Asked by Herosugar
College Physics- Center of Mass?
The objects in the figure below are constructed of uniform wire bent into the shapes shown (Figure 1) . Each peice has a length of 0.15 .
There are two objects that we need to figure out the x- and y-coordinates of the center of mass of the object . Assume the origin to be at the bottom left point.
Help please
1. the first object is like a l_l the wires are connected in this exact shape.
2. is the shape of L
3. is an equilateral triangle.
Remember each side is 0.15m .. help me please
The objects in the figure below are constructed of uniform wire bent into the shapes shown (Figure 1) . Each peice has a length of 0.15 .
There are two objects that we need to figure out the x- and y-coordinates of the center of mass of the object . Assume the origin to be at the bottom left point.
Help please
1. the first object is like a l_l the wires are connected in this exact shape.
2. is the shape of L
3. is an equilateral triangle.
Remember each side is 0.15m .. help me please
Answers
Answered by
drwls
1. On axis of symmetry midway between vertical wires and 0.05 m above bottom wire.
2. Along a 45 degree from the corner of the L. 0.0375 m to the right of and 0.0375 m above the corner.
3. At the center of the triangle. (The center of symnmetry where the angle bisectors intersect.)
2. Along a 45 degree from the corner of the L. 0.0375 m to the right of and 0.0375 m above the corner.
3. At the center of the triangle. (The center of symnmetry where the angle bisectors intersect.)
Answered by
Herosugar
Yes but how do we go about this.
Let me show you what I did for l_l
Let me show you what I did for l_l
Answered by
Elena
r(C) = Σ{r(i) •m(i)}/Σm(i)
Assume the origin to be at the bottom left point and the mass of each part is ’m’
1. X(C) ={ 0+(L/2)+L}m/3m = L/2
Y(C) = { (L/2) +0+(L/2)}m/3m = L/3
C= (L/2; L/3) = (0.075;0.05).
2. X(C) ={ 0+(L/2)}m/2m = L/4
Y(C) = { (L/2) +0}m/2m = L/4
C= (L/4; L/4) = (0.0375;0.0375).
3. X(C) ={ L/2•cos60°)+(L/2)+(L- L/2•cos60°)}m/3m = L/2
Y(C) = { (L/2•sin60°) +0+(L/2•sin60°))}m/3m = 0.29L
C= (L/2; L/3) = (0.075;0.0435).
Assume the origin to be at the bottom left point and the mass of each part is ’m’
1. X(C) ={ 0+(L/2)+L}m/3m = L/2
Y(C) = { (L/2) +0+(L/2)}m/3m = L/3
C= (L/2; L/3) = (0.075;0.05).
2. X(C) ={ 0+(L/2)}m/2m = L/4
Y(C) = { (L/2) +0}m/2m = L/4
C= (L/4; L/4) = (0.0375;0.0375).
3. X(C) ={ L/2•cos60°)+(L/2)+(L- L/2•cos60°)}m/3m = L/2
Y(C) = { (L/2•sin60°) +0+(L/2•sin60°))}m/3m = 0.29L
C= (L/2; L/3) = (0.075;0.0435).
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