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. `vecF cm = 3/8 L hatx + 11/8 L haty`
B. `vecF cm = 11/8 L hatx + 3/8 L haty`
C. `vecF cm = 13/8 L hatx + 5/8 L haty`
D. `vecF cm = 5/8 L hatx + 13/8 L haty`

Answer

C

Explanation

`m_1 = 2m , " " m_2 = m , " " m_3 = m`
We can assume the com of the three rods located at P, Q,
`X_("com") = (m_1x_1 + m_2x_2 + m_3x_3)/(m_1 + m_2 + m_3)`
`= m/(4m) (2L + 2L + (5L)/2) = (13L)/(8)`
`Y_("com") = (m_1Y_1 + m_2Y_2 + m_3Y_3)/(m_1 + m_2 + m_3) , " " Y_("com") = (2m xx L + m xx L//2 + mx xx o)/(m_1 + m_2 + m_3) = m/(4m) (2L + L//2) = (5L)/(8)` .