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. Find the x- and y- coordinates of the center of mass of the object.. Assume the ORIGIN to be at the BOTTOM LEFT POINT.

I see this question has been answered many times but mostly incorrect. I only need the answer for the first object. (1/cosa , ?) The x-value is correct but I cannot figure out the y-value. Thanks for your help.

http://session.masteringphysics.com/problemAsset/1263295/2/yg.8.47.jpg

The objects in the figure below are constructed of uniform wire bent into the shapes shown. Find the x- and y- coordinates of the center of mass of the object.. Assume the ORIGIN to be at the BOTTOM LEFT POINT.

Explanation / Answer

a) the notion of symmetry helps a lot in this case
the CoM must lie on the vertical line through top vertex
so x = L sin(alpha/2)
Now, the centre of mass of a stick lies midway. because both sticks are symmetrically places, both centre of masses are at the same height, so the overall CoM is at the same height
which is y = L cos(alpha/2) /2

b) COM of bottom is at (L/2, 0)
by symmetry of arrangement x = L/2
for y
1 com at 0, 2 at L/2
so CoM = 1/3*(0 +L/2 + L/2)
=L/3

c) There is no difference between x and y and exchanging x and y, since arrangement is symmetrical about x=y line
using the same logic as previously
1 com at x=0
1 com at x = L/2
so x = 1/2 *(0 + L/2) = L /4
ans is (L/4,L/4)

d) again by symmetry x = L/2
here for the two sticks inclined,
y = L/2 * cos (alpha/2) = L/2*cos(60/2) since its an equilateral triangle alpha = 60
so y of the upper two sticks = sqrt(3)*L/4
2 sticks with com at y = sqrt(3)*L/4
1 y = 0
so overall com is y = 1/3*(2*sqrt(3)*L/4+0)
= L/ (2*sqrt(3))

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