A circular ring of mass 6 kg and radius a is placed such that its centre lies at the origin. Two particles of masses 2 kg each are placed at the intersecting points of the circle with +ve x-axis and +ve y-axis. Then the angle made by the position vector of centre of mass of entire system with x-axis is :

Options

A. `45^(@)`
B. `60^(@)`
C. `tan^(-1)(4//5)`
D. `30^(@)`

Answer

A

Explanation

It is clear from the figure that coordinates of the centre of mass C,
`X_(CM)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+_m_(3))=a/5`
`Y_(CM)=(m_(1)y_(1)+m_(2)y_2)+m_(3)y_(3))/(m_(1)+m_(2)+m_(3))=a/5" "therefore(X_(CM), Y_(CM))=(a/5, a/5)`
Hence `bar(OC)=a/5hat(i)+a/5hat(j)`
`therefore` Angle made by `bar(OC)` with x-axis = `tan^(-1)[y/x]=tan^(-1)[(a//5)/(a//5)]=45^(@)`