question_answer

A solid sphere is rolling on a frictionless surface, shown in the figure with a translational velocity v m/s. If it is to climb the inclined surface, then v should be

A) \[\ge \,\sqrt{\frac{10}{7}\,gh}\]               

B) \[\ge \,\sqrt{2\,\,gh}\]

C)             2 gh                             

D) \[\frac{10}{7}\,\,gh\]

Correct Answer: C

Solution :

Kinetic energy is converted to potential energy. From law of conservation of energy, energy can neither be created nor be destroyed but it remains conserved. In the given case the sum of kinetic energy of rotation and translation is converted to potential energy. Potential energy \[I=\frac{2}{5}M{{R}^{2}}\] \[\therefore \,\,\,\,\,\,\,\,\underset{\begin{smallmatrix}  (translational \\  kinetic\,\,energy) \end{smallmatrix}}{\mathop{\frac{1}{2}m{{v}^{2}}}}\,\,\,+\,\,\,\underset{\begin{smallmatrix}  (Rotational \\  energy) \end{smallmatrix}}{\mathop{\frac{1}{2}I{{\omega }^{2}}}}\,\,\,=\,\,\underset{\begin{smallmatrix}  (Potential \\  energy) \end{smallmatrix}}{\mathop{mgh}}\,\] \[\Rightarrow \,\,\,\,\,\frac{1}{2}m{{\nu }^{2}}+\frac{1}{2}\left( \frac{2}{5}M{{R}^{2}} \right)\,\frac{{{\nu }^{2}}}{{{R}^{2}}}=mgh\] where \[\operatorname{v} =\,\,R\omega ,\,\,\,\omega  =angular velocity\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{7}{10}m{{\nu }^{2}}=mgh\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\nu =\sqrt{\frac{10}{7}}\,gh\] Hence, to climb the inclined surface velocity should be greater than \[\sqrt{\frac{10}{7}}\,gh\].