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A thin smooth rod of length L and mass M...

A thin smooth rod of length L and mass M is rotating freely with angular speed `omega_(0)` about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be

A

`(Momega_(0))/(M+6m)`

B

`(Momega_(0))/(M+3m)`

C

`(Momega_(0))/(M+m)`

D

`(Momega_(0))/(M+2m)`

Text Solution

Verified by Experts

The correct Answer is:
A
Applying conservation of angular momentum
`((ML^(2))/(12)+m(0)^(2)+m(0)^(2))omega_(0)=((ML^(2))/(12)+m((L)/(2))^(2)+m((L)/(2))^(2))omega`
implies `(ML^(2))/(12)omega_(0)=((ML^(2))/(12)+(mL^(2))/(2))omega`
we get, `omega=(Momega_(0))/(M+6m)`
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