A ceiling fan is rotating at 1.00 rev/s. When turned off, it slows uniformly to a stop in 10 s.
(a) How many revolutions does it make in this time?
(b) Using the result from part (a), find the number of revolutions the fan must make for its speed to decrease from 1.00 rev/s to 0.50 rev/s.
2007-12-01 11:59:48 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
2 α ( θ - θo) = ω^2 - ωo^2
where
θ - θo is the angular displacement ( what you're looking for )
ωo is the initial angular velocity ( 2π radians / sec )
ω is the final angular velocity ( zero )
α is the angular acceleration ( unknown )
We also know that
ω = ωo + α t
where t is the elapsed time ( 10 s )
Solve this equation for the angular acceleration, then use the first to find the displacement. Divide by 2π to get the number of revolutions.
Part (b) is solved using the same equations and technique.
2007-12-01 12:06:44 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
Turning at 1 revolution per second. When turned off, it slows UNIFORMLY (linearly) to stop in 10seconds. So rps=1-t/10. At
t=0s rps=1, at t=10s rps=1-10/10=0.
Integrate rpsdt from 0 to 10 seconds=
t-tt/20 from 0 to 10 = 10-100/20=5 revs a)
b. find t for rps=.5=1-t/10 which gives t=5s. Evaluating the integral from t=0 to t=5
5-25/20=3.75 revolutions.
2007-12-01 12:46:31 · answer #2 · answered by oldschool 7 · 0⤊ 0⤋
why the heck would i need to know that in life do your home work stuiped
2007-12-01 12:04:20 · answer #3 · answered by Anonymous · 0⤊ 2⤋
I know, but I aint gonna tell ya. Stupid.
2007-12-01 12:02:44 · answer #4 · answered by roadjack1096 2 · 0⤊ 2⤋