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An electric fan is turned off, and its angular velocity decreases uniformly from 570 rev/min to 190 rev/min in 4.50 s . Find the angular acceleration in rev/s2 . Express your answer in revolutions per second squared. Find the number of revolutions made by the motor in the 4.50 s interval. Express your answer as a number of revolutions. How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in ? Express your answer in seconds.

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Final answer:

The angular acceleration of the fan is -1.41 rev/s², and during the 4.50 s interval, the fan makes approximately 28.575 revolutions. It takes an additional 2.24 seconds for the fan to come to rest if the angular acceleration remains constant.

Step-by-step explanation:

To calculate the angular acceleration, we can use the formula \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the change in time. First, convert the angular velocities from revolutions per minute (rev/min) to revolutions per second (rev/s) by dividing by 60. This gives us:

Initial angular velocity, \( \omega_i = \frac{570}{60} = 9.5 \) rev/s
Final angular velocity, \( \omega_f = \frac{190}{60} = 3.17 \) rev/s
Time interval, \( \Delta t = 4.50 \) s

Now apply the equation to find angular acceleration:

\( \alpha = \frac{3.17 - 9.5}{4.50} = -1.41 \) rev/s2

To find the number of revolutions made, use the equation \( \theta = \omega_i \cdot t + 0.5 \cdot \alpha \cdot t^2 \):

\( \theta = 9.5 \cdot 4.50 + 0.5 \cdot (-1.41) \cdot (4.50)^2 \) which calculates to \( \theta = 28.575 \) revolutions.

For the fan to come to rest, we need to find the time when angular velocity is zero (\( \omega_f = 0 \)). Use \( \omega_f = \omega_i + \alpha \cdot t \) and solve for \( t \):

\( 0 = 9.5 + (-1.41) \cdot t \) which gives \( t = \frac{9.5}{1.41} \approx 6.74 \) seconds. However, this is the total time from the initial velocity to rest. Because it already took 4.50 s to slow to 190 rev/min, the additional time required is \( 6.74 - 4.50 \approx 2.24 \) seconds.

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