## Statement

A fan whose blades measure 30cm is located in the roof spinning at 140 r.p.m. A blackout makes the fan stops, after 25 seconds. Calculate:

- Angular acceleration
- Distance traveled by the end of a blade until it stops, and the number of rotations completed
- The value of the linear velocity, tangential, normal and full acceleration at 15 seconds after the power outage

## Solution

**Data**

- Length of the blade:
*R = 30 cm = 0.3 m* - Number of revolutions per minute:
*140* - Time to full stop
*t*_{f}= 25 s - Another time considered:
*t*_{2}=15 s

**Preliminary consideratioins**

We are given as data the number of revolutions per minute, that is the initial angular velocity. We need to make the conversion of revolutions per minute to radians per second knowing that a revolution is 2·π radians and one minute is 60 seconds:

$${\omega}_{0}=\frac{2\xb7\pi}{60}\xb7140=14.66rad/s$$

**Resolution**

**Section 1**

We know that the final angular velocity is 0, so:

$$\omega ={\omega}_{0}+\alpha \xb7t\Rightarrow 0=14.66+\alpha \xb725\Rightarrow \alpha =-0.58rad/{s}^{2}$$

**Section 2**

We are being asked for the final angular position (as number of revolutions):

$$\phi ={\phi}_{0}+{\omega}_{0}\xb7t+\frac{1}{2}\xb7\alpha \xb7{t}^{2}\Rightarrow \phi =\overline{){\phi}_{0}}+14.66\xb725+\frac{1}{2}\xb7\left(-0.58\right)\xb7{25}^{2}=185.25rad$$

We make the conversion by dividing the revolutions by 2·π:

$$\frac{185.25}{2\xb7\pi}=29.48$$

**Section 3**

$$\omega ={\omega}_{0}+\alpha \xb7t\Rightarrow \omega =14.66-0.58\xb715=5.96rad/s;\phantom{\rule{0ex}{0ex}}v=\omega \xb7R=5.96\xb70.3=1.788m/s$$

$${a}_{t}=\alpha \xb7R=-0.58\xb70.3=-0.174m/{s}^{2}$$

$${a}_{n}={\omega}^{2}\xb7R={5.96}^{2}\xb70.3=10.65m/{s}^{2}$$

Finally, we can determine the value of the total acceleration (scalar), according to:

$$a=\sqrt{{{a}_{n}}^{2}+{{a}_{t}}^{2}}=\sqrt{{10.65}^{2}+{\left(-0.174\right)}^{2}}=10.651m/{s}^{2}$$