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.

!

r

rv

v = linear speed

! = angular speed

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. Note though that radians will be an “invisible” unit. For example the units (rad)

sec! ft

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

120rev

min= 120

rev

min! 2"

rad

rev!1

60

min

sec= 4"

rad

sec.

Them we have 4! rad

sec= 4!

rad

sec"180

!

deg

rad= 720º per sec .

(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

sec" 2.5 ft = 10!

ft

sec.

Converting to mph, we obtain

10!ft

sec= 10!

ft

sec" 3600

sec

hr"

1

5280

mile

ft ≈ 21.42 mph.

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

r = 30 miles

hr!1

9 in

"

# $ $

%

& ' ' ! 12 in

ft ! 5280 ft

mile = 211,200 (rad)

hr.

Note that the units actually come out as

1

hr; however, radians are a suppressed unit

with regards to angular velocity. So we write (rad) / hr . How many revolutions per minute are there with this spinning tire? Because one revolution is 2π radians, we have

211,200 (rad)hr

= 211,200 (rad)hr

!

1

2! revrad

!

1

60

hr

min ≈ 560.225 rpm.

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 Rad /Earth's Mass/Earth's Grav/Earth's Rotation Mercury 0.3824 0.0553 0.378 58.646 days Venus 0.9489 0.8150 0.894 243.01 days Earth 1 1 1 23h 56m 4.1s

(Moon) 0.272399 0.0123 0.1653 27.32166 days Mars 0.532036 0.1074 0.379 24h 37m 22.662s

Jupiter 11.194173 317.89 2.54 9h 50m Saturn 9.407131 95.17 1.07 10h 39.9m Uranus 4.0875553 14.56 0.8 17h 14m

Neptune 3.8826366 17.15 1.2 16h 3m Pluto 0.1803 0.002 0.01 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: 10001

m

km, 3.28084

1

ft

m, 1

5280

mile

ft, 1.609344

1

km

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. Solution. (a) We now shall use ! =

2"

time of 1 rev.. The angular velocity of Earth’s spin

is then given by ! = 2!

23 +56

60+

4.1

3600

"

# $

%

& ' hr

≈ 0.262516 (rad)hr

.

We also could say ! =

360!

23 +56

60+

4.1

3600

!

" #

$

% & hr

≈ 15.041º per hour.

(Note that it takes just under 24 hours to rotate 360

! , so every hour the Earth must rotate a little more than 1/24th of a circle, or just over 15

! .) The linear velocity of Earth’ spin at the equator is about

v = ! r ≈ 0.262516 (rad)hr

! 3963.2 miles ≈ 1040.4 mph.

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)hr

! 3963.2 cos(50º) miles ≈ 668.76 mph. 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!

(9 + 50 / 60) hr ! (11.194173 ! 3963.2 miles) ≈ 28,347.65 mph.

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!

1 ! 5 N latitude? (b) Find the linear velocity of the Earth’s rotation at

28!

1 ! 5 N. (c) Find the distance between points at the following coordinates:

28!

1 ! 5 N, 76!

0 ! 8 E and 28!

1 ! 5 N, 53!4 ! 4 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 1. (a) First, ! = 450

rev

min"2# (rad)

rev= 900#

rad

min. Converting to deg per sec, we have

! = 900"rad

min#1

60

min

sec#180

"

deg

rad= 2700º per sec .

(b) v = ! r = 900! rad

min! 2 ft !

60min

hr!

1

5280

mile

ft ≈ 64.26 mph.

2. Because 6 in. = 0.5 feet, we have

! = v !

1

r = 24 miles

hr!

1

0.5 ft

"

# $ $

%

& ' ' ! 5280 ft

mile = 253,440 (rad)

hr.

Then 253,440 (rad)

hr !

1

2! revrad

!

1

60

hr

min ≈ 672.27 rpm.

3. (a) The radius is r ≈ 3963.2

cos(28.25

!) ≈ 3491.146 miles.

(b) The linear velocity is v = ! r = 2!

23 +56

60+4.1

3600

"

# $

%

& '

! 3491.146 miles ≈ 916.48 mph.

(c) We first find the angle between

76!

0 ! 8 E and 53!4 ! 4 E which is given by

75!

6 ! 8 –

53!4 ! 4 =

22!

2 ! 4 = 22.4! . The distance between the points is then s = ! r =

22.4!!

"

180! 3491.146 ≈ 1364.8766 miles.

4. v = ! r = 2! (rad)

24 +37

60+22.662

3600

"

# $

%

& ' hr

( 0.532036 ( 3963.2 miles ≈ 538.05 mph.