Dr. Neal, WKU
MATH 117
Angular Velocity vs. Linear Velocity
Given an object with a fixed speed that is moving in a circle with a fixed radius, we can
define the angular velocity ! of the object. That is, we can determine how fast the radian
measure of the angle is changing as the object moves on its circular path.
v
r
!
v = linear speed
! = angular speed
r
v =!r
We always use radians as the unit of measure when working with angular velocity.
For instance, our angular velocities should be in units such as (rad) / sec or (rad) / hr .
However sometimes we will give the result in other layman’s terms, such as revolutions
per minute or degrees per second, so as to give a better understanding of how fast the angle
is moving.
But in order to do calculations with angular velocity, we always must use radians.
(rad)
! ft
Note though that radians will be an “invisible” unit. For example the units
sec
will yield ft/sec.
Relationship Between Angular Velocity ! and Linear Velocity v
Given a fixed speed v and radius r , then
v = !r
!=
v
r
!=
2"
time of 1 rev.
Example 1. A cylinder with a 2.5 ft radius is rotating at 120 rpm.
(a) Give the angular velocity in rad/sec and in degrees per second.
(b) Find the linear velocity of a point on its rim in mph.
Solution. (a) To convert rpm (revolutions per minute) to radians per second, we first
note that there are 2! radians per revolution. We then have
Dr. Neal, WKU
120
Them we have 4!
rev
rev
rad 1 min
rad
.
= 120
! 2"
!
= 4"
min
min
rev 60 sec
sec
rad
rad 180 deg
= 4!
"
= 720º per sec .
sec
sec
! rad
(b) We use v = ! r , but we must be in the correct units. The angular velocity must use
radians. Thus, the linear velocity is then
4!
rad
ft
.
" 2.5 ft = 10!
sec
sec
Converting to mph, we obtain
10!
ft
ft
sec
1 mile
= 10!
" 3600
"
≈ 21.42 mph.
sec
sec
hr 5280 ft
Example 2. A tire with a 9 inch radius is rotating at 30 mph. Find the angular velocity
of a point on its rim. Also express the result in revolutions per minute.
Solution. We simply use ! = v / r , but we must make sure that v and r have matching
length units. Here we shall use miles in order to put ! in (rad)/hr.
! =v !
1 %
in
1 " miles
ft
(rad)
'' ! 12
!
! 5280
= $$ 30
= 211,200
.
hr
9 in &
ft
r #
mile
hr
1
; however, radians are a suppressed unit
hr
with regards to angular velocity. So we write (rad) / hr .
Note that the units actually come out as
How many revolutions per minute are there with this spinning tire? Because one
revolution is 2π radians, we have
211,200
(rad)
(rad)
1 rev
1 hr
!
!
= 211,200
≈ 560.225 rpm.
hr
hr
2! rad
60 min
Planetary Equatorial Velocities
We also can find the angular velocity and the linear velocity at the equator of a planet
given that we know the radius of the planet r and the time that it takes for the planet to
!
!
make one ( 360 ) rotation on its axis. (Note: One 360 rotation on the axis is not the
same as the actual length of a day due to the planet’s orbital movement relative to the
Sun.)
Dr. Neal, WKU
Planetary Data
Planet
Mercury
Venus
Earth
(Moon)
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
Rad /Earth's
0.3824
0.9489
1
0.272399
0.532036
11.194173
9.407131
4.0875553
3.8826366
0.1803
Mass/Earth's
0.0553
0.8150
1
0.0123
0.1074
317.89
95.17
14.56
17.15
0.002
Grav/Earth's
0.378
0.894
1
0.1653
0.379
2.54
1.07
0.8
1.2
0.01
Rotation
58.646 days
243.01 days
23h 56m 4.1s
27.32166 days
24h 37m 22.662s
9h 50m
10h 39.9m
17h 14m
16h 3m
6d 9h 17m
Radius of Earth ≈ 6378.140 km = 6,378,140 m ≈ 3963.2 miles
Mass of Earth ≈ 5.974383 ! 1024 kg
(Average) Gravity on Earth ≈ 9.80665 m/s2 ≈ 32.174 ft/s2.
1 Earth Day ≈ 24 hours
Conversions:
1 mile 1.609344 km
1000 m 3.28084 ft
,
,
,
1 km
1
m 5280 ft
1
mile
Example 3. The radius of Earth is approximately 3963.2 miles. It takes 23h 56m 4.1s for
the Earth to rotate once on its axis. (a) Find the angular velocity and linear velocity at
the equator. (b) Find the linear velocity at 50º N latitude.
2"
. The angular velocity of Earth’s spin
time of 1 rev.
(rad)
2!
is then given by ! =
≈ 0.262516
.
"
56
4.1 %
hr
+
$ 23 +
' hr
#
60 3600 &
Solution. (a) We now shall use ! =
We also could say ! =
360!
≈ 15.041º per hour.
!
56
4.1 $
+
# 23 +
& hr
"
60 3600 %
!
(Note that it takes just under 24 hours to rotate 360 , so every hour the Earth must
!
rotate a little more than 1/24 th of a circle, or just over 15 .)
The linear velocity of Earth’ spin at the equator is about
v = ! r ≈ 0.262516
(rad)
! 3963.2 miles ≈ 1040.4 mph.
hr
Dr. Neal, WKU
(b) At 50º N latitude, the angular velocity is the same, but the radius is smaller. (Recall:
r = 3963.2 cos(50º ) at 50º N). So the linear velocity at 50º N is now only
0.262516
(rad)
! 3963.2 cos(50º ) miles ≈ 668.76 mph.
hr
Example 4. Jupiter rotates in approximately 9h 50m. Its radius is 11.194173 times that
of Earth’s. Find the linear velocity at the equator of Jupiter.
Solution. We again use v = ! r to obtain
v=!r =
2!
! (11.194173 ! 3963.2 miles) ≈ 28,347.65 mph.
(9 + 50 / 60) hr
Exercises
1. A cylinder with a 2 ft radius is spinning at 450 rpm. (a) Find its angular velocity in
degrees per sec. (b) Find the linear speed on the rim in mph.
2. If a cylinder with a 6 in. radius is spinning at 24 mph, find the angular velocity in
rpm of a point on its rim.
3. (a) What is the radius of the circle at 28! 15! N latitude?
(b) Find the linear velocity of the Earth’s rotation at 28! 15! N.
(c) Find the distance between points at the following coordinates:
28! 15! N, 76! 08! E and 28! 15! N, 53! 44 ! E.
4. The radius of Mars is 0.532036 times that of Earth. One rotation on its axis takes
about 24 hr, 37 min, 22.662 sec. Find the linear speed of its equatorial spin.
Dr. Neal, WKU
Solutions
rev 2# (rad)
rad
. Converting to deg per sec, we have
"
= 900#
min
rev
min
rad
1 min 180 deg
! = 900"
#
#
= 2700º per sec .
min 60 sec
" rad
1. (a) First, ! = 450
(b) v = ! r = 900!
1 mile
rad
min
! 2 ft ! 60
!
≈ 64.26 mph.
5280 ft
min
hr
2. Because 6 in. = 0.5 feet, we have
! =v !
Then 253,440
1
1 " miles
!
= $$ 24
hr
0.5
r #
%
ft
(rad)
'' ! 5280
= 253,440
.
ft &
mile
hr
(rad)
1 rev
1 hr
!
!
≈ 672.27 rpm.
hr
2! rad
60 min
3. (a) The radius is r ≈ 3963.2 cos(28.25! ) ≈ 3491.146 miles.
(b) The linear velocity is v = ! r =
2!
! 3491.146 miles ≈ 916.48 mph.
"
56
4.1 %
23
+
+
$
'
#
60 3600 &
(c) We first find the angle between 76! 08! E and 53! 44 ! E which is given by 75! 68! –
!
53! 44 ! = 22! 24! = 22.4 . The distance between the points is then s = ! r =
"
22.4! !
! 3491.146 ≈ 1364.8766 miles.
180
4. v = ! r =
2! (rad)
( 0.532036 ( 3963.2 miles ≈ 538.05 mph.
"
37 22.662 %
24
+
+
hr
$
'
#
60
3600 &