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jgrins jgrins
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9 years ago
Find the moment of inertia about an axis that passes through masses B and C in Figure 1.

The first parts of the question had me find the coordinates of center of gravity (.06, .04), and I if the axis was coming out of the page at mass A (.002 kgm^2).

But for some reason I'm having trouble finding the last IBC.
I set up my problem as IBC = (.08m)^2 (.2kg) since B and C are along the axis and r=0. However that was wrong.

Then I set up my problem as
IBC = (.2kg)(.1m)^2 + (.2kg)(.1m)^2
But that was also wrong, so I'm not sure where to go from here.
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bio_manbio_man
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9 years ago
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#2
9 years ago
m = 195 g, m' = 100 g, coordinate of A: (6, √(102 - 62)) = (6, 8) cm
a) By symmetry, the coordinates of the center C of mass is on the vertical line through mass A,

so xc = 6 cm,

yc = 195*8/(195 + 100 + 100) = 3.95 cm

C: (0.06, 0.0395) m

b) the moment of inertia about an axis that passes through mass A and is perpendicular to the page

= 0.100*0.062 + 0.100*0.062 + 0.195*02 = 7.20*10-4 kgm2

(c) the moment of inertia about an axis that passes through masses B and C

= 0.100*02 + 0.100*02 + 0.195*0.082 = 1.25*10-3 kgm2
Biology - The only science where multiplication and division mean the same thing.
jgrins Author
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#3
9 years ago
Thanks the image helped a lot. I guess I was just confused because how you find moment of inertia is the sum of the mr^2 so I was trying to square .08m since I knew that was the r for the BC axis...
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