4.2: Circular Motion and Dimensional Analysis
- Page ID
- 961
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Converting between units is essential for mathematics and science in general. Radians are very powerful because they provide a link between linear and angular speed. One radian is an angle that always corresponds to an arc length of one radius. This will allow you to convert the rate at which you pedal a bike to the actual speed you can travel.
1 revolution \(=2 \pi r\)
The gear near the pedals on a bike has a radius of 5 inches and spins once every second. It is connected by a chain to a second gear that has a 3 inch radius. If the second wheel is connected to a tire with a 17 inch radius, how fast is the bike moving in miles per hour?
Circular Motion and Dimensional Analysis
Dimensional analysis just means converting from one unit to another. Sometimes it must be done in several steps in which case it is best to write the original amount on the left and then multiply it by all the different required conversions. To convert 3 miles to inches you write:
\(\frac{3 \text { mile }}{1} \cdot \frac{5280 \text { feet}}{1 \text { mile }} \cdot \frac{12 \text { inches }}{1 \text { foot }}=\frac{3 \cdot 5280 \cdot 12 \text { inches}}{1}=190080\) inches
Notice how miles and feet/foot can be used to convert miles to the desired unit of inches. Units, like numbers, can resolve to one when they exist in both the numerator and denominator \(\left(\frac{3}{3}=1\right.\) and \(\left.\frac{f e e t}{f e e t}=1\right) .\) Also notice each conversion factor is the same distance on the numerator and denominator, just written with different units.
Circular motion refers to the fact that on a spinning wheel points close to the center of the wheel actually travel very slowly and points near the edge of the wheel actually travel much quicker. Angular speed is the ratio of revolutions that occur per unit of time and linear speed is the ratio of distance per unit of time. While the two points have the same angular speed, their linear speed is very different.
Examples
Earlier, you were given the following question: The gear near the pedals on the bike has radius 5 inches and spins once every second. It is connected by a chain to a second gear that has a 3 inch radius. If the second wheel is connected to a tire with a 17 inch radius, how fast is the bike moving in miles per hour?
A bike has pedals that rotate a gear at a circular speed. The gear translates this speed to a linear speed on the chain. The chain then moves a second gear, which is a conversion to angular speed for the rear tire. This tire then converts the angular speed back to linear speed which is how fast you are moving. Instead of doing all these calculations in one step, it is easier to do each conversion in small pieces.
First convert the original gear into the linear speed of the chain.
\(\frac{1 \text { revolution}}{1 \text { second }} \cdot \frac{2 \pi \cdot 5 \text { inches }}{1 \text { revolution }}=10 \pi \frac{i n}{\sec }\)
Then convert the speed of the chain into angular speed of the back gear which is the same as the angular speed of the rear tire.
\(\frac{10 \pi \text { inches}}{1 \text { second }} \cdot \frac{1 \text { revolution}}{2 \pi \cdot 3 \text { inches}}=\frac{10}{6} \frac{\text {rev}}{\text {sec}}\)
Lastly convert the angular speed of the rear tire to the linear speed of the tire in miles per hour.
\(\frac{10 \mathrm{rev}}{6 \mathrm{sec}} \cdot \frac{2 \pi \cdot 17 \mathrm{in}}{1 \mathrm{rev}} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}} \cdot \frac{1 \mathrm{mile}}{5280 \mathrm{ft}} \cdot \frac{60 \mathrm{sec}}{1 \mathrm{~min}} \cdot \frac{60 \mathrm{~min}}{1 \mathrm{hour}}\)
\(=\frac{10 \cdot 2 \cdot \pi \cdot 17 \cdot 60 \cdot 60}{6 \cdot 12 \cdot 5280} \frac{\mathrm{miles}}{\text {hour}}\)
\(\approx 10.1 \frac{\text {miles}}{\text {hour}}\)
Suppose Summit High School has a circular track with two lanes for running. The interior lane is 30 meters from the center of the circle and the lane towards the exterior is 32 meters from the center of the circle. If two people run 4 laps together, how much further does the person on the outside go?
Calculate the distance each person ran separately using 1 lap to be 1 circumference and find the difference.
\(\frac{4 \text { laps }}{1} \cdot \frac{2 \pi \cdot 30 \text { meters }}{1 \text { lap }} \approx 754\) meters
\(\frac{4 \text { laps }}{1} \cdot \frac{2 \pi \cdot 32 \text { meters }}{1 \text { lap }} \approx 804\) meters
The person running on the outside of the track ran about 50 more meters.
Andres races on a bicycle with tires that have a 17 inch radius. When he is traveling at a speed of 30 feet per second, how fast are the wheels spinning in revolutions per minute?
Look for ways to convert feet to revolutions and seconds to minutes.
\(\frac{30 \text { feet}}{1 \text { second }} \cdot \frac{60 \text { seconds }}{1 \text { minute }} \cdot \frac{12 \text { inches }}{1 \text { foot }} \cdot \frac{1 \text { revolution }}{2 \pi \cdot 17 \text { inches }}=\frac{30 \cdot 60 \cdot 12 \text { revolutions}}{2 \pi \cdot 17 \text { minute }} \approx 202.2 \frac{\mathrm{rev}}{\min }\)
When a car travels at 60 miles per hour, how fast are the tires spinning if they have 30 inch diameters?
\(\frac{60 \text { miles }}{1 \text { hour }} \cdot \frac{5280 \text { feet }}{1 \text { mile }} \cdot \frac{12 \text { inches }}{1 \text { foot }} \cdot \frac{1 \text { revolution }}{2 \pi \cdot 15 \text { inches }} \cdot \frac{1 \text { hour }}{60 \text { minute }}\)
\(=\frac{60 \cdot 5280 \cdot 12 \text { revolutions }}{2 \pi \cdot 15 \cdot 60 \text { minute}} \approx 672.3 \frac{\text { rev }}{\min }\)
Mike rides a bike with tires that have a radius of 15 inches. How many revolutions must Mike make to ride a mile?
\(\frac{1 \mathrm{rev}}{2 \pi \cdot 15 \mathrm{in}} \cdot \frac{12 \mathrm{in}}{1 \mathrm{ft}} \cdot \frac{5280 \mathrm{ft}}{1 \mathrm{mi}} \approx 672.3 \frac{\mathrm{rev}}{\text {mile}}\)
Review
For \(1-10,\) use the given values in each row to find the unknown value \((x)\) in the specified units in the row.
| Problem Number | Radius | Angular Speed | Linear Speed |
| 1. | 5 inches | 60 rpm | \(x \frac{i n}{\min }\) |
| 2. | \(x\) feet | 20 rpm | \(2 \frac{i n}{s e c}\) |
| 3. | 15 cm | \(x\) rpm | \(12 \frac{c m}{s e c}\) |
| 4. | \(x\) feet | 40 rpm | \(8 \frac{f t}{\sec }\) |
| 5. | 12 inches | 32 rpm | \(x \frac{i n}{\sec }\) |
| 6. | 8 cm | \(x\) rpm | \(12 \frac{c m}{\min }\) |
| 7. | 18 feet | 4 rpm | \(x \frac{m i}{h r}\) |
| 8. | \(x\) feet | 800 rpm | \(60 \frac{\mathrm{mi}}{\mathrm{hr}}\) |
| 9. | 15 in | \(x\) rpm | \(60 \frac{m i}{h r}\) |
| 10 | 2 in | \(x\) rpm | \(13 \frac{i n}{\sec }\) |
11. An engine spins a wheel with radius 5 inches at 800 rpm. How fast is this wheel spinning in miles per hour?
12. A bike has tires with a radius of 10 inches. How many revolutions must the tire make to ride a mile?
13. An engine spins a wheel with radius 6 inches at 600 rpm. How fast is this wheel spinning in inches per second?
14. Bob has a car with tires that have a 15 inch radius. When he is traveling at a speed of 30 miles per hour, how fast are the wheels spinning in revolutions per minute?
15. A circular track has two lanes. The interior lane is 25 feet from the center of the circle and the lane towards the exterior is 30 feet from the center of the circle. If you jog 6 laps, how much further will you jog in the exterior lane as opposed to the interior lane?