anonymous
  • anonymous
6) The figure below shows a uniform metal plate P of radius 2R from which a disk of radius R has been stamped out (removed) in an assembly line. Using the xy coordinate system shown, locate the center of mass "CM" of the remaining plate.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@experimentX
experimentX
  • experimentX
y = 0, x = find using the weight function

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anonymous
  • anonymous
can you explain me a bit further plz?
experimentX
  • experimentX
|dw:1370495897168:dw|
anonymous
  • anonymous
so...
experimentX
  • experimentX
|dw:1370496097243:dw|
experimentX
  • experimentX
|dw:1370496153111:dw|
experimentX
  • experimentX
put f(x) = sqrt(4R^2 - y^2) - sqrt(R^2 - (x-R)^2) and evaluate the above integral from -2R to +2R
experimentX
  • experimentX
BRB
anonymous
  • anonymous
quite confusing
anonymous
  • anonymous
can you be more specific plz?
experimentX
  • experimentX
Evaluate this integral \[\Large = \frac{\int_{-2R}^{2R} x (\sqrt{4R^2 - x^2} - \sqrt{R^2 - (x-R)^2})dx}{\int_{-2R}^{2R} (\sqrt{4R^2 - x^2} - \sqrt{R^2 - (x-R)^2})dx}\] This will give you the x-coordinate of center of mass. y-coordinate ... it's symmetric, y=0
experimentX
  • experimentX
http://en.wikipedia.org/wiki/Centroid#By_integral_formula
anonymous
  • anonymous
it's incorrect
experimentX
  • experimentX
why? do you have an answer?
anonymous
  • anonymous
yes, answer is (R/3;0)
experimentX
  • experimentX
try this \[ \int _{-2R}^0 x(\sqrt{4R^2 - x^2} - \sqrt{R^2 - (x+R)^2})dx + \int _{0}^{2R} x\sqrt{4R^2 - x^2}dx\]
anonymous
  • anonymous
that integration is so long, is there any other methods? Other problems in that problem sheet don't seem to be so complicated as this one
experimentX
  • experimentX
i don't know any other method, putting R = 2, I am getting x=2/3.
experimentX
  • experimentX
The denominator of the weight function is \[\int _{-2R}^0 (\sqrt{4R^2 - x^2} - \sqrt{R^2 - (x+R)^2})dx + \int _{0}^{2R} \sqrt{4R^2 - x^2}dx \] this is just area under the curve which is \( \frac 1 2 (\pi 4R^2 - \pi R^2) = \frac 3 2 \pi R^2\)
experimentX
  • experimentX
write up the integral as |dw:1370499474985:dw| just substitute 4R^2 - x^2 = u in first integral
experimentX
  • experimentX
|dw:1370499691622:dw|
experimentX
  • experimentX
to evaluate this integral, use trig subs x+R = R sin(theta) \[ \int_{-2R}^0 x \sqrt{R^2 - (x+R)^2}dx = \int_{-\pi/2}^{\pi/2}R (\sin \theta - 1) R \cos \theta R \cos \theta d\theta \] Which i believe you can evaluate it
experimentX
  • experimentX
this can be evaluated as -pi/2 R^3 that - above will make it + and you have 3/2 pi R^2 at denominator which will give R/3

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