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A mass `m` moves in a circles on a smooth horizontal plane with velocity `v_(0)` at a radius `R_(0)`. The mass is atteched to string which passes through a smooth hole in the plane as shown. The tension in string is increased gradually and finally `m` moves in a cricle of radius `(R_(0))/(2)`. the final value of the kinetic energy is A. `(1)/(4) mv_(0^(2)`B. `2 mv_(0)^(2)`C. `(1)/(2) mv_(0^(2)`D. `m_(0)^(2)` |
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Answer» Angular momentum conservation `mv_(0) R_(0) = mv (R_(0))/(2) implies v = 2 v_(0)` `K_(f) = (1)/(2) mv^(2) = (1)/(2) m (2 v_(0))^(2)` `= 2 mv_(0)^(2)` |
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