Center of Mass of a Triangle
Example
Question:
Find the centre of mass of a triangular lamina.
Solution:
The lamina (\( \triangle LMN \)) may be subdivided into narrow strips, each parallel to the base (\( MN \)), as shown in Fig. 6.10.
By symmetry, each strip has its centre of mass at its midpoint. If we join the midpoints of all the strips, we get the median \( LP \).
The centre of mass of the triangle as a whole, therefore, has to lie on the median \( LP \).
Similarly, we can argue that it also lies on the medians \( MQ \) and \( NR \).
This means the centre of mass lies on the point of intersection of the medians (i.e., the centroid of the triangle).
Thus, the centre of mass of a uniform triangular lamina is at the centroid, which is located at the intersection of its medians.
Vertex A
Vertex B
Vertex C
Center of Mass Formula
COM = ( (x₁+x₂+x₃)/3 , (y₁+y₂+y₃)/3 )
The center of mass (COM) is the average position of all the mass in the system. For a uniform triangle, it's located at the intersection point of the medians.



