Solved Problem on Circular Motion
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Two pulleys connected by a belt have radii R1 = 10 cm and R2 = 20 cm.
The first makes 40 rpm. Assuming that the connection belt is not elastic and there is no slip,
Determine:
a) What is the ratio between the speeds of a point on the surface of the first pulley P1 and a point on the surface of the second pulley P2?
b) What is the ratio between frequencies of the pulleys?
c) What is the number of revolutions of the second pulley?
d) What is the angular velocity of each pulley?
a) What is the ratio between the speeds of a point on the surface of the first pulley P1 and a point on the surface of the second pulley P2?
b) What is the ratio between frequencies of the pulleys?
c) What is the number of revolutions of the second pulley?
d) What is the angular velocity of each pulley?
Problem Data:
- Radius of the first pulley: R1 = 10 cm;
- Frequency of the first pulley: f1 = 40 rpm;
- Radius of the second pulley: R2 = 20 cm.
a) The problem says that the belt between the pulleys is non-elastic and there is no slip, so the speed is the same for all points of the belt and the external points of the pulleys. Thus v1 is the speed of the first pulley, v2 the speed of the second pulley, and v1=v2, the relationship between the speeds will be
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\frac{v_{1}}{v_{2}}=1}
\end{gather}
\]
b) The speed as a function of angular velocity and radius is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=\omega r}
\end{gather}
\]
the speeds of points P1 and P2 will be given, respectively, by
\[
\begin{gather}
v_{1}=\omega _{1}R_{1}
\end{gather}
\]
\[
\begin{gather}
v_{2}=\omega _{2}R_{2}
\end{gather}
\]
as item (a) we saw that v1 = v2
\[
\begin{gather}
\omega _{1}R_{1}=\omega _{2}R_{2}
\end{gather}
\]
the angular velocity ω is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\omega =2\pi f}
\end{gather}
\]
substituting, we can rewrite
\[
\begin{gather}
\cancel{2\pi} f_{1}R_{1}=\cancel{2\pi} f_{2}R_{2}\\[5pt]
f_{1}R_{1}=f_{2}R_{2}\\[5pt]
\frac{f_{1}}{f_{2}}=\frac{R_{2}}{R_{1}}
\end{gather}
\]
substituting the values of the radii
\[
\begin{gather}
\frac{f_{1}}{f_{2}}=\frac{20}{10}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\frac{f_{1}}{f_{2}}=2}
\end{gather}
\]
c) Using the value of f1 given in the problem and the previous expression
\[
\begin{gather}
\frac{40}{f_{2}}=2\\[5pt]
f_{2}=\frac{40}{2}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{f_{2}=20\;\text{rpm}}
\end{gather}
\]
d) For the calculation of the angular velocities of the pulleys, we have to convert the units of frequency given at revolutions per minute (rpm) to Hertz (Hz) used in the International System of Units (SI)
\[
\begin{gather}
f_{1}=40\;\frac{\text{revolutions}}{1\;\cancel{\text{min}}}\times\frac{1\;\cancel{\text{min}}}{60\;\text{s}}=\frac{2}{3}\;\text{Hz}\\[5pt]
f_{2}=20\;\frac{\text{revolutions}}{1\;\cancel{\text{min}}}\times\frac{1\;\cancel{\text{min}}}{60\;\text{s}}=\frac{1}{3}\;\text{Hz}
\end{gather}
\]
using the formula above for angular speed, for pulley 1
\[
\begin{gather}
\omega _{1}=2\pi f_{1}\\[5pt]
\omega _{1}=2\pi \times\frac{2}{3}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\omega _{1}=\frac{4}{3}\pi \;\text{rad/s}}
\end{gather}
\]
the angular velocity of pulley 2
\[
\begin{gather}
\omega _{2}=2\pi f_{2}\\[5pt]
\omega _{2}=2\pi \times\frac{1}{3}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\omega _{2}=\frac{2}{3}\pi \;\text{rad/s}}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .