A thin wire is bent into the shape of a semicircle x^2+y^2=4 x>0

If the linear density is 3, find the mass of the wire.

Anyone have a clue how to tackle this problem?

Thanks for your help(Happy)

Possible answers -

18pi

6pi^2

3pi

6pi

2pi

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- Jul 13th 2010, 11:19 AMAlpina540Find mass by linear density? ......SOLVED........
A thin wire is bent into the shape of a semicircle x^2+y^2=4 x>0

If the linear density is 3, find the mass of the wire.

Anyone have a clue how to tackle this problem?

Thanks for your help(Happy)

Possible answers -

18pi

6pi^2

3pi

6pi

2pi - Jul 13th 2010, 11:30 AMAckbeet
If the linear density $\displaystyle \lambda$ is constant, then you can just use the formula $\displaystyle M=\lambda L$, where $\displaystyle M$ is the mass, and $\displaystyle L$ is the length of the wire. Where would you go from there?

- Jul 13th 2010, 11:35 AMAlpina540
Well it looks like now I need to find the length of the curve of wire.

- Jul 13th 2010, 11:42 AMAckbeet
Correct. And how could you find the length of a semicircle?

- Jul 13th 2010, 11:43 AMAlpina540
So I just graphed it and cheated a bit and just ruffed out the length of the curve by the Pythagorean theorem and I got the curve to be ~ 6 so 3*6= 18

So the answer is 6pi?

Thanks again Adrian! - Jul 13th 2010, 11:45 AMAckbeet
You can get the exact answer (which is not $\displaystyle 18\pi$, by the way). What's the length of a wire that's in the shape of a circle?

- Jul 13th 2010, 11:56 AMAlpina540
ok ok you win, lol (duh about the 18pi! thats what happens when 20 vector calc problems are running through your head.)

(180/360)*((2(pi)*2)=6.28

6.28*3=18.84

6*pi=18.84

Thanks for your help, I really appreciate it!

-Anson - Jul 13th 2010, 12:37 PMAckbeet
You're very welcome.

- Jul 13th 2010, 12:39 PMHallsofIvy
By the way, "pie" is what you eat. "pi" is the standard transliteration of $\displaystyle \pi$.

- Jul 13th 2010, 12:43 PMAlpina540
lol........ I knew that I just didn't think about it as I never write pi out in words.