If an airplane propeller rotates at $\mathbf{\text{2000\hspace{0.17em}rev/min}}$while the airplane flies at a speed of$\mathbf{\text{480km/h}}$ relative to the ground, what is the linear speed of a point on the tip of the propeller, at radius $\mathbf{\text{1.5\hspace{0.17em}m}}$, as seen by (a) the pilot and (b) an observer on the ground? The plane’s velocity is parallel to the propeller’s axis of rotation.

i) Frequency of propeller, $f=2000\text{\hspace{0.17em}rev}/\text{min}$

ii) Speed of airplane, $v=480\text{\hspace{0.17em}km}/\text{hr}$

iii) Radius, $r=1.5\text{\hspace{0.17em}m}$

From the given frequency of the propeller, we can find the angular velocity of the propeller. Then, by using the relation between angular velocity and linear velocity we can find the linear speed of a point on the tip of the propeller, at radius$\text{1.5\hspace{0.17em}m,}$as seen by the pilot and as seen by an observer on the ground.

Formulae are as follow:

$\omega =2\pi f$

$v_r=r\omega$

$v=\sqrt{v_r^2+v_g^2}$

Where, v is velocity, f is frequency and $\omega$ is an angular frequency.

For angular velocity of the propeller,

$\omega =2\pi f$

$\omega =\left(2000\frac{\text{rev}}{\text{min}}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\text{\hspace{0.17em}min}}{60\text{\hspace{0.17em}s}}\right)$

$\omega =209\frac{\text{rad}}{\text{s}}$

Now,

$v_r=r\omega$

$v_r=\left(1.5\text{\hspace{0.17em}m}\right)\left(209\frac{\text{rad}}{\text{s}}\right)$

$v_r=313.5\frac{\text{m}}{\text{s}}$

$v_r=3.1\times {10}^2\frac{\text{m}}{\text{s}}$

As speed of any point on the propeller is normal to the direction of the observation of the pilot, the pilot observes the vertical velocity$3.13\times {10}^2\text{m}/\text{s}$normal to these points.

Therefore, the linear speed of a point on the tip of the propeller, at radius$1.5\text{\hspace{0.17em}m},$as seen bythe pilot is $3.1\times {10}^2\text{\hspace{0.17em}m}/\text{s}$.

The given relative speed of the airplane is,

$v_g=480\frac{\text{km}}{\text{hr}}$

$v_g=480\frac{\text{km}}{\text{hr}}\left(\frac{1000\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}km}}\right)\left(\frac{1\text{\hspace{0.17em}hr}}{3600\text{\hspace{0.17em}s}}\right)$

$v_g=133\frac{\text{m}}{\text{s}}$

$v_g=1.33\times {10}^2\frac{\text{m}}{\text{s}}$

Therefore,

$v=\sqrt{v_r^2+v_g^2}$

$v=\sqrt{{(3.13\times {10}^2\frac{\text{m}}{\text{s}})}^2+{(1.33\times {10}^2\frac{\text{m}}{\text{s}})}^2}$

$v=3.4\times {10}^2\frac{\text{m}}{\text{s}}$

Hence, the linear speed of a point on the tip of the propeller, at radius$1.5\text{\hspace{0.17em}m},$as seen by an observer on the ground is$3.4\times {10}^2\text{\hspace{0.17em}m}/\text{s}$.

Therefore, the linear speed of a point on the tip of the propeller can be found, at a given radius, as seen bythe pilot using the relation between frequency and angular velocity of the propeller. Also, the linear speed of a point on the tip of the propeller can be found at a given radius as seen by an observer from the ground.