A) 2 \[\omega \]
B) \[\omega \]
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 in a horizontal circle is given by \[=mr_{1}^{2}=m{{r}^{2}}\] Similarly, the final M.L 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.
You will be redirected in
3 sec
