Ceiling Fan Velocity
March 11, 2011
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Posted by admin
tough physics question?
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity of 0.220 rev/s. The angular acceleration is 0.905 rev/s^2. Its blades form a circle of diameter 0.770 m.
Part A) Compute the angular velocity of the fan after time 0.192 s has passed?
Part B) Through how many revolutions has the blade turned in the time interval 0.192 s from Part A?
Part C) What is the tangential speed v_tan(t) of a point on the tip of the blade at time t = 0.192 s?
Part D) What is the magnitude of the resultant acceleration (a) of a point on the tip of the blade at time t = 0.192 s?
A) ω_f = ω_i + αt
ω_f = final angular velocity
ω_i = initial angular velocity = 0.220 rev/s
α = angular acceleration = 0.905 rev/s²
t = time = 0.192s
ω_f = (0.220 rev/s) + (0.905 rev/s²)(0.192s)
ω_f = 0.394 rev/s
B) Δθ = ωt + (1/2)αt²
Δθ = change in angle
Δθ = (0.220 rev/s)(0.192 s) + (1/2)(0.905 rev/s²)(0.192 s)²
Δθ = 0.059 rev
C) v = rω
v = velocity
r = radius = 0.77/2 = 0.385m
We already have ω from part A, but for the formula to work it has to be in rad/s instead of rev/s. To do that just multiply by 2π.
v = (0.385 m)(0.394 rev/s)(2π rad/rev)
v = 0.953 m/s
D) a_t = rα
a_t = tangential acceleration
a_n = rω²
a_n = normal acceleration
First find tangential acceleration. Again, it has to be in rad/s² so multiply by 2π:
a_t = (0.385m)(0.905 rev/s²)(2π rad/rev)
a_t = 2.189 m/s²
Then find normal (with the same rev→rad deal):
a_n = (0.385 m)[(0.394 rev/s)(2π rad/rev)]²
a_n = 2.359 m/s²
Finally, use the pythagorean theorem to find the resultant:
a = √[(a_n)² + (a_t)²]
a = √(2.359 m/s²)² + (2.189 m/s²)²]
a = 3.22 m/s²
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Tags: ceiling fan velocity, ceiling_fans, fan, fans, floor, home
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