M2 question

Heinemann 2B Q14.

A Uniform wire is bent to form the outline of a sector of a circle, with the wire being doubled along the arc only. Given that the straight sides measure 0.5m and the angle between them is 40°, calculate the distance of the centre of mass of the framework from the center of the circle.

I've tried several times and spent ages on this and got no where so any help would be greatly appreciated!

Thanks!

Can you assume the position of the C of G for the arc of a circle, or do you have to work that from scratch as well?

Use the formula listed on page 40;

(r sin a)/a with the angle of the sector bieng 2a but you'll have to take into account for the doubling

To find the centre of mass you need to use:

$\displaystyle\bar{x}\sum m_i = \sum_i m_i\bar{x}_i$

with the, hopefully, obvious meanings.

So you need to work out the position of the centre of mass and the mass for each component of the figure. I.e. the arc, and each of the sides.

Working out the centre of mass for the arc itself should be in your book, as it's a standard result.

Edit: Scrap that, i understand what you've said.. And it is what i've tried.. If someone wouldn't mind showing a bit of working on how to start? I'll post mine soon for any obvious mistakes.. thanks

okay here you go,

do you agree that from the formula the centre of mass of just the arc is given by;

(0.5sin(pi/9))/(pi/9) = 0.4899m from the centre.
The relative mass of the arc is given by it's legnth
pid * (2pi/9)/(2pi) = pi * (2pi/9)(2pi) = pi/9 = 0.349m however it's doubled over, so the mass doubles = 0.698

can you take it from here? do you see what i've done?

Yes thankyou both. I'll try it from here and see what i get! Thanks!

np ask if you get stuck, on a side note good luck with your application! =P