Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of an equilateral triangle of side b. The coordinates of the centre of mass are :

Options

A. `[0, (7b)/12, (3sqrt(3)b)/12]`
B. `[(3sqrt(3)b)/12, (7b)/12, 0]`
C. `[(7b)/12, (3sqrt(3)b)/12, 0]`
D. `[(7b)/12, 0, (3sqrt(3)b)/12]`

Answer

C

Explanation

The coordinates of the points A, B and C are (0, 0, 0), (b, 0, 0) and `[b/2, (bsqrt(3))/2, 0]` respectively.

Now, as the triangle is in XY plane, ie, z-co-ordinate of all the masses is zero.
Now, `X_(CM)=(1 times 0+ 2times b+3(b//2))(1+2+3)=(7b)/12`
`Y_(CM)=(1 times 0+2 times 0+3sqrt(3)(b//2))/(1+2+3)=(3sqrt(3))/12b`
So, the coordinates of the centre of mass are `[(7b)/12, (3sqrt(3)b)/12, 0]`