A hemisphere of radius R and of mass \[4\,\,m\]is free to slide with its base on a smooth horizontal table. A particle of mass \[m\] is placed on the top of the hemisphere. The angular velocity of the particle relative to centre of hemisphere at an angular displacement \[\theta \] when velocity of hemisphere has become \[v\] is - |
A) \[\frac{5v}{R\cos \theta }\]
B) \[\frac{2v}{R\cos \theta }\]
C) \[\frac{3v}{R\sin \theta }\]
D) \[\frac{5v}{R\sin \theta }\]
Correct Answer: B
Solution :
Let be the velocity of particle relative to hemisphere and v the linear velocity of hemisphere at this moment. Then from conservation of linear momentum, we have |
\[4mv=m\,({{v}_{r}}\cos \theta -v)\]\[or\,\,5v={{v}_{r}}\cos \theta \] \[or\,\,{{v}_{r}}=\frac{5v}{\cos \theta }\] |
\[\therefore \,\,\omega =\frac{{{v}_{r}}}{R}=\frac{5v}{R\cos \theta }\] |
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