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Schu
Aug12-04, 12:54 PM
Here's the question:
the rotor of a helicopter starts from REST and reaches it's operating speed in 7.28s If the angular acceleration of the rotor is 92.5 rad/s^2, what is the angular displacement during that time?

I'm not sure where to start.
If I multiply the time by the acceleration I would get the velocity only in rad/s.
But then I was thinking more like:
Angular position= 1/2 (angular acceleration) *t^2+(initial angular velocity)*t+(initial angular position)

If I plug them into the angular position formula I get 2451.176 rad

Is the angular displacement a fancy way to say angular position?? :confused:
Is this the right track to be on?

arildno
Aug12-04, 01:23 PM
It's definitely the right track!
(displacement=new position minus old position, so by assuming old angular position 0, the angular displacement equals the new angular position)