Reference Frame and Angular Speed Related?

Question

I am given the following problem:

If an airplane propeller rotates at 2000 rev/min while the airplane flies at a speed of 480 km/h relative to the ground, what is the linear speed of a point on the tip of the propeller, at radius 1.5m, 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 was able to solve part a pretty easily by just using the formulas that relate linear speed with angular speed, but I didn’t get part b correct. I thought that the answers would be the same from both perspectives because the speed of the plane does not contribute to the rotation of the propeller, but I’m assuming that is the wrong way to think about it (because that produces the wrong answer). Can someone explain why this is the wrong approach?

The solution to this problem involved noting that "The plane’s velocity $v_p$ and the velocity of the tip $v_t$ (found in the plane’s frame of reference), in any of the tip’s positions, must be perpendicular to each other." How is this relevant to the problem?

Answer 1

The question asks the net linear velocity of the propeller tip and not just the rotational component of that velocity. The net energy of the propeller can be divided into two parts, one the rotational KE because of the angular speed you found in part (a)($\frac12 I \omega^2$), and another the translational kinetic energy due to the translational speed of the centre of mass ($\frac12 m v_p^2$). This two divisionas also hold true for the velocities, i.e. $v_{rot}=\omega r$ and $v_{trans}=v_p$. Therefore, the $v_r$ component stays the same in both the frames but the net velocit, which is the resultant (vector resultant) of both the components changes in the ground frame. Since the net velocity is asked, you have to present the resultant and not just the rotational.

$$\vec v_{net}=\vec{v_p} +\vec{v_{rot}}$$ and $|\vec v_{net}|=\sqrt{v_p^2+v_r^2}$.