**An Electric Ceiling Fan**

Part A = 4.26 * 10^37 kg

Part B = 2.14 * 10^7

Part C = no

Part D = 6.31 * 10^10 m

Part E = yes

**Solution Below:**

Part A

Express your answer numerically in revolutions per second.

Use the formula v = v_{i} + at:

v = v_{i} + at

v = 0.280 + 0.917 * 0.201

v = 0.464317 rev/s (unrounded)

v = 0.464 rev/s

0.464 rev/s

Part B

Express the number of revolutions numerically.

Since we found the ending velocity in Part A, we can use the following formula. The unrounded ending speed was 0.464317 rev/s:

vf^2 = vi^2 + 2a(Δx)

0.464^2 = 0.280^2 + 2 * 0.917 * x

0.2156 = 0.0784 + 1.834x

0.1372 = 1.834x

x = 0.0748 rev

0.0746 rev

Part C

_{tan}(t) of a point on the tip of the blade at time = 0.201 s?

Express your answer numerically in meters per second.

Since the diameter of the fan is 0.700 m, it’s radius is 0.350 m. This gives a circumference of (c = 2πr) 2.20 m. The blade is moving at 0.464 revolutions per second, so the tangential speed is 2.20 * 0.464 = 1.02 m/s

1.02 m/s

Part D

The total acceleration includes the tangential and centripetal accelerations:

α = 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 2.9726 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) = 2πr * a(angular), or a(tan ) = 2.20 * 0.917, which equals 2.0174 m/s^2. Now combine:

α = sqrt(a(tan)^2 + a(cent)^2)

α = sqrt(2.0174^2 + 2.9726^2)

α = sqrt(4.07 + 8.8209)

α = 3.59 m/s^2

3.59 m/s^2