question_answer

A solid sphere and a hollow sphere of equal mass and radius, are placed over a rough horizontal surface after rotating it about its mass centre with same angular velocity\[{{\omega }_{0}}\]. Once the pure rolling starts let \[{{\nu }_{1}}\,\,and\,\,{{\nu }_{2}}\] be the linear speeds of their centres of mass. Then

A) \[{{\nu }_{1}}={{\nu }_{2}}\]

B)   \[{{\nu }_{1}}>{{\nu }_{2}}\]

C) \[{{\nu }_{1}}<{{\nu }_{2}}\]             

D) data is insufficient

Correct Answer: C

Solution :

From conservation of angular momentum about point of contact. \[I{{\omega }_{0}}\,=\,I\omega +mR\omega \] \[I{{\omega }_{0}}\,=\,I\frac{{{\nu }^{2}}}{R}+\,\,mR\nu \] \[\nu =\frac{I{{\omega }_{0}}}{\frac{I}{R}+mR}\]2 \[\nu =\frac{{{\omega }_{0}}}{\frac{1}{R}+\frac{mR}{l}}\] \[Now\,\,{{I}_{solid\,\,sphere}}\,<\,{{I}_{hollow}}\] \[{{\nu }_{solid}}\,<{{\nu }_{hollow}}\] \[{{\nu }_{1}}\,<{{\nu }_{2}}\]