A) 4L
B) 2L
C) \[\frac{L}{2}\]
D) \[\frac{L}{4}\]
Correct Answer: D
Solution :
Let we represent the rotational kinetic energies by \[{{E}_{1}}\]and \[{{E}_{2}}\]. From the relation, \[{{K}_{rot}}=\frac{1}{2}I{{\omega }^{2}}\] \[\left( \begin{align} & \because \,L=I\omega \\ & I=\frac{L}{\omega } \\ \end{align} \right)\] \[E=\frac{1}{2}\frac{L}{\omega }{{\omega }^{2}}\] \[E=\frac{1}{2}L\omega \] Hence, \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{L}_{1}}{{\omega }_{1}}}{{{L}_{2}}{{\omega }_{2}}}\left( \begin{align} & \text{Given:}\,{{E}_{2}}=\frac{{{E}_{1}}}{2} \\ & {{\omega }_{2}}=2{{\omega }_{1}} \\ \end{align} \right)\] \[\frac{{{E}_{1}}}{\frac{{{E}_{1}}}{2}}=\frac{{{L}_{1}}}{{{L}_{2}}}\times \frac{{{\omega }_{1}}}{2{{\omega }_{1}}}\] \[2=\frac{{{L}_{1}}}{{{L}_{2}}}\times \frac{1}{2}\] \[\frac{{{L}_{1}}}{{{L}_{2}}}=4\] \[{{L}_{2}}=\frac{{{L}_{1}}}{4}\]if \[{{L}_{1}}=L\] \[{{L}_{2}}=\frac{L}{4}\]
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