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A thin rod of mass `m` and length `2L` is made to rotate about an axis passing through its center and perpendicular to it. If its angular velocity changes from `O` to `omega` in time `t`, the torque acting on it isA. `(mL^(2)omega)/(12t)`B. `(mL^(2)omega)/(3t)`C. `(mL^(2)omega)/(t)`D. `(4mL^(2)omega)/(6t)` |
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Answer» `I=(m(2L)^(2))/(12)=(mL^(2))/(3)` `tau Delta t=Delta L= I Delta omega` `tau t=(mL^(2))/(3)(omega-0)implies tau=(mL^(2)omega)/(3t)` |
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