A rod of length 4 L is bent into square. Now one side is removed then centre of mass of the system will shift by ?
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see diagram.
Each side is of length L. The center of mass of each side is at its midpoint. Let the mass of each side be equal to m. The center of mass of the rod in the square shape is at the center of square: (L/2, L/2)
Let us say that the side AD is removed. From symmetry, the y coordinate remains at L/2.
x-coordinate of the Center of mass of ABC :
[m* L/2 + m* L + m * L/2] / (m+m+m) = 2 L / 3
The shift in the center of mass = 2L/3 - L/2 = L/6
Each side is of length L. The center of mass of each side is at its midpoint. Let the mass of each side be equal to m. The center of mass of the rod in the square shape is at the center of square: (L/2, L/2)
Let us say that the side AD is removed. From symmetry, the y coordinate remains at L/2.
x-coordinate of the Center of mass of ABC :
[m* L/2 + m* L + m * L/2] / (m+m+m) = 2 L / 3
The shift in the center of mass = 2L/3 - L/2 = L/6
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kvnmurty:
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