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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?

User Dan Powley
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1 Answer

6 votes

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*10^2\text{ rev/min)(}\frac{2(3.14)\text{ rad}}{1\text{ rev}})(\frac{1\text{ min}}{60\text{ s}})(3.00*10^2\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.

User Jeffery Hicks
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