7.54) The three masses represented in the figure are connected by rigid massless rods.
Answer 1
A.
rB = (0.0)
CR=(12.0)cm
For rA, draw a perpendicular AM from A to BC. As AB=AC, M will be the midpoint of BC. So BM = 6cm
So the x coordinate of A will be 6.
For the y coordinate, using Pythagoras thm., AM=sqrt(AB^2 – BM^2) = sqrt(10^-6^2) = sqrt(64) = 8.
Then rA=(6.8) cm
Now the coordinates of the center of gravity
=(200*rA+100*rB+100*rC)/(200+100+100)
=100(2rA+rB+rC)/400
={2(6.8) + (0.0) +(12.0)}/4
=(24.16)/4
=(6.4) centimeters
=(0.06,0.04)m
B
The moment of inertia about an axis passing through A and perpendicular to the plane of the paper is
I=200*0^2 + 100*10^2 +100*10^2 = 0 + 10,000 + 10,000 = 20,000 g*cm^2
=20000*10^-3*10^-4 kg m^2
=2*10^-3 kg m^2
C. Use the same method as in B