A) 6.25 rad/sec
B) 0.625 rad/sec
C) 3.35 rad/sec
D) 0.335 rad/sec
Correct Answer: C
Solution :
From conservation of angular momentum \[mv\,({{r}_{axis}})=\frac{1}{2}l{{\omega }^{2}}\] \[10\times {{10}^{-3}}\times 500\times 0.5=\frac{1}{2}\times \left( \frac{M{{l}^{2}}}{12}+M{{\left( \frac{l}{2} \right)}^{2}} \right)\times {{\omega }^{2}}\] (from parallel axes theorem) \[2.5\times 2=\frac{M{{l}^{2}}}{4}\,\left[ \frac{1}{3}+1 \right]{{\omega }^{2}}\] \[5\times 4=12\times {{(1)}^{2}}\times \frac{4}{3}\times {{\omega }^{2}}\] \[\frac{5\times 3}{12}={{\omega }^{2}}\] \[\frac{15}{12}={{\omega }^{2}}\] \[1.25={{\omega }^{2}}\] \[\omega =1.12\,\,\text{rad/s}\]You need to login to perform this action.
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