The tension in the string is increased gradually and finally m moves in a circle of radius\[\frac{{{R}_{0}}}{2}\]. The final value of the kinetic energy is
A) \[mv_{0}^{2}\]
B) \[\frac{1}{4}mv_{0}^{2}\]
C) \[2\,mv_{0}^{2}\]
D) \[\frac{1}{2}mv_{0}^{2}\]
Correct Answer: C
Solution :
Conserving angular momentum \[{{L}_{i}}={{L}_{t}}\] \[\Rightarrow \,\,\,\,m{{v}_{0}}{{R}_{0}}=mv'\left( \frac{{{R}_{0}}}{2} \right)\] \[\Rightarrow \,\,\,\,\,\,v'=2{{v}_{0}}\] So, final kinetic energy of the particle is \[{{K}_{f}}=\frac{1}{2}\,mv{{'}^{2}}\,=\frac{1}{2}\,m{{(2{{v}_{0}})}^{2}}\] \[=4\frac{1}{2}mv_{0}^{2}=2mv_{0}^{2}\]
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