A thin uniform rod of mass M and length L is rotating about a perpendicular axis passing through its centre with a constant angular velocity `omega`. Two objects each of mass `(M)/(3)` are attached gently to the two end of the rod.
The rod will now rotate with an angular velocity of

Options

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

Answer

A

Explanation

Initial the moment of inertia of the uniform rod.
` I = (1)/(12) ML^(2)`
After attaching the two objects, the moment of inertia,
`I. = (1)/(12) ML^(2) + 2 xx (1)/(3)M ((L)/(2))^(2) = ((1)/(12) + (1)/(6)) ML^(2) = (1)/(4) ML^(2)`
From the law of conservation of the angular momentum,
`I omega = I. omega.`
or, ` omega. = (I)/(I.) omega = ((1)/(12)MN^(2))/((1)/(4)ML^(2)) omega = (1)/(3) omega`