question_answer

A solid sphere is rolling down an inclined plane. Then, the ratio of its translational kinetic energy to its rotational kinetic energy is

A) \[2.5\]                  

B) \[1.5\]

C) \[1\]                    

D) \[0.4\]

Correct Answer: A

Solution :

Translational kinetic energy, \[{{K}_{T}}=\frac{1}{2}m{{v}^{2}}\] Rotational kinetic energy \[{{V}_{1}}\] \[{{K}_{R}}=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\times \frac{2}{5}m{{R}^{2}}\times \frac{{{v}^{2}}}{{{R}^{2}}}=\frac{1}{5}\,\,m{{v}^{2}}\] \[\frac{{{K}_{T}}}{{{K}_{R}}}=\frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{5}m{{v}^{2}}}=\frac{5}{2}=2.5\]