Calculus/mechanics

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Expert: - 10/28/2009


Question
A quarter circle shaped thin lamina of radius r is suspended freely at one of its corners(intersection of radius & circumference). Knowing that for equilibrium position the centroid of the quarter circle will lie directly below the point of suspension, what is the angle made by the top edge(radius)with the vertical? Centroid lies on the axis of symmetry at distance of 2rsin(alpha)/3*alpha from circle centre.Alpha=semi vertical angle=45 degree

Answer
I bet I forgot to include the graph - you'll have to write back for it,
since I can't attach it to add ons.

I drew a full circle and shaded a quarter of it.  The hanging point is at the point at the top of the break in colors.  I then proceeded to find how much weight was in each of the areas, marked I, II, III, and IV.  If you need help with the integrals, I could do that for you as well.

I get the following:

I) ∫(c-x)√(r²-x²)dx, from c downto 0.

II) ∫(c-x)x•tan(Θ)dx, from c downto 0.

III) ∫(x-c)√(r²-x²)dx, from r downto c.

IV) ∫(x-c)x•tan(Θ)dx, from d downto c.

Note the following are things that could be used:
A = πr², where A is the area of the circle;
c² + (y1)² = r²
c² + (y2)² = d²
cos(Θ)=c/r
sin(Θ)=y1/r
area(I+III)=(90-Θ)r²/2
area(II+IV)=Θr²/2

If more help is needed, I will be working more on this tonight.

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