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Two identical particles B and C each of ...

Two identical particles `B` and `C` each of mass `50 g` are connected by a light rod of length `30 CM`. Another particle `A` of same mass moving With a speed `u = 60 CM//s` strikes `B`, in a direction perpendicular to `AB`, and sticks to it. The whole process takes place on a smooth horizontal plane. Find the angular velocity `omega` of the system about its centre of mass, immediately after the impact.

Text Solution

Verified by Experts

Clearly, the centreof mass of the system `(A+B+C)` is at a distance of `10 cm` from along the rod
Initial angualr momentum
`L_(i)=m_(A) v_(A)r_(A)`
`=50xx10^(-3)xx60xx10^(-2)xx10xx10^(-2)`
`=3xx10^(-3)` S.I. unit
Moment of inertia about new centre of mass
`I=(2xx50xx10^(-3))x(10xx10^(-2))^(2)+(50+10^(-3))xx(20xx10^(-2))^(2)`
`=3xx10^(-3)` S.I. unit
From `L_(i)=L_(f)impliesL_(i)=I.omega`
`implies omega=L/i=1rad//s`
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