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Centres of mass

Question posted below.

I don't have much of an idea where to start despite lots of pondering. Bet it is something easy though. I have simply drawn in a line representing the vertical from A to the centre of mass, and put the angle as theta between OA and the vertical.

The centre of mass is correctly worked out as along the axis of symmetry at a distance of 9(2+3)6+πcm\frac{9(2+\sqrt3)}{6+\pi}cm from the origin.

Read the picture on the RHS first. Sorry about smallness. Thanks.
(edited 11 years ago)
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Original post by fayled

Thanks.


You should be able to tell me what the angle AOB is and hence half that angle.

Then cosine rule followed by sine rule.
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fayled
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Original post by ghostwalker
You should be able to tell me what the angle AOB is and hence half that angle.

Then cosine rule followed by sine rule.


Damn it, had all the angles and lengths, thought of using cosine rule, thought of using sine rule.

Did not think of using both.

Thankyou :smile:

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