InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A wheel of moment of inertia `0.500 kg-m^2` and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel. |
|
Answer» Here, `I_(1) = 0.5 kg m^(2), r = 20 cm = (20)/(100)m, omega_(1) = 20.0 rad//s, omega_(2) = ?` As `I_(1) = (1)/(2)m_(1)r_(1)^(2) :. m_(1) = (2I_(1))/(r_(1)^(2)) = (2 xx 0.5)/((1 xx 5)^(2)) = 25 kg` Mass picked up, `m_(2) = 200g = 0.2 kg` New moment of inertia `I_(2) = (1)/(2)m_(1)r^(2) + m_(2)r^(2) = (1)/(2)(m_(1) + m_(2))r^(2) = (1)/(2) (25 + 0.2) ((1)/(5))^(2) = (25.2)/(50) kg m^(2)` According to the principle of conservation of angular momentum, `I_(2) omega_(2) = I_(1) omega_(1)` `omega_(2) = (I_(1) omega_(1))/(I_(2)) = (0.5 xx 20 xx 50)/(25.2) = 19.84 rad//s` |
|