question_answer

A small steel sphere of mass m is tied to a string of length r and is whirled in a horizontal circle with a uniform angular velocity 2\[\omega \]  The string is suddenly pulled, so that radius of the circle is halved. The new angular velocity will be

A)  2\[\omega \]                                   

B)  4\[\omega \]

C)   6\[\omega \]                                  

D)  8\[\omega \]

Correct Answer: D

Solution :

                 When no external torque is acting upon a body rotating about an axis, then angular momentum of the body remains constant.                                 Therefore, \[J=I\omega =\] constant where              \[I=m{{r}^{2}}\]                (moment of inertia) \[\therefore \]  \[{{m}_{1}}r_{1}^{2}\omega ={{m}_{2}}r_{2}^{2}\,{{\omega }_{2}}\] \[\Rightarrow \]               \[{{\omega }_{2}}=\frac{{{m}_{1}}\,r_{1}^{2}\,{{\omega }_{1}}}{{{m}_{2}}\,r_{2}^{2}}\] Given,   \[{{m}_{1}}=m\],       \[{{r}_{1}}=r\],      \[{{\omega }_{1}}=2\omega \]                 \[{{m}_{2}}=m,\,{{r}_{2}}=\frac{r}{2}\] \[\therefore \]  \[{{\omega }_{2}}=\frac{m{{r}^{2}}(2\,\omega )}{m\,{{(r/2)}^{2}}}=\frac{m{{r}^{2}}}{m{{r}^{2}}}8\omega \] \[\Rightarrow \]               \[{{\omega }_{2}}=8\omega \]