A) \[2\omega \]
B) \[4\omega \]
C) \[6\omega \]
D) \[8\omega \]
Correct Answer: D
Solution :
Given: Mass of sphere\[=m\] Initial radius of path\[{{r}_{1}}=r\] Final radius of path \[{{r}_{2}}=\frac{r}{2}\] The initial M.I. of a particle of mass \[m\]rotating m a horizontal circle is given by \[=mr_{1}^{2}=m{{r}^{2}}\] Similarly, the final M.I. of a particle rotating in a circle is \[=mr_{2}^{2}=m{{\left( \frac{r}{2} \right)}^{2}}\] \[=\frac{m{{r}^{2}}}{4}\] Applying law of conservation of momentum \[m{{r}^{2}}\times 2\omega =\frac{m{{r}^{2}}}{4}\times {{\omega }_{2}}\] \[\therefore \] \[{{\omega }_{2}}=4\times 2\omega =8\omega \]You need to login to perform this action.
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