Two particles of masses `m_1 and m_2` are joined y a light rigid rod of length r. The system rotates at an angular speed `omega` about an axis through the centre of mass of the systemand perpendicular to the rod. Shwo thast the angular momentum of the system `is L=mur^2omega where mu` is the reduced mass of the system defined as `mu=(m_1m_2)/(m_1+m_2)`

Options

A. `mu=m_(1)+m_(2)`
B. `mu=(m_(1)m_(2))/(m_(1)+m_(2))`
C. `mu= sqrt(m_(1)m_(2))`
D. `mu=(m_(1)+m_(2))/(m_(1)m_(2))`

Explanation

Let `x_(1)` and `x_(2)` are the distances of particles from centre of mass as shown in figure.

`x_(1)=(m_(2)l)/((m_(1)+m_(2)))` and `x_(2)=(m_(1)l)/((m_(1)+m_(2)))`
Moment of inertia of the system about axis passing through the centre of mass can be written as follows:
`I=m_(1)[(m_(2)l)/((m_(1)+m_(2)))]^(2)+m_(2)[(m_(1)l)/((m_(1)+m_(2)))]^(2)`
`impliesI=[(m_(1)m_(2))/(m_(1)+m_(2))]l^(2)`
Angular momentum can be written as
`L=IomegaimpliesL=((m_(1)m_(2))/(m_(1)+m_(2)))l^(2)omega=mul^(2)omega` Comparing it with the given result we get `mu=(m_(1)m_(2))/(m_(1)+m_(2))`.