Precalculus

Angular and linear speed

Two points ride the same turning wheel, one near the center and one at the rim. They turn together, so why does the rim point move so much faster?

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Radian measurePrecalculus Rate and unit rateGrade 6

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Height on a turning wheelPrecalculus

Where you are 4 in Angles and their measure

Trigonometry7

Angles and their measure4

Angles in standard position and coterminal angles Radian measure

Angular and linear speed

Arc length and sector area in radians

Why this exists

A pottery wheel spins at 150 revolutions per minute. That one number says how fast the wheel turns, but not how fast the clay moves under the potter’s hands: clay near the center barely creeps, while clay at the rim of a wide plate races past. A turning object needs two speeds. Angular speed is how fast it turns, the angle swept per unit of time, and it is the same for every point of the object. Linear speed is how fast one point travels along its circle, distance per unit of time, and it grows with the distance from the center.

Radians join the two. Because a radian measures an angle by arc over radius (radian measure), turning speed times radius gives travel speed with no stray factor. The same link answers a practical question: when a belt or chain connects two wheels of different sizes, how fast does the second one turn?

Definition

An object turning about a fixed center sweeps an angle \theta in a time t. Its angular speed is the angle per unit of time:

\omega = \frac{\theta}{t}

The letter \omega is the Greek omega. Angular speed may be given in radians per second, degrees per second, or revolutions per minute (\text{rpm}), the unit printed on most motors. One revolution is a full turn, 2\pi radians or 360^\circ.

A point of the object at distance r from the center travels along an arc of length s in the same time. Its linear speed is the distance per unit of time:

v = \frac{s}{t}

Both are rates: a quantity per one unit of time. When the turning is steady, every time interval gives the same rate; when it is not, \displaystyle \frac{\theta}{t} is the average angular speed over that interval.

Intuition

In the time the wheel turns through \theta, every point on it turns through that same \theta, because the wheel is one rigid piece. The arcs they trace are not the same length.

Figure 1Every point turns through the same angle \theta, but the outer point travels farther.

A point at radius r_1 traces s_1 = r_1\theta and a point at radius r_2 traces s_2 = r_2\theta. Same angle in the same time, so the same angular speed; the outer arc is longer, so the outer point is faster. Triple the radius and the arc triples, and so does the speed. At a fixed angular speed, linear speed is proportional to the radius, and the center itself, at r = 0, does not travel at all.

Results

Linear speed from angular speed. With \omega in radians per unit of time,

v = r\omega

The speed v comes out in the length unit of r per the time unit of \omega. A radian is a ratio of two lengths and has no unit, so “radians” drops out of the answer.

Revolutions to radians. One revolution is 2\pi radians, so an angular speed of n rpm is 2\pi n radians per minute, and dividing by 60 gives radians per second.

Belts and chains. A belt that wraps two pulleys without slipping or stretching moves at one speed along its whole length, and the rim of each pulley moves with it. So the two rims have the same linear speed:

r_1\omega_1 = r_2\omega_2

which is the same as

\frac{\omega_2}{\omega_1} = \frac{r_1}{r_2}

Figure 2The belt moves at one speed, so both rims do too. The small pulley must turn faster to keep up.

The small pulley turns faster, by exactly the ratio of the radii: the angular speeds are inversely proportional to the radii. The ratio is the same whatever unit the angular speeds use, so rpm can go in and come out unchanged. A chain over toothed wheels (sprockets) works the same way with the number of teeth in place of the radius: the chain moves one link per tooth, so a wheel’s tooth count is proportional to its circumference.

Two wheels on one shaft turn together, so they share one angular speed, not one linear speed.

Why it is true

By the definition of radian measure, \displaystyle \theta = \frac{s}{r}, so the arc a point travels is s = r\theta (the arc length formula in radians). Divide both sides by the time t:

v = \frac{s}{t} = \frac{r\theta}{t} = r \cdot \frac{\theta}{t} = r\omega

The radius comes out of the fraction because it does not change while the point travels. The step needs \theta in radians: in degrees the arc of n^\circ is \displaystyle \frac{\pi r n}{180}, not rn, and the formula would carry a stray \displaystyle \frac{\pi}{180}.

For the belt: in a time t some length L of belt passes over each pulley. The rim of pulley 1 moves with the belt, so L = r_1\theta_1, and likewise L = r_2\theta_2. Then r_1\theta_1 = r_2\theta_2, and dividing by t gives r_1\omega_1 = r_2\omega_2.

Worked examples

Example 1

Angular speed in three units

Gulmira’s pottery wheel turns at 150 \text{ rpm}. Find its angular speed in radians per minute, in radians per second and in degrees per second.

Solution

Each revolution is 2\pi radians, so multiply by 2\pi radians per revolution; the revolutions cancel:

\begin{aligned} 150 \cdot 2\pi &= 300\pi \text{ rad/min} \\ &\approx 942.48 \text{ rad/min} \end{aligned}

A minute is 60 seconds, so divide by 60:

\begin{aligned} \omega &= \frac{300\pi}{60} = 5\pi \text{ rad/s} \\ &\approx 15.708 \text{ rad/s} \end{aligned}

In degrees, each revolution is 360^\circ: 150 \cdot 360^\circ = 54{,}000^\circ per minute, which is 900^\circ per second. All three describe one speed; radians per unit of time is the form v = r\omega needs.

Example 2

Two points on one wheel

On the same wheel Gulmira throws a pot of radius 12 cm. Find the linear speed of the clay at the pot’s rim and of a point 4 cm from the center.

Solution

Both points share \omega = 5\pi rad/s from Example 1, since the wheel and the clay turn as one piece. With v = r\omega:

\begin{aligned} v_{\text{rim}} &= 12 \cdot 5\pi = 60\pi \\ &\approx 188.50 \text{ cm/s} \\ v_{\text{inner}} &= 4 \cdot 5\pi = 20\pi \\ &\approx 62.83 \text{ cm/s} \end{aligned}

The rim moves about 1.88 meters every second, three times as fast as the inner point, because its radius is three times as large.

Example 3

Angular speed from a measurement

Kanat’s turntable should spin a record at \displaystyle 33\frac{1}{3} \text{ rpm}. He puts a chalk mark on the record and finds that in 18 seconds it makes 10 full turns and a further 120^\circ. Find the angular speed in radians per second and in rpm. Is the turntable right?

Solution

Change the whole angle to radians. Ten turns are 20\pi, and \displaystyle 120^\circ = \frac{2\pi}{3}:

\theta = 20\pi + \frac{2\pi}{3} = \frac{62\pi}{3}

Divide by the time:

\begin{aligned} \omega &= \frac{62\pi}{3} \div 18 = \frac{31\pi}{27} \text{ rad/s} \\ &\approx 3.607 \text{ rad/s} \end{aligned}

To get rpm, multiply by 60 seconds per minute and divide by 2\pi radians per revolution:

\frac{31\pi}{27} \cdot \frac{60}{2\pi} = \frac{310}{9} \approx 34.44 \text{ rpm}

That is more than \displaystyle 33\frac{1}{3}. The ratio of the two speeds is \displaystyle \frac{310}{9} \div \frac{100}{3} = \frac{31}{30}, so the turntable runs about 3.3\% fast.

Example 4

From a speed limit to a setting

Madina is throwing a plate of radius 18 cm and wants the clay at its rim to move no faster than 1.5 meters per second. What is the highest whole-number rpm she can set?

Solution

Put both lengths in one unit: 1.5 m/s is 150 cm/s. Solve v = r\omega for \omega:

\begin{aligned} \omega &= \frac{v}{r} = \frac{150}{18} = \frac{25}{3} \text{ rad/s} \\ &\approx 8.333 \text{ rad/s} \end{aligned}

Change to rpm:

\frac{25}{3} \cdot \frac{60}{2\pi} = \frac{250}{\pi} \approx 79.58 \text{ rpm}

Faster turning means faster clay, so she rounds down to 79 \text{ rpm}. Rounding up fails: at 80 \text{ rpm} the rim moves \displaystyle 18 \cdot \frac{80 \cdot 2\pi}{60} = 48\pi \approx 150.80 cm/s.

Example 5

A belt from motor to wheel

The motor of Gulmira’s wheel turns at 1{,}125 \text{ rpm} and carries a pulley of radius 2 cm. A belt runs from it to a pulley of radius 15 cm under the wheel head, as in the belt figure. Find the speed of the belt and the angular speed of the wheel.

Solution

The belt moves with the rim of the motor pulley. That pulley turns at 1{,}125 \cdot 2\pi = 2250\pi rad/min, so

\begin{aligned} v &= 2 \cdot 2250\pi = 4500\pi \text{ cm/min} \\ &= 75\pi \text{ cm/s} \\ &\approx 235.62 \text{ cm/s} \end{aligned}

The rim of the large pulley moves with the same belt, so its angular speed is

\omega_2 = \frac{v}{r_2} = \frac{4500\pi}{15} = 300\pi \text{ rad/min}

which is \displaystyle \frac{300\pi}{2\pi} = 150 \text{ rpm}. The ratio rule gives it in one line: \displaystyle 1{,}125 \cdot \frac{2}{15} = 150 \text{ rpm}. The fast motor and the belt together turn the wheel at a steady working speed.

Example 6

Two belts and a shared shaft

Gulmira’s second wheel, used for trimming, runs slower. Its motor turns at 1{,}800 \text{ rpm} with a pulley of radius 1.5 cm, belted to a pulley of radius 9 cm on a middle shaft. A pulley of radius 2 cm on that same shaft is belted to a pulley of radius 5 cm under the wheel head, whose rim has radius 16 cm. Find the wheel’s rpm and the linear speed of the wheel head’s rim.

Solution

Across the first belt the rims share a linear speed, so the angular speeds are in the inverse ratio of the radii:

\omega_{\text{shaft}} = 1{,}800 \cdot \frac{1.5}{9} = 300 \text{ rpm}

The 2 cm pulley sits on the same shaft, so it also turns at 300 \text{ rpm}: through a shared shaft the angular speed carries over, not the linear speed. Across the second belt:

\omega_{\text{wheel}} = 300 \cdot \frac{2}{5} = 120 \text{ rpm}

Overall the motor speed is multiplied by \displaystyle \frac{1.5}{9} \cdot \frac{2}{5} = \frac{1}{15}. For the rim, 120 \text{ rpm} is \displaystyle \frac{120 \cdot 2\pi}{60} = 4\pi rad/s, so

v = 16 \cdot 4\pi = 64\pi \approx 201.06 \text{ cm/s}

Example 7

Constant turning, changing groove speed

A record’s groove spirals inward from a radius of 14.5 cm to 6 cm while the turntable holds \displaystyle 33\frac{1}{3} \text{ rpm}. Find the speed at which the groove passes under the needle at the start and at the end. If the groove speed were instead held at its starting value, how fast would the record have to turn at the end?

Solution

The angular speed is \displaystyle \frac{100}{3} \cdot \frac{2\pi}{60} = \frac{10\pi}{9} rad/s. With v = r\omega:

\begin{aligned} v_{\text{start}} &= 14.5 \cdot \frac{10\pi}{9} = \frac{145\pi}{9} \\ &\approx 50.61 \text{ cm/s} \\ v_{\text{end}} &= 6 \cdot \frac{10\pi}{9} = \frac{20\pi}{3} \\ &\approx 20.94 \text{ cm/s} \end{aligned}

The groove slows as the needle moves in, by the factor \displaystyle \frac{6}{14.5}. To hold \displaystyle \frac{145\pi}{9} cm/s at radius 6 cm the record would need

\omega = \frac{v}{r} = \frac{145\pi}{9} \div 6 = \frac{145\pi}{54} \text{ rad/s}

which is \displaystyle \frac{145\pi}{54} \cdot \frac{60}{2\pi} \approx 80.56 \text{ rpm}, the same as \displaystyle 33\frac{1}{3} \cdot \frac{14.5}{6}. Holding the linear speed fixed forces the angular speed up as the radius shrinks, since \displaystyle \omega = \frac{v}{r}.

Boundaries

Radians in v = r\omega. The formula multiplies the radius by a radian measure. With \omega in rpm the answer is short by the factor 2\pi; in degrees per second it is off by the factor \displaystyle \frac{180}{\pi}. Convert first.

Matching units. v takes its length unit from r and its time unit from \omega. A radius in centimeters with \omega per minute gives centimeters per minute, and a speed in meters per second must be changed before it is divided by a radius in centimeters.

Rigid turning about a fixed center. All points share one \omega only when the object turns as one piece about a fixed axis. Something sliding across the wheel has an angular speed of its own.

An average over an interval. \displaystyle \frac{\theta}{t} is the average angular speed over the time t. A wheel that is speeding up has a different speed at each moment, and one interval does not give it.

Speed, not direction. v says how fast a point moves. Its direction turns with the wheel all the time, always along the tangent to its circle.

No slip, no stretch. r_1\omega_1 = r_2\omega_2 holds only while the belt grips both pulleys and keeps its length. A slipping belt turns the driven pulley slower than the rule says. A crossed belt, run in a figure eight, obeys the same speed rule but turns the second pulley the other way.

Where this goes next

Once \omega is known, the angle swept after a time t is \theta = \omega t. Placed in standard position, that angle says where a point on a turning wheel is at every moment, and its height above the ground is the subject of height on a turning wheel, the first model of motion that repeats. The belt rule, a product r\omega held fixed, is another case of inverse variation, and the same reasoning runs gear trains, where meshing teeth play the part of the belt.

Exercises

Angular speed

Kanat’s turntable also plays old records at 78 \text{ rpm}. Find the angular speed in radians per minute and in radians per second.
A pottery wheel turns at 240 \text{ rpm}. Find its angular speed in degrees per second and in radians per second.
A wheel head makes 5 revolutions in 4 seconds. Find its angular speed in radians per second and in rpm.
Write an angular speed of 12 rad/s in rpm.
A platter turns through 1000^\circ in 2.5 seconds. Find its angular speed in degrees per second, in radians per second and in rpm.

Linear speed

Madina centers a pot of radius 9 cm on a wheel turning at 90 \text{ rpm}. Find the linear speed of the pot’s rim.
A record turns at 45 \text{ rpm}. Find the speed of the groove under the needle when the needle is 10 cm from the center.
A wheel head of diameter 30 cm turns at 200 \text{ rpm}. Find the linear speed of its rim in centimeters per second and in meters per second.
A point 8 cm from the center of a wheel moves at 2 meters per second. Find the angular speed in radians per second and in rpm.
The wheel turns at 60 \text{ rpm} and the rim of Gulmira’s bowl moves at 90 cm/s. Find the bowl’s radius.
A wheel head of radius 17 cm turns at 100 \text{ rpm} for 3 minutes. How far does a point on its rim travel, in meters?

Belts, chains and shafts

A motor pulley of radius 3 cm turns at 1{,}500 \text{ rpm} and is belted to a pulley of radius 25 cm. Find the larger pulley’s rpm and the speed of the belt in cm/s.
A turntable’s belt runs around the platter’s rim, of radius 13 cm, and around a pulley on a motor turning at 250 \text{ rpm}. What radius must the motor pulley have for the platter to turn at \displaystyle 33\frac{1}{3} \text{ rpm}? For 45 \text{ rpm}?
Show that the ratio of the two pulley radii in Exercise 13 is 1.35, and that it would be the same for any platter size and any motor speed.
A motor at 1{,}440 \text{ rpm} has a 2 cm pulley belted to an 8 cm pulley on a shaft. A 3 cm pulley on that shaft is belted to a 12 cm pulley on the wheel. Find the wheel’s rpm.
A wheel’s pulley has radius 12 cm and must turn at 160 \text{ rpm}. The motor runs at 1{,}920 \text{ rpm}. What radius of motor pulley is needed?
An old pottery wheel is driven by a chain. The motor’s sprocket has 15 teeth and turns at 640 \text{ rpm}; the wheel’s sprocket has 48 teeth. Find the wheel’s rpm.
A belt moves at 3 meters per second around a pulley of radius 5 cm. Find the pulley’s angular speed in radians per second and in rpm.

Reasoning

Explain why every point on a turning pottery wheel has the same angular speed but not the same linear speed.
Kanat computes the rim speed of a 10 cm wheel at 30 \text{ rpm} as 10 \cdot 30 = 300 cm per minute. Explain what he left out, and give the correct speed.
Two pulleys are joined by a crossed belt, run in a figure eight. Explain why r_1\omega_1 = r_2\omega_2 still holds, and what is different about the second pulley.
Madina doubles the radius of both pulleys in a belt drive. Explain why the driven pulley’s angular speed does not change, and what does change.
A record turns at a constant rate. Explain why the groove passes the needle faster near the edge than near the label, and find by what factor faster at radius 14 cm than at 7 cm.

Answers

156\pi rad/min, and \displaystyle \frac{13\pi}{5} \approx 8.168 rad/s. 78 \cdot 2\pi = 156\pi, then divide by 60.
1440^\circ per second, and 8\pi \approx 25.13 rad/s. 240 \cdot 360^\circ = 86{,}400^\circ per minute, divided by 60; and 240 \cdot 2\pi = 480\pi rad/min, divided by 60.
\displaystyle \frac{5\pi}{2} \approx 7.854 rad/s, and 75 \text{ rpm}. Five revolutions are 10\pi radians, and \displaystyle \frac{10\pi}{4} = \frac{5\pi}{2}; 5 turns in 4 seconds is 75 in 60 seconds.
\displaystyle \frac{360}{\pi} \approx 114.59 \text{ rpm}. 12 \cdot 60 = 720 rad/min, divided by 2\pi radians per revolution.
400^\circ per second, \displaystyle \frac{20\pi}{9} \approx 6.981 rad/s, and \displaystyle 66\frac{2}{3} \text{ rpm}. \displaystyle \frac{1000}{2.5} = 400; \displaystyle 400^\circ = \frac{20\pi}{9} rad; 400 \cdot 60 = 24{,}000^\circ per minute, divided by 360^\circ.
27\pi \approx 84.82 cm/s. 90 \text{ rpm} is \displaystyle \frac{90 \cdot 2\pi}{60} = 3\pi rad/s, and v = 9 \cdot 3\pi.
15\pi \approx 47.12 cm/s. 45 \text{ rpm} is \displaystyle \frac{3\pi}{2} rad/s, and \displaystyle v = 10 \cdot \frac{3\pi}{2}.
100\pi \approx 314.16 cm/s, about 3.14 m/s. The radius is 15 cm and \displaystyle \omega = \frac{200 \cdot 2\pi}{60} = \frac{20\pi}{3} rad/s, so \displaystyle v = 15 \cdot \frac{20\pi}{3}.
25 rad/s, and \displaystyle \frac{750}{\pi} \approx 238.73 \text{ rpm}. In centimeters, \displaystyle \omega = \frac{200}{8}; then 25 \cdot 60 = 1500 rad/min, divided by 2\pi.
\displaystyle \frac{45}{\pi} \approx 14.32 cm. 60 \text{ rpm} is 2\pi rad/s, and \displaystyle r = \frac{v}{\omega} = \frac{90}{2\pi}.
About 320.44 meters. In 3 minutes the wheel makes 300 turns, 600\pi radians, so s = 17 \cdot 600\pi = 10{,}200\pi \approx 32{,}044 cm.
180 \text{ rpm}, and 150\pi \approx 471.24 cm/s. \displaystyle 1{,}500 \cdot \frac{3}{25} = 180; the belt moves at 3 \cdot 1{,}500 \cdot 2\pi = 9000\pi cm/min, divided by 60.
\displaystyle \frac{26}{15} \approx 1.73 cm, and 2.34 cm. From r_1\omega_1 = r_2\omega_2 with rpm on both sides, \displaystyle r_1 = \frac{13 \cdot \frac{100}{3}}{250} and \displaystyle r_1 = \frac{13 \cdot 45}{250}.
\displaystyle \frac{2.34}{26/15} = 1.35. Each pulley radius is \displaystyle \frac{13\,\omega_{\text{platter}}}{250}, so their ratio is \displaystyle \frac{45}{100/3} = \frac{27}{20} = 1.35: the platter radius and the motor speed cancel, and only the two platter speeds remain.
90 \text{ rpm}. \displaystyle 1{,}440 \cdot \frac{2}{8} = 360 \text{ rpm} on the shaft, shared by the 3 cm pulley, then \displaystyle 360 \cdot \frac{3}{12} = 90.
1 cm. \displaystyle r_1 = \frac{r_2\omega_2}{\omega_1} = \frac{12 \cdot 160}{1{,}920}.
200 \text{ rpm}. Teeth play the part of radius: \displaystyle 640 \cdot \frac{15}{48} = 200.
60 rad/s, and \displaystyle \frac{1800}{\pi} \approx 572.96 \text{ rpm}. \displaystyle \omega = \frac{300}{5} in centimeters; then 60 \cdot 60 = 3600 rad/min, divided by 2\pi.
The wheel is rigid, so in any time every point turns through the same angle \theta and \displaystyle \omega = \frac{\theta}{t} is shared. But a point at radius r travels the arc r\theta, so its distance, and its speed v = r\omega, grow with r.
He used rpm in place of radians per minute. Each revolution is 2\pi radians, so \omega = 60\pi rad/min and v = 10 \cdot 60\pi = 600\pi \approx 1885 cm per minute, 2\pi times his answer.
The belt still moves at one speed and grips both rims, so the rims share a linear speed and r_1\omega_1 = r_2\omega_2. Crossing the belt makes it leave one pulley’s top and meet the other’s bottom, so the second pulley turns the opposite way.
The rule gives \displaystyle \omega_2 = \omega_1 \cdot \frac{r_1}{r_2}, and \displaystyle \frac{2r_1}{2r_2} = \frac{r_1}{r_2}, so \omega_2 is unchanged. The belt speed r_1\omega_1 doubles, because the motor pulley’s rim is twice as far from its center.
Every point of the record shares one \omega, so the groove speed v = r\omega is proportional to the radius. At 14 cm it is \displaystyle \frac{14}{7} = 2 times the speed at 7 cm.