A thin uniform rod of mass .M. and length .L. rotates with constant angular velocity `.omega.` about perpendicular axis passing through its centre. Two bodies each of mass `(M)/(3)` are attached to its two ends. What is its angular velocity?

Options

A. `(2)/(3)omega`
B. `(1)/(7)omega`
C. `(1)/(6)omega`
D. `(1)/(2)omega`

Answer

A

Explanation

Initially angular momentum
`L_(1)=Iomega`
but `I=(ML^(2))/(12)`
`therefore L_(1)=(ML^(2)omega)/(12)….(1)`
Angular momentum in final (other) position

`L_(2)=Iomega.+M^(1)R^(2)omega.+M^(1)R^(2)omega.`
`=((ML^(2))/(12)+(M)/(3)xx(L^(2))/(4)+(M)/(3)xx(L^(2))/(4))omega.`
`=((ML^(2))/(12)+(ML^(2))/(12)+(ML^(2))/(12))omega.`
`L_(2)=3(ML^(2))/(12)omega.....(2)`
Angular momentum is conserved.
`therefore L_(1)=L_(2)`
`therefore (ML^(2)omega)/(12)=(3ML^(2)omega.)/(12)`
`therefore omega=3omega.`
`therefore omega.=(omega)/(3)`