Angular Velocity vs. Linear Velocity


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1 MATH 7 Angular Velocity vs. Linear Velocity Dr. Neal, WKU Given an object with a fixed speed that is moving in a circle with a fixed ius, we can define the angular velocity of the object. That is, we can detere how fast the ian measure of the angle is changing as the object moves on its circular path. r v v = linear speed = angular speed r v = r We always use ians as the unit of measure when working with angular velocity. For instance, our angular velocities should be in units such as () / sec or () /. However sometimes we will give the result in other layman s terms, such as olutions per ute 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 ians. Note though that ians will be an invisible unit. For example the units () sec will yield /sec. Relationship Between Angular Velocity and Linear Velocity v Given a fixed speed v and ius r, then v = r = v r = 2 time of. Example. A cylinder with a 2.5 ius is rotating at 20 rpm. (a) Give the angular velocity in /sec and in degrees per second. (b) Find the linear velocity of a point on its rim in mph. Solution. (a) To convert rpm (olutions per ute) to ians per second, we first note that there are ians per olution. We then have
2 Them we have 4 sec 20 = 20 = 4 sec 80 deg 2 60 sec = 720º per sec. = 4 sec. (b) We use v = r, but we must be in the correct units. The angular velocity must use ians. Thus, the linear velocity is then Converting to mph, we obtain 4 sec 2.5 = 0 sec. 0 sec = 0 sec 3600 sec mph. Example 2. A tire with a 9 inch ius is rotating at 30 mph. Find the angular velocity of a point on its rim. Also express the result in olutions per ute. Solution. We simply use = v /r, but we must make sure that v and r have matching length units. Here we shall use s in order to put in ()/. = v r s = 30 % ' 2 in # 9 in& 5280 = 2,200 (). Note that the units actually come out as ; however, ians are a suppressed unit with regards to angular velocity. So we write () /. How many olutions per ute are there with this spinning tire? Because one olution is 2π ians, we have 2,200 () = 2,200 () 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 ius 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.)
3 Planetary Data Planet Rad /Earth's Mass/Earth's Grav/Earth's Rotation Mercury days Venus days Earth 23h 56m 4.s (Moon) days Mars h 37m s Jupiter h 50m Saturn h 39.9m Uranus h 4m Neptune h 3m Pluto d 9h 7m Radius of Earth km = 6,378,40 m s Mass of Earth kg (Average) Gravity on Earth m/s /s 2. Earth Day 24 hours Conversions: 000 m km, m,, km Example 3. The ius of Earth is approximately s. It takes 23h 56m 4.s 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 Solution. (a) We now shall use =. The angular velocity of Earth s spin time of. is then given by = (). % ' # 3600& We also could say = # & 3600% 5.04º per hour. (Note that it takes just under 24 hours to rotate 360, so every hour the Earth must rotate a little more than /24th of a circle, or just over 5.) The linear velocity of Earth spin at the equator is about v = r () s mph.
4 (b) At 50º N latitude, the angular velocity is the same, but the ius is smaller. (Recall: r = cos(50º) at 50º N). So the linear velocity at 50º N is now only () cos(50º) s mph. Example 4. Jupiter rotates in approximately 9h 50m. Its ius is.9473 times that of Earth s. Find the linear velocity at the equator of Jupiter. Solution. We again use v = r to obtain v = r = ( / 60) ( s) 28, mph. Exercises. A cylinder with a 2 ius 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. ius is spinning at 24 mph, find the angular velocity in rpm of a point on its rim. 3. (a) What is the ius of the circle at 28 5 N latitude? (b) Find the linear velocity of the Earth s rotation at 28 5 N. (c) Find the distance between points at the following coordinates: 28 5 N, E and 28 5 N, E. 4. The ius of Mars is times that of Earth. One rotation on its axis takes about 24, 37, sec. Find the linear speed of its equatorial spin.
5 . (a) First, = 450 2# () = 900 # 60 sec # 80 deg Solutions = 900#. Converting to deg per sec, we have = 2700º per sec. (b) v = r = mph. 2. Because 6 in. = 0.5 feet, we have = v r s % = 24 ' 5280 # 0.5 & = 253,440 (). Then 253,440 () rpm. 3. (a) The ius is r cos(28.25 ) s. (b) The linear velocity is v = r = % ' # 3600& s mph. (c) We first find the angle between E and E which is given by = = The distance between the points is then s = r = s v = r = () % ( ( s mph. ' # 3600 &
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