Center of MassHardBloom L3

Question

Centre of mass of two thin uniform rods of same length but made up of different materials & kept as shown, can be, if the meeting point is the origin of co-ordinates

Options

A.(L/2, L/2)
B.(2L/3, L/2)
C.(L/3, L/3)
D.(L/3, L/6)

Solution

Sol.

$x_{cm} = \frac{m_{1}.\frac{L}{2}}{m_{1} + m_{2}} = \frac{(\lambda_{1}.L).{L/}2}{(\lambda_{1}.L + \lambda_{2}L)} = \left( \frac{\lambda_{1}}{\lambda_{1} + \lambda_{2}} \right).\frac{L}{2}$

$y_{cm} = \frac{m_{2}.\frac{L}{2}}{m_{1} + m_{2}} = \left( \frac{\lambda_{2}.L}{\lambda_{1}.L + \lambda_{2}L} \right)\frac{L}{2} = \left( \frac{\lambda_{2}}{\lambda_{1} + \lambda_{2}} \right).\frac{L}{2}$

$x_{cm} + y_{cm} = \left( \frac{\lambda_{1} + \lambda_{2}}{\lambda_{1} + \lambda_{2}} \right).\frac{L}{2}$

⇒ $x_{cm} + y_{cm} = \frac{L}{2}$

Create a free account to view solution

View Solution Free
Topic: Center of Mass·Practice all Center of Mass questions

More Center of Mass Questions