A ceiling fan is turned on and reaches an angular speed of 132 rev/min in 20 s.It is then turned off and coasts to a stop in an additional 40 s.The ratio of the average angular acceleration for the first 20 s to that for the last 40 s is which of the following?
A)6.6
B)6.4
C)−0.15
D)−2.0

Question

Quizplus · Accepted Answer

The average angular acceleration during the first 20 seconds ($\alpha_1$) can be calculated using 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. The angular speed of 132 rev/min is equivalent to $132 \times 2\pi$ rad/min, or $132 \times 2\pi / 60$ rad/s, since there are $2\pi$ radians in a revolution and 60 seconds in a minute. This gives an initial angular acceleration of $\frac{132 \times 2\pi / 60}{20}$.For the deceleration phase ($\alpha_2$), the fan stops in 40 seconds from the same speed, so the average angular deceleration is $\frac{-132 \times 2\pi / 60}{40}$, negative because it's a deceleration.The ratio of the average angular acceleration for the first 20 s to that for the last 40 s is $\frac{\alpha_1}{\alpha_2} = \frac{\frac{132 \times 2\pi / 60}{20}}{\frac{-132 \times 2\pi / 60}{40}} = \frac{40}{-20} = -2.0$.

A Ceiling Fan Is Turned on and Reaches an Angular