Angular momentum and moment of intertia
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1) The position of the center of mass for a system of point masses located at positions r_{i} and masses m_{i} is by definition:
R = 1/M Sum_{i} m_{i} r_{i} (1.1)
where M is the total mass:
M = Sum_{i} m_{i} (1.2)
We want to prove that this formula also holds for a collection of extended masses if you take the r_{i} to be the locations of the center of masses. If you reorder the summation in (1.1) so that you sum first over the point masses that constitute extended body nr. 1 and then nr. 2 etc. (in your problem there are only two external bodies), then you can write:
R = 1/M Sum_{i} m_{i} r_{i} = 1/M [S1 + S2 + ...] (1.3)
where S1, S2 etc. are given by:
S1 = Sum_{i over body1} m_{i} r_{i} (1.4)
S2 = Sum_{i over body2} m_{i} r_{i} (1.5)
etc.
We can apply formula (1.1) to the summations in S1, S2, etc. to express these in terms of the center of masses of the extended masses. According to (1.1):
R1 = 1/M1 Sum_{i over body 1} m_{i} r_{i} (1.6)
R2 = 1/M2 Sum_{i over body 2} m_{i} r_{i} (1.7)
etc.
Using (1.4) and (1.5) you can rewrite (1.6) and (1.7) as:
R1 = 1/M1 S1
R2 = 1/M2 S2
So, the partial summations S1 and S2 are given by:
S1 = M1 R1
S2 = M2 R2
Inserting this in (1.3) gives:
R = 1/M [M1 R1 + M2 R2 + ...]
M is the total mass which is obviously the sum of the masses of the extended bodies and thus given by M1 +M2+...
2)
Let's put the semicircle on the upper half plane of the xy-plane. Introduce polar coordinates so that theta is the angle with the positive x-axis and a move in the counterclockwise direction corresponds to increasing theta. So, the positive y-axis is at theta = 90 degrees = pi/2 and the negative x-axis is at theta = 180 degrees = pi.
The semicircle covers the region from theta = 0 to theta = pi and r = 0 to r = R. By symmetry the center of mass will be on the y-axis, so we only have to find the y-coordinate of the center of mass. Let's introduce a surface mass density defined as the ...
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