The objects in the figure below are constructed of uniform wire bent into the sh

Question

The objects in the figure below are constructed of uniform wire bent into the shapes shown (Figure 1) . Each peice has a length of 0.45m .

A.) Find the x- and y- coordinates of the center of mass of the object (a). The angle shown is 12degrees . Assume the origin to be at the bottom left point.

B.) Find the x- and y-coordinates of the center of mass of the object (b). Assume the origin to be at the bottom left point.

C.) Find the x- and y-coordinates of the center of mass of the object (c). Assume the origin to be at the bottom left point.

D.) Find the x- and y-coordinates of the center of mass of the object (d). Assume the origin to be at the bottom left point.

Explanation / Answer

When you have a group of masses m1,m2,..etc with knowm CM's (x1,y1),(x2,y2),.. etc you get the CM of the group from;
X = (1/M)[m1x1 + m2x2 + ...]
Y = (!/M)[m1y1 + m2y2 + ...]

So treat each straight segment as a separate mass with its CM at its center, since its uniform & straight.
For example, the figure I_I is three masses. the CM of the first verticle is at (0,L/2) The Horizontal segment's CM is at (L/2,0) and the last verticle is at (L,L/2) . Then the CM of the group is at;
X = (1/3m)[m(0) + m(L/2) + m(L)] = L/2
Y = (1/3m)[m(L/2) + m(0) + m(L/2)] = L/3

The second figure is the same as the above, with the last mass missing;
X = (1/2m)[m(0) + m(L/2)] = L/4
Y = (1/2m)[m(L/2) + m(0)] = L/4

You'll have to use some trig to find the CM location's of each of the sides of the triangle (The bottom side is at (L/2,0) ), but otherwise the procedure is the same

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