InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A man stands on a rotating platform with his arms stretched holding a `5 kg` weight in each hand. The angular speed of the platform is `1.2 rev s^-1`. The moment of inertia of the man together with the platform may be taken to be constant and equal to `6 kg m^2`. If the man brings his arms close to his chest with the distance `n` each weight from the axis changing from `100 cm` to`20 cm`. The new angular speed of the platform is.A. `2 rev s^(-1)`B. `3 rev s^(-1)`C. `5 rev s^(-1)`D. `6 rev s^(-1)` |
|
Answer» Correct Answer - b Initial moment of inertia, `I_(1) = 6 +2 xx 5 xx (1)^(2)= 16 kg m^(2)` Initial angular velocity, `omega_(1) = 1.2 rev s^(-1)` Initial angular momentum, `L_(1) = I_(1)omega_(1)` Final moment of inertia, `I_(2) = 6+2 xx 5 xx(0.2)^(2) = 6.4 kg m^(2)` Final angular momentum , `L_(2) = I_(2)omega_(2)` According to law of conservation of angualr momentum, `L_(1) = L_(2)` `omega_(2) = (I_(1)omega_(1))/(I_(2)) = 16 kg m^(2)(1.2 rev s^(-1))/(6.4 kg ms^(2)) = 3 rev s^(-1)` |
|