Question

2)The rotor on a helicopter turns at an angular velocity of 3.2x102 revolutions per minute. If the rotor has a radius of 2.00 m, what arclength does the tip of the blade trace out in 3.00x102 s?

Answer 1

The angular displacement of the rotor in the given time is,

$\Delta\theta=\omega t$

Substitute the known values,

$\begin{gathered} \Delta\theta=(3.2\times10^2\text{ rev/min)(}\frac{2(3.14)\text{ rad}}{1\text{ rev}})(\frac{1\text{ min}}{60\text{ s}})(3.00\times10^2\text{ s)} \\ =10050\text{ rad} \end{gathered}$

The arc length traced by tip of the blade is,

$l=r\Delta\theta$

Substitute the known values,

$\begin{gathered} l=(2.00\text{ m)(10050 rad)} \\ =20100\text{ m} \end{gathered}$

Thus, the arc length traced by the tip of blade is 20100 m.