A) 4/3
B) 1/3
C) 1/2
D) 2/3
Correct Answer: C
Solution :
Linear kinetic energy \[{{K}_{l}}=\frac{1}{2}m{{\upsilon }^{2}}\] \[=\frac{1}{2}m{{\upsilon }^{2}}{{\omega }^{2}}=\frac{1}{2}l{{\omega }^{2}}\] Rotational kinetic energy\[{{K}_{r}}=\frac{1}{2}l{{\omega }^{2}}\] Total kinetic energy \[K={{K}_{l}}+{{K}_{r}}\] \[\begin{align} & =\frac{1}{2}l{{\omega }^{2}}+\frac{1}{2}l{{\omega }^{2}} \\ & =l{{\omega }^{2}} \\ \end{align}\] \[\frac{{{K}_{r}}}{K}=\frac{1}{2}\]
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