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A small bead of mass `m` moving with velocity `v` gets threaded on a stationary semicircular ring of mass `m` and radius `R` kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system? A. `v/R`B. `(2v)/R`C. `v/(2R)`D. `(3v)/R` |
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Answer» Correct Answer - C Using conservation of angular momentum about `O` we get `mvR=(mR^(2)+mR^(2))omega, omega=v/(2R)` |
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