A) 24 rad/s
B) 36 rad/s
C) 15 rad/s
D) 26 rad/s
Correct Answer: A
Solution :
Key Idea: Angular momentum in absence of any external torque remains constant. If no external torque acts on a system of particles, then angular momentum of the system remains constant, that is \[\tau =0\] \[\Rightarrow \] \[\frac{dL}{dt}=0\] \[\Rightarrow \] \[L=I\omega =\] constant \[\Rightarrow \] \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] \[\therefore \] \[\frac{1}{2}M{{r}^{2}}{{\omega }_{1}}=\frac{1}{2}(M+2m){{r}^{2}}{{\omega }_{2}}\] ... (i) Here, \[M=2\,kg,\,m=0.25\,kg,\,r=0.2\,m\], \[{{\omega }_{1}}=30\,rad/s\] Hence, we get after putting the given values in Eq.(i) \[\frac{1}{2}\times 2\times {{(0.2)}^{2}}\times 30\] \[=\frac{1}{2}\times (2+2\times 0.25){{(0.2)}^{2}}\times {{\omega }_{2}}\] \[\Rightarrow \] \[1.2=0.05\,{{\omega }_{2}}\] \[\Rightarrow \] \[{{\omega }_{2}}=24\,rad/s\]
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