A) \[\sqrt{2\,gh}\]
B) \[\sqrt{\frac{2\,mgh}{I+2mr}}\]
C) \[\sqrt{\frac{2\,mgh}{I+m{{r}^{2}}}}\]
D) \[\sqrt{\frac{2\,gh}{I+mr}}\]
Correct Answer: C
Solution :
[c] Loss of potential energy = mgh Gain of kinetic energy\[=\frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}m{{\omega }^{2}}{{r}^{2}}={{\omega }^{2}}\left[ \frac{m{{r}^{2}}+I}{2} \right]\] Now, \[{{\omega }^{2}}\left[ \frac{m{{r}^{2}}+I}{2} \right]=mgh\] \[\omega ={{\left[ \frac{2mgh}{I+m{{r}^{2}}} \right]}^{1/2}}\]
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