08.1 An earring is cut from a circular plate of radius 10 cm with uniform thickn

Question

08.1 An earring is cut from a circular plate of radius 10 cm with uniform thickness and density. A circle of radius 9.0 cm has been cut out, where the edges of the two circles meet at the top (see diagram). The center of the larger circle is at point A, and the center of the smaller circle is at point B. Where is the center of mass of the earring relative to point A? [Hint: Use the subtraction method.

I've seen a bunch of different answers on Chegg, not sure which is correct. Thanks for the help.

08 1 An earring is cut from a cireular plate of 08.1 An earring is cut from a circular plate of radius 10 cm with uniform thickness and density. A circle of radius 9.0 cm has been cut out, where the edges of the two circles meet at the top (see diagram). The center of the larger circle is at point A, and the center of the smaller circle is at point B. Where is the center of mass of the earring relative to point A? [Hint: Use the subtraction method. 9 cm 10 cm

Explanation / Answer

given that

radius of big circle   is R = 10 cm

radius of small circle is    r = 9 cm

we know that density of big circle is

rho = M/(pi*R^2)

M = rho *pi*R^2

than relative center of mass is

(rcm) = -rho (pi*r^2)(2(R - r)) / [M - rho(pi*r^2)]

(rcm) = -rho(pi*r^2) (2(R - r)) / [rho (pi*R^2) - rho(pi*r^2)]

(rcm) = -2(r^2)(R - r) / (R^2 - r^2)

(rcm) = -2(r^2) / (R + r)

(rcm) = -2 ((9.0*10^(-2))^2) / (10.0*10^(-2) + 9.0*10^(-2) )

(rcm) = -8.53*10^(-2) (negative sign show that relative center of mass is opposite side of A as B)

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