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A bullet of mass m collides inelasticall...

A bullet of mass m collides inelastically at the periphery of a disc of mass M and radius R, with a speed v as shown. The disc rotates about a fixed horizontal axis. Find the angular velocity of the disc bullet system just after the impact.

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In this impact, the torque caused by the impulsive forces acting about the centr O is zero as for the system of bullet and disc impact force is internal force.
That is why we can conserve the angular momentum of the system about the axis passing through O just before and after the impact. The angular momentum of the system about O, is `mvR = (I_(0)) omega" "...(1)`

where `I_(0)` = M.I. of the system about `O = (I_("disc"))_(0) + (I_("bullet"))_(0)`
`I_(0) = (1)/(2) MR^(2) + mR^(2) = ((M + 2m)/(2))R^(2)" "...(2)`
Using (A1) & (2) we obtain
`omega = (v)/(R((M)/(2m)+1))`
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