the position vector of the centre of mass lt
`to`
r cm of an asymmetric uniform bar of negligible area of cross - section as shown in figure is :

Options

A. `vec( r) cm = (3)/(8)L hat(x)+(11)/(8)L hat(y)`
B. `vec( r) cm =(11)/(8) L hat(x)+(3)/(8)Lhat(y)`.
C. `vec( r) cm = (13)/(8)L hat(x)+(5)/(8)Lhat(y)`.
D. `vec( r) cm =(5)/(8) L hat(x)+(13)/(8) Lhat(y)`.

Answer

B

Explanation

`m_1=2m`
`m_2=m`
`m_3=m`
We can assume the com of the three rods located at P,Q,R
`X_("com")=(m_1x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
`=(m)/(4m) (2L+2L+(5L)/(2))=(13L)/(8)`
`Y_("com")=(m_1 Y_(1)+m_(2)Y_(2)+m_(3)Y_(3))/(m_(1)+m_(2)+m_(3))`
`Y_("com")=(2m xxL+mxxL//2+mxxxO)/(m_1+m_2+m_3)=(m)/(4m)(2L+L//2)=(5L)/(8)`
.