3 particles of masses m1 = 1.5 kg , m2 = 2.5 kg and m3 = 3.5 kg form an equilateral triangle of edge length l= 135 cm. Determine the center of mass of the system.
To determine the center of mass of the system, we need to find the coordinates of the center of mass. We'll start by finding the coordinates of each particle.
Let's assume the equilateral triangle lies in the xy-plane and one of the vertices is located at the origin (0, 0). Let's denote the other two vertices as A and B.
The length of the equilateral triangle's sides is given as l = 135 cm. Since it is an equilateral triangle, all sides are equal. The distance from the origin to vertex A is half the length of a side, which is l/2 = 135/2 = 67.5 cm.
Let's denote the coordinates of vertex A as (x1, y1). Since it is equidistant from the origin, we have x1^2 + y1^2 = (67.5)^2.
The coordinates of vertex B can be obtained by rotating vertex A by 60 degrees counterclockwise. Let's denote the coordinates of vertex B as (x2, y2). We can find their relationship using a rotation matrix:
x2 = x1 * cos(60°) - y1 * sin(60°)
y2 = x1 * sin(60°) + y1 * cos(60°)
Now, let's calculate the coordinates of vertex A:
x1^2 + y1^2 = (67.5)^2
x1^2 + x1^2 = (67.5)^2 (since y1 = x1)
2x1^2 = (67.5)^2
x1^2 = [(67.5)^2]/2
x1^2 ≈ 2281.25
x1 ≈ √2281.25 ≈ 47.78 cm
Now, let's use the rotation matrix to find the coordinates of vertex B:
x2 = x1 * cos(60°) - y1 * sin(60°)
x2 = 47.78 * cos(60°) - 47.78 * sin(60°) ≈ 78.89 cm
y2 = x1 * sin(60°) + y1 * cos(60°)
y2 = 47.78 * sin(60°) + 47.78 * cos(60°) ≈ 27.29 cm
Now that we have the coordinates for each vertex, we can calculate the coordinates of the center of mass using the formula:
X_com = (m1 * x1 + m2 * x2 + m3 * x3) / (m1 + m2 + m3)
Y_com = (m1 * y1 + m2 * y2 + m3 * y3) / (m1 + m2 + m3)
Plugging in the given masses and coordinates, we get:
X_com = (1.5 kg * 0 cm + 2.5 kg * 78.89 cm + 3.5 kg * 67.5 cm) / (1.5 kg + 2.5 kg + 3.5 kg)
X_com ≈ 59.46 cm
Y_com = (1.5 kg * 0 cm + 2.5 kg * 27.29 cm + 3.5 kg * 0 cm) / (1.5 kg + 2.5 kg + 3.5 kg)
Y_com ≈ 10.38 cm
Therefore, the center of mass of the system is approximately located at (59.46 cm, 10.38 cm).