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?

A thin uniform rod of mass .M. and length .L. rotates with constant an

1.	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?
Answer» `(2)/(3)omega` `(1)/(7)omega` `(1)/(6)omega` `(1)/(2)omega` Solution : 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)`