Non-u.c.m. in the blades of a fan

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: t₂=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:

$${\omega }_0=\frac{2·\pi }{60}·140=14.66rad/s$$

Resolution

Section 1

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

$$\omega ={\omega }_0+\alpha ·t\Rightarrow 0=14.66+\alpha ·25\Rightarrow \alpha =-0.58rad/s^2$$ 

Section 2

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

$$\phi ={\phi }_0+{\omega }_0·t+\frac{1}{2}·\alpha ·t^2\Rightarrow \phi =\cancel{{\phi }_0}+14.66·25+\frac{1}{2}·\left(-0.58\right)·{25}^2=185.25rad$$

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

$$\frac{185.25}{2·\pi }=29.48$$

Section 3

$$\begin{gathered} \omega ={\omega }_0+\alpha ·t\Rightarrow \omega =14.66-0.58·15=5.96 rad/s; \\ v=\omega ·R=5.96·0.3=1.788m/s \end{gathered}$$

$$a_t=\alpha ·R=-0.58·0.3=-0.174m/s^2$$

$$a_n={\omega }^2·R={5.96}^2·0.3=10.65m/s^2$$

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

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