A) 2 : 7
B) 10 : 7
C) 7 : 10
D) 2 : 5
Correct Answer: A
Solution :
Key Idea: Total energy of a rolling sphere is the sum of translational and rotational kinetic energy. Translational kinetic energy of rolling sphere \[{{K}_{T}}=\frac{1}{2}M{{v}^{2}}\] Rotational kinetic energy of rolling sphere \[{{K}_{R}}=\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}\left( \frac{2}{5}M{{R}^{2}} \right){{\omega }^{2}}\] \[\left( \because I=\frac{2}{5}M{{R}^{2}} \right)\] where R is the radius of sphere and M its mass. Thus, total energy \[K={{K}_{T}}+{{K}_{R}}\] \[=\frac{1}{2}M{{v}^{2}}+\frac{1}{5}M{{R}^{2}}{{\omega }^{2}}\] \[=\frac{1}{2}M{{R}^{2}}{{\omega }^{2}}+\frac{1}{5}M{{R}^{2}}{{\omega }^{2}}\] \[=\frac{7}{10}M{{R}^{2}}{{\omega }^{2}}\] Hence, \[\frac{{{K}_{T}}}{K}=\frac{(1/5)\,M{{R}^{2}}{{\omega }^{2}}}{(7/10)\,M{{R}^{2}}{{\omega }^{2}}}=\frac{2}{7}\]
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