The formulas radian measure was built to produce: arc length as radius times angle with no conversion constant, sector area as half the radius squared times the angle, and the two velocities of circular motion — angular velocity as the rate the angle changes and linear velocity as the rate the position changes, linked by v equals r omega. Includes ordinary frequency, angular frequency and period, and the unit discipline that makes every one of these formulas either work or fail.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.1 Angles and their Measure, pp. 706-711
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 706-711 — the pages these objectives are drawn from
Warm-up
You already know the circumference of a circle, and you already know what a fraction of a circle looks like. Those two facts are the entire lesson.
Discussion prompt
A circle has radius 10. What length of arc does a quarter of the circle cut off? Now do it for one twelfth. What did you actually compute both times?
Hint: You almost certainly took a fraction of the circumference. Write down the fraction you used each time.
Answer:
\[ C = 2\pi r = 20\pi \qquad \tfrac{1}{4}(20\pi) = 5\pi \qquad \tfrac{1}{12}(20\pi) = \tfrac{5\pi}{3} \]
Both times you took the fraction of a revolution, times the whole circumference. That is the only idea in this lesson. Everything below is that sentence written with the fraction replaced by theta over two pi, which is exactly what the previous lesson showed the fraction of a revolution to be.
Concept
Radian measure was defined as arc length divided by radius. Multiply both sides by the radius and you have the arc length. There is no new content here at all, which is the point worth noticing.
\[ \theta = \frac{s}{r} \quad\Longleftrightarrow\quad s = r\theta \]
The absence of a conversion constant is not a convenience. It is the entire reason radian measure was introduced in the previous lesson, and it is why every physics and calculus formula involving an angle assumes radians without saying so.
Figure (svg): A circle of radius r with a central angle theta and its subtended arc s highlighted, beside the definition of radian measure rearranged to give s equals r times theta
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 701-701 — the definition this rearrangement comes from
Section
Section 1
Concept
For a central angle of theta radians in a circle of radius r, the subtended arc has length r times theta. The formula is true only in radians, and if you hand it a degree measure it will quietly return a wrong number rather than an error.
subtended arc — The portion of the circle lying between the two rays of a central angle. Its length is the radius times the radian measure of the angle.
\[ s = r\theta, \quad \theta \text{ in radians} \]
A sanity check that costs nothing: the arc can never be longer than the circumference, so s must never exceed two pi r for an angle inside one revolution.
Figure (svg): A circle of radius r with a central angle theta and its subtended arc s highlighted, beside the definition of radian measure rearranged to give s equals r times theta
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 706-706
Picture it
Almost nothing in this lesson goes wrong arithmetically. It goes wrong in the units.
Figure (svg): Two columns showing which quantities in the circular motion formulas carry units and which do not, and what goes wrong if the angle is left in degrees
Before computing anything below, get every angle into radians and every time into one consistent unit. Then the formulas are one multiplication each.
Worked example
The angle arrives in degrees, so the first move is forced.
\[ \text{A circle has radius } 12 \text{ cm. Find the length of the arc subtended by a central angle of } 150^\circ. \]
Convert the angle to radians
Why: The arc length formula is only valid in radians; using 150 directly would be off by a factor of about 57.
\[ 150 ^\circ = 5 \pi / 6 \]
Write the formula
Why: Radius times angle, with the angle now in the right unit.
\[ s = r \theta \]
Substitute and multiply
Why: Twelve times five pi over six is ten pi.
\[ s = 12(5 \pi / 6) = 10 \pi \]
Attach the unit and approximate
Why: The radius was in centimetres, and radians contribute no unit, so the arc is in centimetres.
\[ 10 \pi = 31.4 \text{cm} \]
Figure (svg): The solution to Worked example arc length from a degree measure shown as a ladder of expressions, one row per legal move
\[ s = 12 \cdot \frac{5\pi}{6} = 10\pi \approx 31.4 \text{ cm} \]
Verify: compare with the circumference
Why: The full circumference is two pi times 12, which is 24 pi. The angle is 150 out of 360 degrees, which is five twelfths, and five twelfths of 24 pi is 10 pi. The proportion route gives the same answer.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 706-706
Prediction
A circle has radius 6 and a central angle measures 2 radians.
Predict first
Is the arc longer or shorter than the radius, and by roughly how much?
Correct: About twice the radius.
Why: Radian measure counts radius-lengths of arc, so an angle of 2 radians cuts off exactly two radius-lengths — the arc is twice the radius, which is 12 here. This is the fastest possible sanity check on any arc length calculation: the answer divided by the radius should equal the angle in radians. Notice that the numeral 6 never entered the reasoning at all.
Worked example
The same formula, solved for a different letter. Nothing new is needed.
\[ \text{An arc of length } 8\pi \text{ metres is subtended by a central angle of } \frac{2\pi}{3}. \text{ Find the radius.} \]
Write the formula and identify the unknown
Why: The arc and the angle are given, so the radius is what must come out.
\[ s = r \theta,\text{ solve for } r \]
Divide both sides by the angle
Why: The angle is nonzero, so this is a legal move.
\[ r = \frac{s}{\theta} \]
Substitute
Why: Dividing by a fraction is multiplying by its reciprocal.
\[ r = 8 \pi \times 3 / (2 \pi) \]
Cancel the pi and simplify
Why: The pi appears top and bottom and disappears, which is the usual sign that the setup was right.
\[ r = 12 \]
Figure (svg): The solution to Worked example find the radius from the arc shown as a ladder of expressions, one row per legal move
\[ r = \frac{s}{\theta} = \frac{8\pi}{2\pi/3} = 8\pi \cdot \frac{3}{2\pi} = 12 \text{ m} \]
Verify: substitute back
Why: With r equal to 12 and theta equal to two pi over three, the arc is 12 times two pi over three, which is 8 pi — the arc we were given.
Trap
\[ r = 5, \quad \theta = 60^\circ \quad\Longrightarrow\quad s = 5(60) = 300 \]
Substitute the degree measure straight into s equals r theta
Why: The formula does not carry a warning label, and 60 is a perfectly good number to multiply by.
An arc of length 300 on a circle of radius 5 is impossible: the entire circumference is only about 31.4. The answer is not slightly wrong, it is nearly ten times the whole circle.
\[ 60^\circ = \frac{\pi}{3} \quad\Longrightarrow\quad s = 5 \cdot \frac{\pi}{3} = \frac{5\pi}{3} \approx 5.24 \]
Convert to radians first, then multiply
Why: The formula was derived from theta equals s over r, and that definition measures theta in radians.
The check that catches this every time: compare the answer with the circumference. Here two pi times 5 is about 31.4, and one sixth of that is about 5.24 — which is what the correct calculation gave.
Fill the middle
A circle of radius 9 has a central angle of 40 degrees.
Fill in the blanks
40^\circ \cdot \frac2 pi/92 pi = ___ \qquad s = 9 \cdot ___ = ___
Why: Forty over 180 reduces to two ninths, so the angle is two pi over nine radians. Multiplying by the radius 9 cancels the nine in the denominator entirely, leaving two pi, about 6.28. The cancellation is a good sign: it happens whenever the radius and the denominator of the reduced angle share a factor, which is common in textbook problems.
Sorting
The circle has radius 4, so its circumference is 8 pi, about 25.1.
Sort into buckets
Sort each proposed arc length by whether it is possible for a central angle within one revolution.
Socratic
The area of a circle is pi r squared and the circumference is two pi r. Both carry a constant. Arc length is just r theta.
Discussion prompt
Explain why the arc length formula escapes having a constant in it, and what that tells you about where the constants in the other two formulas come from.
Hint: Ask what unit theta is measured in, and what one unit of it means.
Answer:
The constant is hiding inside theta. Radian measure is defined as arc over radius, so multiplying the radian measure by the radius gives back the arc by construction — there is nothing left to convert.
The two pi in the circumference formula is exactly the radian measure of a full revolution. Writing s equals r theta and putting theta equal to two pi is the circumference formula, so the constant there is not a separate fact; it is the radian measure of one turn.
The same is true of the sector area formula later in this lesson. Every constant in circle geometry turns out to be an angle in disguise, once angles are measured the circle's own way.
Section
Section 2
Concept
An object moving on a circular path is changing two things at once: its position along the path, and the angle its radius makes. Each has its own rate, and they are not the same quantity or even the same kind of quantity.
angular velocity — The rate of change of the angle with respect to time, written omega and measured in radians per unit time. It says how fast the object is turning, and says nothing about how big the circle is.
\[ \overline{v} = \frac{s}{t} \qquad \overline{\omega} = \frac{\theta}{t} \]
Linear velocity is a signed quantity: positive for counterclockwise motion and negative for clockwise. Its absolute value is the speed. Angular velocity carries the same sign convention for the same reason.
Figure (svg): An object moving along a circular path of radius r from a point P to a point Q, sweeping a central angle theta and travelling an arc of length s during a time t
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 706-707
Picture it
Every point on a rigid rotating body turns through the same angle in the same time. That is what makes it rigid.
Figure (svg): Two points on the same rotating disc, one near the centre and one near the rim, showing that they share an angular velocity but the outer point has a much larger linear velocity
The radius is the magnification factor. Double the distance from the centre and you double the speed, with the turning rate untouched.
Worked example
This derivation is three lines and it is worth doing once rather than memorising the result.
\[ \text{Show that } v = r\omega \text{ for motion at constant angular velocity.} \]
Start from the definition of linear velocity
Why: Velocity is displacement over time.
\[ v = \frac{s}{t} \]
Replace the arc with radius times angle
Why: This is the arc length formula from the first section, valid because theta is in radians.
\[ v = \frac{r \theta}{t} \]
Regroup so the angle sits over the time
Why: The radius is a constant and can be pulled out of the quotient.
\[ v = r(\frac{\theta}{t}) \]
Recognise the bracket
Why: The angle over the time is the definition of angular velocity.
\[ v = r \omega \]
Figure (svg): The solution to Worked example derive the link between the two shown as a ladder of expressions, one row per legal move
\[ v = \frac{s}{t} = \frac{r\theta}{t} = r \cdot \frac{\theta}{t} = r\omega \]
Verify: check the units
Why: The left side has length over time. The right side has length times radians over time, and since radians are dimensionless that reduces to length over time as well. The units balance, which they would not if theta had been in degrees.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 707-707
Discrimination
Each quantity below is one or the other. Deciding which is most of the work in these problems.
Sort into buckets
Sort each described quantity.
Worked example
Example 10.1.5. Lakeland Community College sits at 41.628 degrees north latitude, where the radius of the circle it traces is about 2960 miles.
\[ \text{The Earth completes one revolution in about } 24 \text{ hours. Find the linear velocity of a point at radius } 2960 \text{ miles.} \]
Convert the rotation rate into radians per hour
Why: One revolution is two pi radians, so the angular velocity is two pi over 24.
\[ \omega = 2 \pi / 24 = \frac{\pi}{12}\text{ per hour} \]
Write the velocity formula
Why: Radius times angular velocity, with the angular velocity now in radians per hour.
\[ v = r \omega \]
Substitute
Why: Two thousand nine hundred sixty times pi over twelve.
\[ v = 2960(\frac{\pi}{12}) \]
Evaluate and attach units
Why: The radius was in miles and the time in hours, so the velocity is in miles per hour.
\[ v = 775 \text{mph}\text{ approx} \]
Figure (svg): The solution to Worked example the linear velocity of a point on the Earth shown as a ladder of expressions, one row per legal move
\[ \omega = \frac{2\pi}{24 \text{ h}} = \frac{\pi}{12 \text{ h}} \qquad v = 2960 \cdot \frac{\pi}{12} \approx 775 \text{ mph} \]
Verify: check by the direct route
Why: In 24 hours the point travels one full circumference, two pi times 2960, which is about 18,600 miles. Dividing by 24 hours gives about 775 miles per hour — the same answer without ever using omega.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 707-708
Error analysis
A wheel of radius 0.3 metres turns at 60 revolutions per minute. A student finds the speed of a point on the rim.
Annotate
On: \( v = r\omega = 0.3 \cdot 60 = 18 \text{ m/min} \)
The error is a factor of two pi, about 6.28, and it is silent — nothing in the arithmetic looks wrong. Whenever a rotation rate is given in revolutions, or in hertz, or as 'so many times per second', converting to radians is the first move and not an optional one.
Faded example
A drill turns at 1200 revolutions per minute. Express its angular velocity in radians per second.
Fill in the blanks
\frac2 pi rad}60} \cdot \frac40 pi}___} \cdot \frac___}___ \text___} = ___ \text___
Why: Two conversion factors are needed, one for the angle unit and one for the time unit, and each is written so the unit being removed cancels. Twelve hundred times two pi is 2400 pi radians per minute, and dividing by 60 seconds per minute gives 40 pi radians per second, about 125.7. Writing the units into the fractions makes both factors self-checking.
Prediction
Two children sit on a roundabout, one at 1 metre from the centre and one at 3 metres.
Predict first
Compare their angular velocities and their linear speeds.
Correct: Same angular velocity; the outer child is three times faster.
Why: The roundabout is rigid, so both children sweep the same angle in the same time and share an angular velocity. Linear speed is r times omega, and with omega fixed the speed is proportional to the radius, so tripling the distance from the centre triples the speed. This is why the outside of a roundabout feels dramatically faster than the middle, and why the tip of a helicopter rotor can approach the speed of sound while the hub is nearly stationary.
Edge cases
The relation v equals r omega says the speed grows without limit as the radius grows, for a fixed rotation rate.
Discussion prompt
A neutron star spins at about 700 revolutions per second. Taking its radius as 10 kilometres, compute the speed of a point on its equator, and say what physical fact the formula is failing to warn you about.
Hint: Compute it, then compare with the speed of light, which is about 300,000 kilometres per second.
Answer:
\[ \omega = 700 \cdot 2\pi \approx 4400 \text{ rad/s} \qquad v = 10 \cdot 4400 = 44\,000 \text{ km/s} \]
That is about fifteen percent of the speed of light — enormous, but physically possible, and real millisecond pulsars do reach comparable figures.
What the formula fails to warn you about is that it is a statement of geometry, not of physics. Push the radius or the rotation rate high enough and v equals r omega will happily return a speed exceeding the speed of light, which cannot happen. The formula has no idea that relativity exists; it only knows that arc equals radius times angle. A formula's domain of validity is never inside the formula, and that is worth remembering every time one returns an answer that seems too large.
Section
Section 3
Concept
The quantity written as one revolution per twenty-four hours in the Earth example has a name: it is the ordinary frequency of the motion. It measures how often a complete cycle happens, and it is related to angular velocity by a factor of two pi.
period — The time taken to complete one full cycle of the motion, written T. It is the reciprocal of the ordinary frequency.
\[ \omega = 2\pi f \qquad T = \frac{1}{f} \qquad \omega = \frac{2\pi}{T} \]
Ordinary frequency counts cycles per unit time; angular frequency counts radians per unit time. Because one cycle is two pi radians, converting between them is exactly the same conversion you have been doing all lesson.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 708-708
Picture it
Fixing the frequency fixes how long one trip round takes. Points further out must cover more ground in that same time.
Figure (svg): Two points on the same rotating disc, one near the centre and one near the rim, showing that they share an angular velocity but the outer point has a much larger linear velocity
This is the sentence the book uses: points farther from the centre need to travel faster to maintain the same angular frequency, since they have farther to travel to make one revolution in one period's time.
Worked example
A period is given rather than a frequency. One extra step, then the usual route.
\[ \text{A Ferris wheel of radius } 20 \text{ m completes one revolution every } 50 \text{ seconds. Find the speed of a seat.} \]
Identify the period
Why: One revolution every 50 seconds is a period of 50 seconds per cycle.
\[ T = 50 s \]
Convert the period to angular velocity
Why: One cycle is two pi radians, so omega is two pi over the period.
\[ \omega = 2 \pi / 50 = \frac{\pi}{25} \]
Apply the velocity relation
Why: Radius times angular velocity.
\[ v = 20(\frac{\pi}{25}) \]
Simplify and attach units
Why: Twenty over twenty-five reduces to four fifths.
\[ v = 4 \pi / 5 = 2.51 m / s \]
Figure (svg): The solution to Worked example from period to linear speed shown as a ladder of expressions, one row per legal move
\[ \omega = \frac{2\pi}{50} = \frac{\pi}{25} \text{ rad/s} \qquad v = 20 \cdot \frac{\pi}{25} = \frac{4\pi}{5} \approx 2.51 \text{ m/s} \]
Verify: check by circumference over period
Why: One trip round is two pi times 20, about 125.7 metres, taken in 50 seconds. Dividing gives about 2.51 metres per second, matching. That is a comfortable walking pace, which is about right for a Ferris wheel.
Matching
Every one of these needs the same conversion, applied carefully.
Match the pairs
Why: For a frequency, multiply by two pi: one revolution per second gives two pi radians per second. For a period, divide two pi by it: a 4 second period gives two pi over 4, which is pi over 2. Thirty revolutions per minute is half a revolution per second, so pi radians per second. A 24 hour period gives two pi over 24, which is pi over 12 per hour — the Earth figure from the worked example.
Worked example
The speed is specified and the rotation rate must be found. Same relation, different unknown.
\[ \text{A centrifuge rotor of radius } 0.08 \text{ m must reach a rim speed of } 100 \text{ m/s. Find the required frequency in revolutions per minute.} \]
Solve the velocity relation for angular velocity
Why: Divide the speed by the radius.
\[ \omega = \frac{v}{r} \]
Substitute
Why: One hundred divided by eight hundredths.
\[ \omega = \frac{100}{0.08} = 1250 \text{rad} / s \]
Convert radians per second to revolutions per second
Why: Divide by two pi, since one revolution is two pi radians.
\[ f = \frac{1250}{2 \pi} = 199 r e v / s \]
Convert to revolutions per minute
Why: Multiply by 60 seconds per minute.
\[ 199(60) = 11, 900\text{ rpm approx} \]
Figure (svg): The solution to Worked example work backwards to a required frequency shown as a ladder of expressions, one row per legal move
\[ \omega = \frac{100}{0.08} = 1250 \text{ rad/s} \qquad f = \frac{1250}{2\pi} \approx 199 \text{ rev/s} \approx 11\,900 \text{ rpm} \]
Verify: run it forwards
Why: At 11,900 revolutions per minute the frequency is 198.3 per second, so omega is 198.3 times two pi, about 1246 radians per second. Times the radius 0.08 gives about 99.7 metres per second, which rounds back to the required 100. The small gap is the rounding, not an error.
Trap
\[ f = 5 \text{ Hz} \quad\Longrightarrow\quad \omega = 5 \text{ rad/s} \]
Read hertz as radians per second
Why: Both are rates measured per second, and both get called frequency in ordinary speech, so they blur together.
Five hertz means five complete cycles every second. Five radians per second is less than one complete cycle every second, since one cycle is about 6.28 radians. The two differ by a factor of two pi.
\[ f = 5 \text{ Hz} \quad\Longrightarrow\quad \omega = 2\pi f = 10\pi \approx 31.4 \text{ rad/s} \]
Multiply the ordinary frequency by two pi
Why: Each cycle contributes two pi radians, so cycles per second times radians per cycle gives radians per second.
The units make the conversion self-checking: cycles per second, times radians per cycle, gives radians per second, with the cycles cancelling. Any time a rate is quoted in hertz, revolutions, orbits, or beats, it is an ordinary frequency and needs the factor of two pi before it can enter v equals r omega.
Estimation
A car tyre has radius about 0.32 metres and the car is travelling at 30 metres per second.
Predict first
Roughly how many revolutions per second is the tyre making?
Correct: About 15.
Why: The circumference is two pi times 0.32, about 2.01 metres, so one revolution carries the car about two metres. At 30 metres per second that is about 15 revolutions per second. Going through omega gives the same thing: omega is 30 divided by 0.32, about 94 radians per second, and dividing by two pi gives about 15 revolutions per second. The distractor 94 is exactly the angular velocity in radians per second, which is the number you get if you stop one step early.
Ranking
All four are rotating. Put every rate into the same unit before comparing.
Put in order
Why: Convert each to radians per hour. The hour hand completes a revolution in 12 hours, giving pi over 6, about 0.52. The Earth takes 24 hours, giving pi over 12, about 0.26. The minute hand takes 1 hour, giving two pi, about 6.28. The record makes 33 revolutions a minute, which is 1980 an hour, giving about 12,440. So the Earth is the slowest, then the hour hand, then the minute hand, then the record — making the order d, a, b, c. The surprise is that the Earth turns more slowly than an hour hand, because the hour hand goes round twice a day and the Earth only once.
Counterexample
A student proposes: if two rotating objects have the same linear rim speed, they have the same angular velocity.
Discussion prompt
Find two objects that make this claim false, and then state the correct relationship.
Hint: Fix the speed and vary the radius.
Answer:
\[ v = r\omega \quad\Longrightarrow\quad \omega = \frac{v}{r} \]
Take a bicycle wheel of radius 0.35 metres and a car wheel of radius 0.32 metres, both with a rim speed of 10 metres per second. Their angular velocities are 10 over 0.35, about 28.6 radians per second, and 10 over 0.32, about 31.3. Same rim speed, different angular velocity.
The correct relationship is that with the speed fixed, angular velocity is inversely proportional to the radius. A small wheel must spin faster to keep up with a large one, which is exactly why the small gears in a bicycle's cassette are the fast ones.
Section
Section 4
Concept
A sector is the pie-slice region bounded by two radii and the arc between them. It occupies theta over two pi of the whole circle, so it has that share of the circle's area — and simplifying gives a formula with no pi in it at all.
circular sector — The region of a disc bounded by two radii and the arc they cut off. Its area is half the square of the radius times the radian measure of its central angle.
\[ A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{1}{2}r^2\theta \]
Notice that the pi cancels. A sector formula written in degrees keeps the pi and adds a 360, which is again the arbitrary constant doing nothing but undoing itself.
Figure (svg): A circular sector shaded, formed by a central angle theta in a circle of radius r, with the proportion argument that gives its area as one half r squared theta
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 710-711 — Exercise 56 sets this derivation; Exercises 57-62 drill it
Picture it
Do not memorise it. Derive it in the two lines the picture shows, every time, until it sticks.
Figure (svg): A circular sector shaded, formed by a central angle theta in a circle of radius r, with the proportion argument that gives its area as one half r squared theta
Both the arc length formula and the sector area formula are the same sentence: this sector is theta over two pi of the circle, so give it that fraction of the circumference, or that fraction of the area.
Worked example
The angle is in degrees again, so the first move is again forced.
\[ \text{Find the area of a sector of radius } 6 \text{ cm with central angle } 120^\circ. \]
Convert the angle to radians
Why: The sector formula, like the arc formula, is stated in radians.
\[ 120 ^\circ = 2 \pi / 3 \]
Write the formula
Why: Half the radius squared times the angle.
\[ A = (\frac{1}{2}) r ^{2} \theta \]
Substitute
Why: Six squared is 36, and half of that is 18.
\[ A = (\frac{1}{2}) (36) (2 \pi / 3) \]
Simplify
Why: Eighteen times two pi over three is twelve pi.
\[ A = 12 \pi \]
Figure (svg): The solution to Worked example the area of a sector shown as a ladder of expressions, one row per legal move
\[ A = \tfrac{1}{2}(6)^2\left(\tfrac{2\pi}{3}\right) = 12\pi \approx 37.7 \text{ cm}^2 \]
Verify: check the fraction of the circle
Why: The whole circle has area pi times 36, which is 36 pi. The sector is 120 out of 360 degrees, or one third, and one third of 36 pi is 12 pi. The proportion route agrees.
Comparison
Fill the blanks from memory. The two formulas share a derivation and differ in exactly one way.
Comparison matrix
| Arc length | Sector area | |
|---|---|---|
| What fraction of the circle | theta over two pi | theta over two pi |
| Fraction of what | the circumference, two pi r | the area, pi r squared |
| Formula | s = r theta | A = one half r squared theta |
| Powers of r | one | two |
| Units | length | length squared |
The only structural difference is which whole-circle quantity the fraction is taken of. Everything else follows.
Worked example
Solved for theta this time. The same formula, rearranged.
\[ \text{A sector of a circle of radius } 10 \text{ has area } 25\pi. \text{ Find its central angle in radians and in degrees.} \]
Write the formula with the unknown isolated
Why: Multiply both sides by 2 and divide by r squared.
\[ \theta = 2 A / r ^{2} \]
Substitute
Why: Twice 25 pi is 50 pi, and 10 squared is 100.
\[ \theta = 50 \pi / 100 \]
Simplify
Why: Fifty over one hundred is one half.
\[ \theta = \frac{\pi}{2} \]
Convert to degrees for the second part
Why: Multiply by 180 over pi.
\[ \frac{\pi}{2} = 90 ^\circ \]
Figure (svg): The solution to Worked example find the angle from the area shown as a ladder of expressions, one row per legal move
\[ \theta = \frac{2A}{r^2} = \frac{50\pi}{100} = \frac{\pi}{2} = 90^\circ \]
Verify: check the fraction
Why: A quarter circle of radius 10 has area a quarter of 100 pi, which is 25 pi — the area we were given. And a quarter revolution is indeed pi over two, or 90 degrees.
Error analysis
A student finds the area of a sector of radius 8 with central angle pi over 4.
Annotate
On: \( A = r\theta = 8 \cdot \frac{\pi}{4} = 2\pi \)
Count the powers of r. Anything that is a length carries one r; anything that is an area carries two. That single check separates these two formulas without remembering which is which.
Fill the middle
Derive the sector area formula from the proportion.
Fill in the blanks
A = \fracpi r^22 \cdot (1/2) r^2 theta = \frac______} = ___
Why: The sector is theta over two pi of the circle, and the circle has area pi r squared. Multiplying, the pi in the numerator cancels the pi in the denominator, leaving theta r squared over 2. That is the formula, and deriving it this way each time is more reliable than recalling whether the constant is a half or a quarter.
Prediction
A sector has a fixed central angle. Its radius is doubled.
Predict first
What happens to its arc length and to its area?
Correct: Arc length doubles, area quadruples.
Why: Arc length is r theta, linear in r, so doubling the radius doubles it. Area is half r squared theta, quadratic in r, so doubling the radius multiplies the area by four. This is the same scaling behaviour every length and area have under a dilation, and counting the powers of r in the formula predicts it without any computation.
Two truths and a lie
Rule out the statements that are true. The survivor is the false one.
Eliminate the wrong options
One of these statements about sectors is wrong.
Survives elimination: B
Why: Statement B is the false one, and it is false in a way that units cannot rescue. The formula half r squared theta was derived by writing the fraction of the circle as theta over two pi, which is only the correct fraction when theta is in radians. Feeding it degrees overstates the area by a factor of about 57, and no choice of length unit for the radius can compensate, because the error is in a dimensionless quantity. The degree version needs a different formula: theta over 360 times pi r squared.
Section
Section 5
Concept
Almost every question in this section gives you two of the quantities radius, angle, arc, area, time, angular velocity and linear velocity, and asks for a third. Knowing which formula links which is the whole skill.
Every row assumes the angle is in radians and every time is in one consistent unit. Getting those two things right before touching a formula removes almost every error available in this lesson.
| Given | Wanted | Use |
|---|---|---|
| radius and angle | arc length | s = r theta |
| radius and angle | sector area | A = one half r squared theta |
| arc and radius | angle | theta = s over r |
| angle and time | angular velocity | omega = theta over t |
| radius and angular velocity | linear velocity | v = r omega |
| frequency or period | angular velocity | omega = two pi f = two pi over T |
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 706-708
Picture it
This is the checklist to run before substituting anything into anything.
Figure (svg): Two columns showing which quantities in the circular motion formulas carry units and which do not, and what goes wrong if the angle is left in degrees
If a rotation rate arrives in revolutions, hertz, orbits, or beats per minute, it is an ordinary frequency and needs multiplying by two pi. If an angle arrives with a degree symbol, it needs multiplying by pi over 180. Do both first.
Worked example
Three formulas in sequence. Take them one at a time and label every intermediate quantity.
\[ \text{A pulley of radius } 15 \text{ cm turns at } 120 \text{ rpm. Find the speed of the belt, and the length of belt passing over it in } 4 \text{ seconds.} \]
Convert the rotation rate to radians per second
Why: One hundred twenty revolutions per minute is two per second, and each is two pi radians.
\[ \omega = 2(2 \pi) = 4 \pi \text{rad} / s \]
Find the belt speed
Why: The belt moves with the rim, so its speed is the rim's linear velocity.
\[ v = 15(4 \pi) = 60 \pi \text{cm} / s \]
Find the length in 4 seconds
Why: Distance is speed times time.
\[ s = 60 \pi(4) = 240 \pi \text{cm} \]
Convert to a sensible unit
Why: Two hundred forty pi centimetres is about 754 centimetres.
\[ \text{about } 7.54\text{ metres} \]
Figure (svg): The solution to Worked example a multi-step problem shown as a ladder of expressions, one row per legal move
\[ \omega = 4\pi \text{ rad/s} \qquad v = 60\pi \approx 188 \text{ cm/s} \qquad s = 240\pi \approx 754 \text{ cm} \]
Verify: count the revolutions independently
Why: At 120 revolutions per minute the pulley makes 2 per second, so 8 revolutions in 4 seconds. Each revolution passes one circumference of belt, two pi times 15, which is 30 pi. Eight times 30 pi is 240 pi — the same answer, reached without omega at all.
Sorting
Reading the question for what is given and what is wanted is most of the work.
Sort into buckets
Sort each question by the formula it needs first.
Worked example
This one wants an area, so watch the powers of the radius.
\[ \text{A wiper blade runs from } 20 \text{ cm to } 55 \text{ cm from the pivot and sweeps } 110^\circ. \text{ Find the area it clears.} \]
Convert the angle
Why: One hundred ten over 180 reduces to eleven eighteenths.
\[ 110 ^\circ = 11 \pi / 18 \]
Recognise the shape as a difference of two sectors
Why: The blade clears the big sector minus the small one it never reaches.
Write the difference and factor
Why: Both sectors share the angle, so it factors out along with the half.
\[ A = (\frac{1}{2}) \theta(55 ^{2} - 20 ^{2}) \]
Evaluate
Why: Fifty-five squared is 3025 and 20 squared is 400, a difference of 2625.
\[ A = (\frac{1}{2}) (11 \pi / 18) (2625) \]
Figure (svg): The solution to Worked example a windscreen wiper shown as a ladder of expressions, one row per legal move
\[ A = \tfrac{1}{2}\left(\tfrac{11\pi}{18}\right)(3025 - 400) = \tfrac{28875\pi}{36} \approx 2520 \text{ cm}^2 \]
Verify: sanity-check the size
Why: The region is roughly a rectangle 35 cm wide, the blade's length, by the arc it travels at its middle radius of 37.5 cm. That arc is 37.5 times 11 pi over 18, about 72 cm. So roughly 35 times 72, about 2520 square centimetres — matching closely, since the shape really is close to a curved rectangle.
Trap
\[ \omega = 300 \text{ rpm}, \quad r = 0.2 \text{ m} \quad\Longrightarrow\quad v = 0.2(300)(2\pi) = 377 \text{ m/s} \]
Convert revolutions to radians but leave the time in minutes
Why: The factor of two pi was applied correctly, so the calculation looks finished.
The answer is in metres per minute, not per second, but it has been labelled per second. As a speed it is over 1300 kilometres per hour, which for a 20 centimetre pulley is absurd.
\[ 300 \text{ rpm} = 5 \text{ rev/s} = 10\pi \text{ rad/s} \quad\Longrightarrow\quad v = 0.2(10\pi) \approx 6.28 \text{ m/s} \]
Convert both the angle unit and the time unit before substituting
Why: Two conversions are needed here, not one, and skipping the second is silent.
Six and a bit metres per second is about 23 kilometres per hour, which is a believable rim speed. Always name the unit of the answer out loud before writing it down — the absurdity check only works if you have said what the number is measuring.
Notation
This line appeared in a physics worked solution. Read what each symbol is doing.
Annotate
On: \( v = r\omega = r \cdot 2\pi f = \frac{2\pi r}{T} \)
The last form is worth remembering on its own. When a problem gives a period, going straight to two pi r over T skips both conversions and is almost impossible to get wrong.
Reverse engineer
Here is a complete solution with the question removed.
Fill in the blanks
\omega = \frac241670} \qquad v = 6378 \cdot \omega \approx ___ \text___
Why: The radius 6378 kilometres is the Earth's equatorial radius, and a period of 24 hours is one rotation, so this is the speed of a point on the equator: two pi over 24 is about 0.262 radians per hour, times 6378 gives about 1670 kilometres per hour. The question must have been to find how fast a point on the equator moves. Notice that the Lakeland example in the book got about 775 miles per hour, roughly 1250 kilometres per hour, at latitude 41.6 degrees — slower, because the circle traced there is smaller.
Constraint
You are designing a playground roundabout. Safety rules cap the rim speed at 3 metres per second.
Discussion prompt
If the roundabout has radius 1.8 metres, what is the fastest it may be pushed, in revolutions per minute? Then say what happens to that limit if a larger model of radius 2.5 metres is used instead, and why the answer is not obvious to a child pushing it.
Hint: Solve the velocity relation for omega, then convert to revolutions per minute.
Answer:
\[ \omega = \frac{v}{r} = \frac{3}{1.8} \approx 1.67 \text{ rad/s} \;\Longrightarrow\; \frac{1.67}{2\pi}(60) \approx 15.9 \text{ rpm} \]
\[ r = 2.5: \quad \omega = \frac{3}{2.5} = 1.2 \text{ rad/s} \;\Longrightarrow\; \approx 11.5 \text{ rpm} \]
The larger roundabout must be turned more slowly, about 11.5 revolutions per minute against 15.9, to stay under the same rim speed.
It is not obvious to whoever is pushing, because a person pushing feels the angular rate — how quickly the thing comes round to them — while the danger depends on the linear speed at the rim. The bigger roundabout feels lazier at exactly the rotation rate that makes it more dangerous, and v equals r omega is the whole explanation.
Comparison
Fill the blanks from memory. Every one of them assumes radians.
Comparison matrix
| Quantity | Formula | Units |
|---|---|---|
| Arc length | s = r theta | length |
| Sector area | A = one half r squared theta | length squared |
| Angular velocity | omega = theta over t | radians per unit time |
| Linear velocity | v = r omega | length per unit time |
| From frequency | omega = two pi f | radians per unit time |
| From period | omega = two pi over T | radians per unit time |
Count the powers of r to tell arc from area, and check whether time appears to tell a static formula from a motion one. Those two questions classify every formula in the table.
Pattern
Whether the question asks for a length, an area or a speed, the same five moves cover it.
Steps one and two together prevent nearly every mistake available in this lesson. The mathematics here is one multiplication; the difficulty is entirely in the units.
Check
Arc length. Do it on paper before you click.
Check your understanding
A circle has radius 15 cm. What is the length of the arc subtended by a central angle of 0.8 radians?
Answer: A
Why: Arc length is radius times angle, so 15 times 0.8 is 12 centimetres. The angle is already in radians, so no conversion is needed. As a check, 0.8 radians is a little under a radian, so the arc should be a little under one radius length — and 12 is a little under 15.
Check
Circular motion. Watch the units.
Check your understanding
A wheel of radius 0.5 metres turns at 240 revolutions per minute. What is the speed of a point on its rim?
Answer: B
Why: Two hundred forty revolutions per minute is 4 revolutions per second, which is 8 pi radians per second, about 25.1. Multiplying by the radius 0.5 gives about 12.6 metres per second. Equivalently, each revolution covers a circumference of pi metres, and 4 of those per second is about 12.6 metres per second.
Check
Sector area. Count the powers of the radius.
Check your understanding
What is the area of a sector of radius 8 with central angle pi over 3?
Answer: C
Why: The area is half of 8 squared times pi over 3, which is half of 64 times pi over 3, giving 32 pi over 3, about 33.5. As a check, pi over 3 is a sixth of a revolution, and a sixth of the full disc area of 64 pi, about 201, is indeed about 33.5.
Real world
A hard drive platter spins at 7200 revolutions per minute. Data sits in concentric tracks, and the read head is a fixed distance above the surface. The innermost track is 15 millimetres from the centre and the outermost is 45 millimetres.
Discussion prompt
Compute the linear speed of the surface passing under the head at both radii. Then explain why early drives stored the same number of bits on every track and wasted most of the outer ones, and what changed when manufacturers started varying it.
Hint: Both tracks share an angular velocity. They do not share a linear one.
Answer:
\[ \omega = 7200 \cdot \frac{2\pi}{60} = 240\pi \approx 754 \text{ rad/s} \]
\[ v_{\text{in}} = 0.015(754) \approx 11.3 \text{ m/s} \qquad v_{\text{out}} = 0.045(754) \approx 33.9 \text{ m/s} \]
The outer track passes under the head three times as fast, because it is three times further out and the platter is rigid. In one revolution it presents three times as much surface.
Early drives used constant angular velocity with a fixed sector count: every track held the same number of sectors, so the outer tracks stored their bits stretched out over three times the length and wasted roughly two thirds of their capacity. The fix, called zoned bit recording, groups tracks into zones and gives outer zones more sectors — which is why an outer-zone read is faster as well as denser, and why a disk benchmark shows throughput falling as it fills up.
The whole design question is the sentence from this lesson: on a rigid rotating body, omega is shared and v is not.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A formula gives the arc length as s equals r theta. What happens if you substitute an angle measured in degrees?
Correct: The answer comes out too large by a factor of about 57.
\[ \frac{60}{\pi/3} = \frac{180}{\pi} \approx 57.3 \]
Why: A degree measure is a much larger number than the corresponding radian measure, since one radian is about 57.3 degrees. Substituting 60 where pi over three, about 1.047, belongs multiplies the answer by 60 over 1.047, which is about 57.3. Nothing returns an error, and that is the dangerous part: the formula is a bare multiplication that will accept any number you hand it and produce a plausible-looking result. The only defence is checking the answer against the circumference, which an arc 57 times too long fails instantly.
Explain it
They have just learned that the circumference is two pi r and the area is pi r squared, and they are being asked about a pie slice.
Discussion prompt
In no more than five sentences and using no formula they have not met, explain how to find both the crust length and the area of a slice. Then tell them the one question to ask before doing any arithmetic.
Hint: Everything is a fraction of the whole.
Answer:
A usable answer: work out what fraction of the whole pie the slice is — a quarter, a sixth, whatever it is. The crust length is that fraction of the whole circumference, and the area is that same fraction of the whole area. That is all a sector formula ever does.
The question to ask first is what fraction of a full turn is this angle? If the angle is in degrees, divide by 360; if it is in radians, divide by two pi. Getting that fraction right is the entire problem, and the formulas s equals r theta and A equals half r squared theta are just that fraction with the division already carried out.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Arc against area is fixed by counting powers of the radius: one for a length, two for an area. The rpm conversion is fixed by writing both conversion factors with their units so the cancelling is visible. Angular against linear is fixed by asking whether the quantity mentions a distance: if it does not, it is angular. The derivation is fixed by writing the fraction theta over two pi first and multiplying it by the whole-circle quantity, which produces both formulas in two lines. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Draw a large circle with its centre marked, and shade a sector of roughly a fifth of the circle. Label the radius r, the central angle theta, and the arc s. To the right of the circle, write the two static formulas, arc length and sector area, and beside each one write the fraction theta over two pi and the whole-circle quantity it multiplies, so the derivation is visible rather than just the result. Below the circle, draw the same circle again with a dot on the rim and an arrow showing motion, and write the three motion relations: angular velocity as angle over time, linear velocity as arc over time, and the link between them. In the bottom left corner write the three conversions you must do before using any of these: degrees to radians, revolutions to radians, and any time unit to a single consistent one. Finally, in the bottom right corner, work one complete example of your own invention that uses at least three of the formulas in sequence, and label the units of every intermediate quantity.
Check your page against one test: every formula on it should contain theta in radians, and none of them should contain the number 360 or 180. If a 180 has appeared anywhere except in the conversion corner, something has been written in degrees that should not have been.
Recap
Five things, and the last one is what makes the first four reliable rather than lucky.
| If the question says | Your first move is |
|---|---|
| Find the arc length | Check the angle is in radians, then multiply by r |
| Find the area of the sector | Write theta over two pi times pi r squared, then simplify |
| Given in revolutions per minute | Multiply by two pi, then fix the time unit |
| Two points on the same wheel | They share omega; compute each v separately |
| Given the period | Go straight to two pi r over T for the rim speed |
That completes Section 10.1. Everything so far has been about the angle itself — measuring it, drawing it, and computing lengths and speeds from it. The next section asks the question the whole subject is named for: given an angle, where exactly is the point at the end of the terminal side?
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.1 Angles and their Measure §10.1, pp. 706-711 — everything on these slides traces back here
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