A circular pizza of radius R has a circular piece of radius R /2 removed from on

Question

A circular pizza of radius R has a circular piece of radius R/2 removed from one side. Clearly the center of gravity has moved fromC to C' along the x-axis. Show that the distance from C to C' is R/6. (Assume that the thickness and density of the pizza are uniform throughout.)

A circular pizza of radius R has a circular piece of radius R/2 removed from one side. Clearly the center of gravity has moved fromC to C' along the x-axis. Show that the distance from C to C' is R/6. (Assume that the thickness and density of the pizza are uniform throughout.)

Explanation / Answer

A circular pizza of radius R has a circular piece of radius R/2 removed from one side. Clearly the center of gravity has moved fromC to C' along the x-axis

then

Big Circle:

A = pi*r^2
Xc = r
A*Xc = pi*r^3

small cirlce

A = -pi*(r/2)^2 * here the area is negative, b/c it a hole

Xc = r/2
A*Xc = -(pi*r^3)/8

then added both eq

Total area: (pi*r^2) + (-pi*(r/2)^2) = (3*pi*r^2)/4
Total A*Xc: (pi*r^3) + (-(pi*r^3)/8) = (7*pi*r^3)/8

from above eq

[(7*pi*r^3)/8] / [(3*pi*r^2)/4] = 7r/6

then Before the hole was removed

the center of gravity = r

then

(7r/6) - (r) = (r/6)

therefore it is r/6

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