1) A straight, thin rod PQ of length 2a of constant cross sectional area A is of variable density, r.
Given r = ro (1 + 2x 2 / a 2) where x is the distance along the rod from the end of P and ro is a constant. Find the position of the center of mass.
Solution:
We make use of the sketch below with blackened element dx and distance x from P.
The volume element can then be expressed:
dV= A dr x
and the mass element: dm = Adx r
Where r is the density. Then we can write:
dm = A dx ro (1 + 2x 2 / a 2)
The x -coordinate of the center of mass of dm is x, so therefore:
x =
ò 2a 0 x. Adx ro ( 1 + 2x 2 / a 2) / ò 2a 0 Adx ro ( 1 + 2x 2 / a 2)
x = A ro ò 2a 0 ( 1 + 2x 2 / a 2) dx /A ro ò 2a 0 ( 1 + 2x 2 / a 2) dx
= [x 2 /2 + 2x4 /4a 2] 2a 0 / [x + 2x3 /3a 2] 2a 0
= (2a 3 + 8 a 2) / (2a + 16a/3) = 15a/ 11 (from end of P)
2) Find the center of mass of a solid hemisphere of radius r if the density at any point P is proportional to the distance of P from the base of the hemisphere. (Hint: Add another lower latitutde layer to the upper latitude circle below to form a thin slice of thickness dy. Then imagine the whole hemisphere cut into slices of thickness dy, i.e. by planes perpendicular to the y-axis)
Solution:
Below I show the hemisphere again with the lower latitude circle added and slice thickness dimension dy indicated, which will provide the basis for integration:
Note: We imagine the entire solid cut into slices of thickness dy so can write:
dV = A(y) dy as the volume of the representative slice at distance y above the base of the hemisphere (which lies in the x-z plane). The area of a face of the slice dV is:
A(y) = p x 2 where: x 2 + y 2 = r 2
And so: dV = A(y) dy = p (r 2 - y 2) dy
With mass element: dm = kp (r 2 - y 2) y dy
The center of mass of the 'ground' slice may then be taken at the geometrical center (x, y, z) = (0, y, 0) or:
y = ò y dm / ò dm
= ò r 0 kp (r 2 - y 2) y dy / ò r 0 kp (r 2 - y 2) dy
= kp [r 2 y3 /3 - y5 /5 ] r 0 / kp [r 2 y2 /2 - y4 /4 ] r 0
y = 8r/ 15
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