A) \[\frac{5}{2}R\]
B) \[\frac{5}{2}(R-r)\]
C) \[\frac{25}{10}(R-r)\]
D) \[\frac{27}{10}R-\frac{17r}{10}\]
Correct Answer: D
Solution :
[d] The minimum velocity at P, top of the loop, should be\[v=\sqrt{g(R-r)}\] , if the sphere keeps on rolling at top \[v=\omega R\] \[mgh=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}+mg(2R-r)\] \[=\frac{1}{2}mg(R-r)+\frac{1}{2}\left( \frac{2}{5} \right)m{{R}^{2}}2{{\omega }^{2}}+mg(2R-r)\] \[=\frac{7}{10}mg(R-r)+mg(2R-r)\left[ \omega R=v=\sqrt{g(R-r)} \right]\] \[=\frac{mg}{10}(27R-17r)\,\,or\text{ }h=\frac{1}{10}(27R-17r)\]You need to login to perform this action.
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