An electric ceiling fan is rotating about a fixed axis with an initial angular v

Question

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.260 rev/s . The magnitude of the angular acceleration is 0.883 rev/s2 . Both the the angular velocity and angular accleration are directed clockwise. The electric ceiling fan blades form a circle of diameter 0.790 m .

Compute the fan's angular velocity magnitude after time 0.193 s has passed

Through how many revolutions has the blade turned in the time interval 0.193 s from Part A?

What is the tangential speed vtan(t) of a point on the tip of the blade at time t = 0.193 s ?

What is the magnitude a of the resultant acceleration of a point on the tip of the blade at time t = 0.193 s ?

Explanation / Answer

1) Use the formula v = vi + at:

v = vi + at
v = 0.260 + 0.883 * 0.193
v = 0.43 rev/s

2) Since we found the ending velocity in Part A, we can use the following formula. The unrounded ending speed was 0.43 rev/s:

vf^2 = vi^2 + 2a(x)
0.43^2 = 0.260^2 + 2 * 0.883 * x
0.1173 = 1.766x
x = 0.0664 rev

3) Since the diameter of the fan is 0.790 m, it’s radius is 0.395 m. This gives a circumference of (c = 2r) 2.48 m. The blade is moving at 0.43 revolutions per second, so the tangential speed is 2.48 * 0.43 = 1.07 m/s

4) The total acceleration includes the tangential and centripetal accelerations:

= sqrt(a(tan)^2 + a(cent)^2)

Using the tangential velocity from Part C, we can calculate the centripetal acceleration (a = v^2/r) as 2.9 m/s^2. The tangential acceleration is found by converting the angular acceleration to m/s^2 (similar to what we did in Part C for velocity). The tangential acceleration is therefore: a(tan) = 2r * a(angular), or a(tan ) = 2.48 * 0.883, which equals 2.2 m/s^2. Now combine:

= sqrt(a(tan)^2 + a(cent)^2)
= sqrt(2.2^2 + 2.9^2)
= 3.64 m/s^2

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