The position vector of the center of mass r → cm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

Step-by-step solution

The position vector of center of mass r → cm is given as r → cm = X cm x ^ + Y cm   y ^ Where, X cm = x - coordinate of center of mass and Y cm = y - coordinate of center of mass r → cm = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3   x ^   + m 1 y 1 + m 2 y 2 + m 3 y 3 m 1 + m 2 + m 3   y ^ r → cm = 2 m × L + m × 2 L + m 5 L 2 2 m + m + m x ^ + 2 m × L + m × L 2 + m × 0 2 m + m + m y ^ r → cm = 13 L 8 x ^ + 5 L 8 y ^