How To Calculate Linear Speed

Article objectives

To calculate the speed and angular velocity of objects.

To understand the human relationship between linear and angular speed.

Radian measure out and arc length tin can be applied to the study of round motion. In physics the average speed of an object is divers as:

$$\text{average speed} = \frac{\text{distance traveled}}{\text{time elapsed}}$$

And then suppose that an object moves along a circle of radius r, traveling a distance south over a period of time t, as in Effigy i. So it makes sense to define the (average) linear speed ν of the object as:

$$v = \frac{s}{t} \; \; \; \; (1)$$

Let θ be the angle swept out by the object in that flow of time. And then we define the (boilerplate) athwart speed ω of the object equally:

$$ω = \frac{θ}{t} \; \; \; \; (2)$$

Angular speed gives the rate at which the primal bending swept out by the object changes as the object moves around the circumvolve, and it is thus measured in radians per unit of measurement fourth dimension. Linear speed is measured in distance units per unit time (eastward.yard. feet per 2d). The word linear is used because straightening out the arc traveled by the object forth the circumvolve results in a line of the aforementioned length, so that the usual definition of speed equally distance over fourth dimension tin can exist used. We will ordinarily omit the word average when discussing linear and athwart speed here.

Since the length s of the arc cut off by a central angle θ in a circumvolve of radius r is s = r θ, we see that

$$v = \frac{southward}{t} = \frac{r θ}{t} = \frac{θ}{t} . r $$

and so that we get the following relation between linear and angular speed:

$$v = ω r \; \; \; \; (3)$$

Example 1

An object sweeps out a fundamental angle of $\frac{π}{three}$ radians in 0.5 seconds as it moves forth a circle of radius 3m. Find its linear and angular speed over that time catamenia.

Solution: Here we have t = 0.five sec, r = three m, and θ = $\frac{π}{iii}$ rad. So the athwart speed ω is

$$ ω = \frac{θ}{t} = \frac{\frac{π}{iii} rad}{0.5 sec} ⇒ ω = \boxed{\frac{2π}{3} rad/sec},$$

and thus the linear speed ν is

$$v = ω r = \left(\frac{2π}{3} rad/sec \correct) (3 m) ⇒ \boxed{v = 2π m/sec}$$

Note that the units for ω are rad/sec and the units of ν are m/sec. Remember that radians are really unitless, which is why in the formula ν = ωr the radian units disappear.

Instance 2

An object travels a distance of 35 ft in 2.7 seconds as information technology moves along a circle of radius ii ft. Find its linear and athwart speed over that time menstruation.

Solution: Here we have t = ii.vii sec, r = ii ft, and s = 35 ft. So the linear speed ν is

$$v = \frac{due south}{t} = \frac{35 feet}{two.7 sec} ⇒ \boxed{12.96 ft/sec} , $$

and thus the angular speed ω is given by

$$five = ω r = 12.96 ft/sec = ω(2 ft) ⇒ \boxed{ω = half-dozen.48 rad/sec}$$

Example 3

An object moves at a constant linear speed of 10 k/sec around a circle of radius four m. How big of a central angle does it sweep out in 3.one seconds?

Solution: Here we accept t = 3.1 sec, ν = ten chiliad/sec, and r = 4 m. Thus, the bending θ is given by

$$due south = rθ ⇒ θ = \frac{s}{r} = \frac{vt}{r} = \frac{(10 yard/sec)(3.1 sec)}{4 one thousand} = \boxed{7.75 rad}. $$

In many physical applications angular speed is given in revolutions per minute, abbreviated as rpm. To convert from rpm to, say, radians per second, notice that since there are 2π radians in one revolution and 60 seconds in one minute, we can convert Due north rpm to radians per second by "canceling the units" every bit follows:

$$N rpm = North \frac{rev}{min} . \frac{2π}{1 rev} . \frac{1 min}{threescore sec} = \frac{North . 2π}{60} rad/sec$$

This works because all we did was multiply by one twice. Converting to other units for angular speed works in a like way. Going in the opposite direction, say, from rad/sec to rpm, gives:

$$Due north rad/sec = \frac{N . sixty}{2π} rpm$$

Example iv

A gear with an outer radius of r ₁ = 5 cm moves in the clockwise direction, causing an interlocking gear with an outer radius of r ₂ = 4 cm to move in the counterclockwise management at an athwart speed of ω ₂ = 25 rpm. What is the angular speed ω _i of the larger gear?

Solution: Imagine a particle on the outer radius of each gear. Later on the gears accept rotated for a period of fourth dimension t > 0, the circular deportation of each particle volition be the aforementioned. In other words, s ₁ = southward ₂, where s ₁ and s _ii are the distances traveled by the particles on the gears with radii r ₁ and r _two, respectively. But s _ane = ν ₁ t and due south _ii = ν _two t, where ν_ane and ν_ii are the linear speeds of the gears with radii r₁ and r_ii, respectively. Thus,