Post a New Question

College physics - Center of mass

posted by .

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

  • College physics - Center of mass -

    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.)

  • College physics - Center of mass -

    Yes but how do we go about this.

    Let me show you what I did for l_l

  • College physics - Center of mass -

    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).

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question