Three identical spheres of mass M are placed at the corners of eqvilateral triangle of sides 2m taking one of the corner as origin the position vector of the centre of mass is
I have no idea how your triangle is oriented with respect to the x and y axes. You only have said that one corner is at the origin.
The center of mass is at the center of symmetry, where the angle bisectors intersect.
To find the position vector of the center of mass of three identical spheres placed at the corners of an equilateral triangle, you can use the concept of the centroid.
The centroid of any triangle is the point of intersection of its medians. A median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.
In this case, you have an equilateral triangle, so all its sides have equal length. Let's assume that the side length of the triangle is "a" (given as 2m).
To find the position vector of the centroid, follow these steps:
1. Label the three corners (A, B, and C) of the equilateral triangle.
- Let A be the corner with the origin (0, 0).
- B will be at coordinates (a, 0).
- C can be found by using trigonometry. Since it is an equilateral triangle, the angle at each corner is 60 degrees. So, by using the cosine rule, the coordinates of C will be (a/2, √3a/2).
2. Find the coordinates of the midpoint of each side:
- The midpoint of side AB will be M1: (a/2, 0).
- The midpoint of side BC will be M2: (3a/4, √3a/4).
- The midpoint of side CA will be M3: (a/4, √3a/4).
3. Calculate the position vector of the center of mass (centroid):
- The x-coordinate of the centroid (Cx) can be found by taking the average of the x-coordinates of the midpoints: (CxAvg = (a/2 + 3a/4 + a/4)/3)
- The y-coordinate of the centroid (Cy) can be found by taking the average of the y-coordinates of the midpoints: (CyAvg = (√3a/4 + √3a/4 + √3a/2)/3)
4. Simplify the coordinates to get the final position vector:
- The position vector of the centroid is: (Cx, Cy)
Thus, the position vector of the center of mass is (Cx, Cy), where Cx = (5a/8) and Cy = (√3a/8).