A) \[\frac{{{K}^{2}}}{{{K}^{2}}+{{R}^{2}}}\]
B) \[\frac{{{R}^{2}}}{{{K}^{2}}+{{R}^{2}}}\]
C) \[\frac{{{K}^{2}}+{{R}^{2}}}{{{R}^{2}}}\]
D) \[\frac{{{K}^{2}}}{{{R}^{2}}}\]
Correct Answer: A
Solution :
Key Idea: In rolling without slipping, total energy of ball is the sum of its translational and rotational energy. Kinetic energy of rotation is \[{{K}_{rot}}=\frac{1}{2}\,I\,{{\omega }^{2}}=\frac{1}{2}\,M\,{{K}^{2}}\frac{{{v}^{2}}}{{{R}^{2}}}\] where K is radius of gyration Kinetic energy of translation is \[{{K}_{trans}}.=\frac{1}{2}\,M{{v}^{2}}\] Thus, total energy \[E={{K}_{rot}}+{{K}_{trans.}}\] \[=\frac{1}{2}M{{K}^{2}}\,\frac{{{v}^{2}}}{{{R}^{2}}}+\frac{1}{2}M{{v}^{2}}\] \[=\frac{1}{2}M{{v}^{2}}\left( \frac{{{K}^{2}}}{{{R}^{2}}}+1 \right)\] \[=\frac{1}{2}\,\,\frac{M{{v}^{2}}}{{{R}^{2}}}({{K}^{2}}+{{R}^{2}})\] Hence, \[\frac{{{K}_{rot}}.}{{{K}_{trans.}}}=\frac{\frac{1}{2}M{{K}^{2}}\frac{{{v}^{2}}}{{{R}^{2}}}}{\frac{1}{2}\frac{M{{v}^{2}}}{{{R}^{2}}}({{K}^{2}}+{{R}^{2}})}=\frac{{{K}^{2}}}{{{K}^{2}}+{{R}^{2}}}\]
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