012-12-29 05:33:58

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

'm wondering why can't one directly find the moment of curve CB.........

Well spotted. You can. I had already slogged through the first calculation before I realised that .... so I left it in two parts.

finally we need the centroid of the whole of boomerang ..........

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.

So the top half has an x coordinate as calculated and, without further calculation, the bottom half has the same x coord.

So the combined halves have the same x coordinate as the two halves separately.

So the whole shape has centroid at (113 , 100).

Bob

Maiya · 2012-12-29 20:36:31

Hello bob;
sorry about that -ve sign question .............
now that i worked out i found it myself.........

Last edited by Maiya (2012-12-29 21:11:59)

Maiya · 2012-12-29 20:41:44

ya for that curve CB.......
i arrived at the same answer even when i did i for the whole curve so,i just clarified .....
anyway thank you for your help BOB............:):D

Last edited by Maiya (2012-12-29 20:42:47)

Bob · 2012-12-30 05:10:25

Excellent to get the same answer!

Glad to help.

Bob

Math Is Fun Forum

#26 2012-12-21 00:46:40

Re: centroids

#27 2012-12-23 01:50:01

Re: centroids

#28 2012-12-26 03:05:15

Re: centroids

#29 2012-12-26 03:42:21

Re: centroids

#30 2012-12-26 04:00:42

Re: centroids

#31 2012-12-26 04:12:57

Re: centroids

#32 2012-12-26 23:02:43

Re: centroids

#33 2012-12-26 23:39:16

Re: centroids

#34 2012-12-27 01:42:13

Re: centroids

#35 2012-12-28 00:00:45

Re: centroids

#36 2012-12-28 00:40:49

Re: centroids

#37 2012-12-28 01:28:28

Re: centroids

#38 2012-12-28 01:33:45

Re: centroids

#39 2012-12-28 02:07:43

Re: centroids

#40 2012-12-29 03:36:09

Re: centroids

#41 2012-12-29 03:41:02

Re: centroids

#42 2012-12-29 05:33:58

Re: centroids

#43 2012-12-29 20:36:31

Re: centroids

#44 2012-12-29 20:41:44

Re: centroids

#45 2012-12-30 05:10:25

Re: centroids

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