A) \[\frac{mvr}{{{I}_{0}}+m{{r}^{2}}}\]
B) \[\frac{2mvr}{{{I}_{0}}}\]
C) \[\frac{v}{2r}\]
D) \[\frac{mvr}{2{{I}_{0}}}\]
Correct Answer: A
Solution :
[a] Given that \[{{I}_{0}}\] is the moment of inertia of table and gun and m the mass of bullet. Initial angular momentum of system about centre \[{{L}_{i}}({{I}_{0}}+m{{r}^{2}}){{\omega }_{0}}\] ? (i) Let w be the nagular velocity of table after the bullet is fired. Final angular momentum \[Lf=({{I}_{0}}\omega -m(v-r\omega )r\] ? (ii) Where (\[v-r\omega \]) is absolute velocity of bullet to the right. Equations (i) and (ii); we get \[(\omega -{{\omega }_{0}})=\frac{mvr}{{{I}_{0}}+m{{r}^{2}}}\]
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