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Three particles of masses 1 kg, (3)/(2) ...

Three particles of masses `1 kg, (3)/(2) kg, and 2 g` are located the vertices of an equilateral triangle of side `a`. The `x, y` coordinates of the centre of mass are.

A

`(5a)/(9) , (2a)/(3sqrt 3)`

B

`(2a)/(3sqrt (3)) , (5a)/(9)`

C

`(5a)/9 , (2a)/(sqrt 3)`

D

`(2a)/(sqrt(3)) , (5a)/(9)`

Text Solution

Verified by Experts

The correct Answer is:
a
Let the masses 1 kg, `3/2`, kg and 2kg are located at the vertices, A,B and C as shown in figure. The coordinates of points A,B and C are (0,0), (a,0), `(a/2, (sqrt(3)a)/(2))` respectively. The coordinates of center of mass are
`X_(CM) = (m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
`=[(1 xx 0 +3/2 xx a+2 xx a/2)]/([1 + 3/2 +2])=(5a)/(9)`
`Y_(CM) =(m_(1)y_(1)+m_(2)y_(2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))`
`=([1 xx 0 +3/2 xx 0 + 2 xx (sqrt(3)a)/(2)])/[(1+3/2+2)]= (2sqrt(3)a)/(9)=(2a)/(3sqrt(3))`
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