Solved Problem on Circular Motion

An airplane propeller has blades 2 m in length and rotates with a frequency of 1200 rpm. Calculate:
a) The frequency in hertz;
b) The period of rotations;
c) The angular speed of the propeller;
d) The speed of a point situated at the tip of one of the propeller blades;
e) The magnitude of the centripetal acceleration.

Problem data:

Length of the propeller (radius of the circle): r = 2 m;
Frequency of rotation: f = 1200 rpm.

Problem diagram:

Taking a point at the tip of the propeller, we have the velocity, \( \vec{v} \) , it will be tangent to the trajectory, and the propeller rotates with angular speed ω (Figure 1).

Figure 1

Solution

a) Converting the frequency given in rpm to hertz

\[ \begin{gather} f=1200\frac{\text{rotations}}{1\;\mathrm{\cancel{min}}}\times\frac{1\;\mathrm{\cancel{min}}}{60\;\mathrm{s}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {f=20\;\mathrm{Hz}} \end{gather} \]

Note: rotations is a dimensionless number, which is why it does not appear in the final result.

b) The period as a function of frequency will be

\[ \begin{gather} \bbox[#99CCFF,10px] {T=\frac{1}{f}} \end{gather} \]

\[ \begin{gather} T=\frac{1}{20\;\mathrm{Hz}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T=0.05\;\mathrm s} \end{gather} \]

c) The angular velocity as a function of frequency is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\omega =2\pi f} \end{gather} \]

using the frequency given in the problem

\[ \begin{gather} \omega=2\times\pi\times\left(20\;\mathrm{Hz}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\omega=40 \pi \;\mathrm{rad/s}} \end{gather} \]

Note: we can obtain the angular speed using the period calculated in item (b). The expression for the angular speed as a function of period will be

\[ \begin{gather} \omega=\frac{2\pi}{T}\\[5pt] \omega=\frac{2\pi}{0.05\;\mathrm s}=40\pi\;\mathrm{rad/s} \end{gather} \]

d) Letting \( \pi=3.14 \) the speed will be

\[ \begin{gather} \bbox[#99CCFF,10px] {v =\omega r} \end{gather} \]

\[ \begin{gather} v=\left(40\times 3.14\;\mathrm{\frac{rad}{s}}\right)\times\left(2\;\mathrm m\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v \approx 251\;\mathrm{m/s}} \end{gather} \]

e) The magnitude of centripetal acceleration is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {|\;a_{cp}\;|=\frac{v^2}{r}} \end{gather} \]

\[ \begin{gather} |\;a_{cp}\;|=\frac{\left(251\;\mathrm{\frac{m}{s}}\right)^2}{2\;\mathrm m}\\[5pt] |\;a_{cp}\;|=\frac{63001\;\mathrm{\frac{m^{\cancel 2}}{s^2}}}{2\;\mathrm{\cancel m}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {|\;a_{cp}\;|\approx 31501\;\mathrm{m/s^2}} \end{gather} \]

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