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# 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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A detailed solution is given.

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