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Chapter 5A qs 7 Heninemann(M3)

FInd the centre of mass of a uniform lamina in the shape of the quadrant of a circle of radius r.
Reply 1
Avatar for dvs
dvs
10
x² + y² = r²

Mass of the lamina:
M = (p/4)(pi r²)

x-coord of the center of mass:
M xˉ\bar{x} = p 0∫r xy dx
= p 0∫r x sqrt[r² - x²] dx (Note that the derivative of (r²-x²)3/2 is -3x(r²-x²)1/2.)
= -p/3 0∫r -3x(r²-x²)1/2 dx
= -p/3 [(r²-x²)3/2]0r_{0}^{r}
= (pr³)/3

xˉ\bar{x} = (pr³)/3M
= (pr³)/[3(p/4)(pi r²)]
= 4r/3pi

y-coord:
M yˉ\bar{y} = 0.5p 0∫r y² dx
= 0.5p0∫r r²-x² dx
= 0.5p [r²x-x³/3]0r_{0}^{r}
= (pr³)/3

yˉ\bar{y} = (pr³)/3M
= 4r/3pi

So the coords of the center of mass are (4r/3pi, 4r/3pi).
Reply 2
Avatar for Fermat
Fermat
8
ziya_the_king
FInd the centre of mass of a uniform lamina in the shape of the quadrant of a circle of radius r.

Just got beat :frown:
Anyway here's a pic.
ziya.jpg90.7KB

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