A) \[\sqrt{\frac{\mu g}{r}}\]
B) \[\sqrt{\frac{2\mu g}{2}}\]
C) \[\sqrt{\frac{\mu g}{2r}}\]
D) \[2\sqrt{\frac{\mu g}{r}}\]
Correct Answer: A
Solution :
In absence of any sliding, net force on the cube in the frame of reference rotating with disc, will be zero. We find two forces in the plane of disc-frictional force and centrifugal force. Hence, \[M{{w}^{2}}f=1\] But \[f\le \mu \,mg\] Hence, \[w\le \sqrt{\mu g/r}\] \[\Rightarrow \] \[w\le \sqrt{\mu g/r}\] \[\Rightarrow \] \[{{w}_{\max }}=\sqrt{\frac{\mu g}{r}}\]You need to login to perform this action.
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