Question

1. a helicopter blade spins at exactly 100 revolutions per minute. its tip is 5.00 m from the center of rotation. (a) calculate the average speed of the blade tip in the helicopter’s frame of reference. (b) what is its average velocity over one revolution?

Explanation:

Step1: Convert rotational frequency to angular frequency

The rotational frequency $f = 100\ \text{rev/min}=\frac{100}{60}\ \text{rev/s}=\frac{5}{3}\ \text{rev/s}$. The angular frequency $\omega=2\pi f$. So $\omega = 2\pi\times\frac{5}{3}=\frac{10\pi}{3}\ \text{rad/s}$.

Step2: Calculate the average speed in part (a)

The formula for the linear - speed $v = r\omega$, where $r = 5.00\ m$ and $\omega=\frac{10\pi}{3}\ \text{rad/s}$. Then $v=5\times\frac{10\pi}{3}=\frac{50\pi}{3}\approx52.4\ m/s$.

Step3: Analyze the average velocity in part (b)

The displacement of the blade - tip over one revolution is $\Delta\vec{x}=\vec{0}$ (since it returns to its starting point). The average velocity $\vec{v}_{avg}=\frac{\Delta\vec{x}}{\Delta t}$. Since $\Delta\vec{x} = 0$ over one revolution, $\vec{v}_{avg}=0\ m/s$.

Answer:

(a) $\frac{50\pi}{3}\ m/s\approx52.4\ m/s$
(b) $0\ m/s$