5.09 Linear and angular speed

 

We can extend the concepts of radian measure and arc length to the study of circular motion, or motion in a circular path.

Recall that for an object traveling over a linear path, we can define the speed of an object as the ratio of the distance the object travels in a given time.

$\text{speed }$speed

$=$=

$\frac{\text{distance}}{\text{time}}$distancetime​

 

This same relationship applies to an object following a curved path. However, we now have two different ways to define the speed of the object.

Suppose we use the length of the circlular arc, $s$s,  to measure the distance traveled. In that case we are finding the linear speed of the object. (Linear because we can straighten out the arc of a circle into a line of the same length). The linear speed, $v$v, of an object is defined as

$v$v

$=$=

$\frac{s}{t}$st​

 

If we let $\theta$θ stand for the angle swept by the object in that period of time, then we can define the angular speed, $\omega$ω, of the object as

$\omega$ω

$=$=

$\frac{\theta}{t}$θt​

 

Because the linear and angular speed are calculated using different measurements, they are reported in different units. The angular speed is measured in radians per unit time (like radians / second). However, the linear speed is measured in linear units per unit of time (like meters / second).

Since the length $s$s of the arc cut off by a central angle $\theta$θ in a circle of radius $r$r is $s=r\theta$s=rθ, substitution gives us the following relation between linear and angular speed:

$v$v

$=$=

$\omega r$ωr

 

Worked examples

Question 1

Solve: Consider the moon as it orbits the earth. The moon's orbit is really an ellipse but its eccentricity is low so that it can be approximated by a circular orbit. The radial distance from the earth is, on average, $384000$384000 km and each complete orbit takes $27.32$27.32 days. What is the instantaneous speed of the moon?

Think/Do: The length of one complete orbit must be $2\times\pi\times384000\approx2413000$2×π×384000≈2413000 km. This distance is covered in $27.32$27.32 days, which is $24\times27,32\approx656$24×27,32≈656 hours. Therefore, relative to the earth, the moon has an orbital speed of $2413000\div656\approx3680$2413000÷656≈3680 km/h.

Reflect: This instantaneous speed is about $1.02$1.02 kilometers per second in magnitude.

Question 2

Solve: A cyclist is riding at $14$14km/h on a bicycle with wheels whose radius is $33$33 cm. How many times does the wheel turn in one second? What is the angular speed of the wheel relative to the hub?

Think/Do: The speed, $14$14 km/h, is $14\times1000\times100=1400000$14×1000×100=1400000 cm/h or $388.9$388.9 cm/s. The circumference of the wheel is $33\times2\times\pi=207.3$33×2×π=207.3 cm. Therefore, in one second the wheel turns $388.9\div207.3\approx1.87$388.9÷207.3≈1.87 times.

The angular speed could be given as $1.87$1.87 revolutions per second or, in terms of an angle, $1.87\times360^\circ=675.3$1.87×360°=675.3 degrees per second. In radian measure it would be $1.87\times2\pi=11.75$1.87×2π=11.75 radians per second.

 

Practice questions

Question 3

Question 4

Question 5