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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

Hello guys;

i was wondering if anyone would help me find the centroid of a boomerang.......???

*Last edited by Maiya (2012-12-09 16:56:00)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Maiya,

Do you want a practical, experiment based approach or a mathematical model.

experimental approach: if you could balance the shape on a knife edge, then the C of G will lie above that edge. The tricky bit for a boomerang is that I suspect the C of G lies outside the surface so how do you balance it? You might be able to get a good approximation by placing it on a sheet of card and then trying the balance trick. What you'd actually get is the C of G of the combined sheet + boomerang system, but, if the card is light and the boomerang heavy, the result would be close. Symmetry would help too.

You could also try fixing a string to one end and letting the boomerang hang freely down. When it stops swinging, the C of G would be exactly below the point of suspension. Do this from several different points to help eliminate experimental error.

model: In order to do the theoretical calculation, you need to have a formula for the shape. The most complicated I have done is a rod that gets heavier as you track from one end to the other, so a boomerang would be a tough one to do. Any idea what the formula is?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

Hello;

'm actually trying it in a theoretical way............

here's the problem.........

consider a square with its side equal to 200 units

from both the ends of the base two quarter circle of radii 200 units are inscribed inside the square and a semi circle with a radius of 100 units is drawn joining the sides of the base.....

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Maiya,

That's very specific. Brilliant. Give me a little time and I think I can do that.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

Hello bob bundy.....

that's not very much of a boomerang .....

but the centroid of the area that looks like a boomerang is required.......

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

Please check I've got the right shape.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

yup bob ....

that's it .....

how can the centroid of that shaded area be located.......

*Last edited by Maiya (2012-12-10 01:16:12)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

I'll have to make some formulas:

the C of a semicircle, the C of a quadrant.

I recommend you come back later .... maybe much later ......

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

Here's the one for a quadrant.

Imagine a circle centred on (0,0) with radius r.

equation

Divide the quadrant into infinitesimally thin strips, y tall and dx wide

Assume the mass is proportional to area so I'll just work with areas.

If the centroid is X units from the origin then moments about y axis for whole shape is

and for the thin strips (each at distance x from the y axis) and added up using integration

Equating these two expressions for the total moment:

Check in Wiki. tick.

You can regard a semi circle as two quadrants fitted together, so the position is the same.

So now I have the required Cs for the components.

Got to do some chores so I'll come back to finish the job this evening ≈ 8pm GMT.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Maiya,

It is just as well that I didn't try to finish it earlier. What I did is correct, but it isn't what we need.

The actual task is much more difficult. I am fairly sure I can do it but it is going to take more time.

Have a look at my revised diagram. I have turned it through 90 to simplify the working.

My method will be this:

(i) Find the coordinates of B

(ii) By splitting the shape into two parts at line AD, find the area of the left and right parts (just down to the line of symmetry)

(iii) Using the moments technique find the moment of the left and right bits and add them together.

(iv) Hence find the distance of the centroid from the y axis. (It'll be on the line AB by symmetry).

I've done (i) so far.

Back later.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

Progress report.

(i) I have B = (100, 100√3) this looks right by inspecting the diagram.

(ii) I have 9585. This doesn't look right as an estimate based on square counting gives 4000<area<5000

Have to check my integration.

Edit: As I looked at this post my eyes fell on the diagram above it and I realised I've not truncated one area at the line of symmetry. That would account for the error. So I may yet get this right

Bob

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

Hello bob;

i know how to locate the centroids of those individual figures...........

if u want i can give it myself but 'm having trouble locating the area of that shaded portion..........

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Maiya,

It will take a little time to put into LaTex but I have the area calculations.

Shall I post what I have and if you want to try it you can?

Bob

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

hi bob;

its okay if its its in regular text

later you can post in latex....

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

Here's what I've got:

refer to the diagram in post 10.

O is the origin, C the point (0,200) and A is (100,100).

There are three curves, all part of circles.

curve joining C to A

curve joining C to B

curve joining O to B

substitution into CB gives B is the point (100√3,100)

Area of half boomerang (CDBA) is found by integrating CD minus CA from 0 to 100, and adding integral DB less rectangle below red line.

Here's one of these (the rest are similar)

Substitute x = 200 sinu, dx/du = 200 cosu, x = 0 => u = 0, x = 100 => u = pi/6

Phew! that took some typing, so it would be great if you were able to do the rest of this stage.

If you post back when you've got an answer I'll show you how to finish it off.

Bob

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

Hello bob;

sorry but 'm wondering how'd u get that equation for curve CA..........???:o

*Last edited by Maiya (2012-12-12 23:03:45)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Maiya,

Every point (x,y) on the circumference obeys the rule

That's using Pythagoras.

see diagram for right angled triangle.

Bob

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

hello bob;

i've never seen that before....

can u be more precise...............

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Maiya,

Have a look at

http://www.mathsisfun.com/algebra/circle-equations.html

Hopefully that will cover it.

I'm busy doing shows for the next three days. After which I'll come back to this and do all the steps for you.

Bob

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

Hello bob;

how'd u find the co-ordinates of the point "B".......

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi

It's on the line of symmetry so y = 100.

Then it is on the circle CB, so put y = 100 and solve for x.

Bob

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

hello bob;

can you please complete it.........

'm completely lost...........

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

First a correction:

Post 15. Change minus sign to plus

Check estimate by square counting = about 46 squares at 20 x 20 each = 18400

so far so good.

This is the area between the curve CD and the x axis.

Now for the area between the curve CA and the x axis.

I could use integration, but as this area is a quarter circle plus a rectangle I might as well do it that way.

Just a little less than the previous answer seems good.

Now for the area between DB and the line of symmetry.

Note. the curve is the same as the first so I can write the integral straight away but with new limits.

Check estimate by square counting = 8 x 400 = 3200

So the area of half a boomerang is first answer minus second answer plus third answer

= 4429.7153521131

Back at post 11 I estimated this answer would be between 4000 and 5000 so I'm happy with that.

That's step 2 complete.

I'll start step 3 in a new post.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

The centroid is found from this formula:

For a uniform shape the weight can be replaced by the area:

In step 2 I calculated the area of the whole shape, so now I have to work out the RHS.

first component under CD:

second component quarter circle CA (I'll use the formula I did way back in post 9

third component 100 x 100 square under this

fourth component under DB

fifth component rectangle under this

Now the required total is 1st - 2nd - 3rd + 4th - 5th = 500000

Finally for step 4,

So let's say 113 from the y axis on the line of symmetry.

That looks reasonable so it may be right (meaning not obvious that it is wrong).

I'll check it at the weekend.

Bob

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**Maiya****Member**- Registered: 2011-08-11
- Posts: 124

hello bob;

its a quarter circle,yes......

but how is it a rectangle..........

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