Before we begin our discussion, we need to define a new unit for measuring angles, the radian (see figure 2.35).
A radian is defined as the central angle subtending a length of arc equal to the radius of the circle.
A radian is approximately equal to 57.3°. The conversion factors for angle units are:
1 revolution = 360° radians or degrees. We call this angle the angular displacement of the point and use the Greek letter theta (θ) to represent this angular displacement.
Figure 2.35
If the point moves with constant speed it also has a constant angular velocity. That is, the line drawn from the point to the centre of the circle sweeps out a definite number of revolutions, radians, or degrees each second or minute. The symbol used to represent angular velocity is the Greek letter omega (ω).
Angular velocity can be expressed in different units, such as,
It is also possible that the point is not moving with constant angular velocity. It may be increasing or decreasing its angular velocity. When a CD starts rotating in a CD drive the angular velocity increases until it reaches a constant value. After the reject button is pushed the angular velocity decreases until the CD comes to rest.
In both of the above cases we say that the point has an angular acceleration. The Greek letter alpha (α) is used for angular acceleration. Note that a is positive if the angular velocity is increasing and negative if the angular velocity is decreasing.
Angular acceleration can also be expressed in different units,
Now as a body moves in a circular path four similar equations hold as in the case of a body moving in a straight-line path. Both sets of equations will be shown below. It is important to re-memorize the equations for straight-line motion. In this way the other four equations will also be known, since they are exactly analogous.
Figure 2.36: A point moving in a circle
s= θ=
v=u+at
s= ut+ ½ at2 θ=ω1t+ ½ αt2
v2=u2+2as ω2
2=ω1
2 + 2 αθ
Study these equations carefully and note that the set to the right, the "rotational analogy" are easily remembered if the left set is well known. We recall that the subscripts "u" and "v" indicate "initial" and "final".
These four rotational equations help us to solve many practical problems dealing with rotating bodies.
EXAMPLE:
A rotating machine part increases in angular velocity from 3 rev./min. to 35 rev./min. In 3.5 minutes. What is its angular acceleration?
We use the following equation and solve it for a.
ω2 = ω1+αt
=α We now substitute our known values.
α=
= 9.14 rev /min2
EXAMPLE:
A propeller starts from an angular velocity of 900 rev./min. and accelerates at 100 rev./min.2 for 5 minutes.
Through how many revolutions has it turned?
θ = ω1t+ ½ αt2
θ=(900rev./min.)(5min) + ½ (100rev./min.2)(5min.)2 θ = 5,750 revolutions
EXAMPLE:
A propeller starts at 1,000 rev./min. and accelerates at 100 rev./min.2 through 2,000 revolutions. What is its final angular velocity?
ω2 2 = ω1
2 +2αθ ω2
2 = (1,000 rev./min.)2 +2 (100 rev./min.2)(2,000 rev.) ω2 =1,180 rev./min.
We note that there is an acceleration of the body "in the path", called the tangential acceleration. The body is increasing or decreasing its speed, or traversing the circle. We recall also that when a body moves in a circle there is also a centripetal acceleration, V2/R, that is always directed toward the centre of the circular path.
Thus when a body is increasing speed as it moves in a circular path there are two acceleration vectors, one tangential to the path, and the other directed to the centre of the path (centripetal acceleration). In figure 2.37, the body is increasing speed in the counter-clockwise sense. The directions of the two acceleration vectors are shown.
Figure 2.37: Tangential acceleration (at) and Centripetal acceleration (ac) Radian Measure
In figure 2.38, 's' is the length along the path. We would like to relate this distance to the size of the central angle (6) and the radius (R) of the circular path. In our preceding discussion, the angle (θ) was measured in any of three different units, degrees, revolutions, or radians.
Figure 2.38: s, R and θ
The equation that relates s to θ and R is a very simple one if we limit the angular unit to radians. This equation is:
S = Rθ
We see that this equation is true if we look at figure 2.38. We note, by measuring, that the equation is satisfied. We also see that it would not be true if the angle θ was in revolutions or degrees
We now have a new problem to deal with in our treatment of rotational motion. There is a limit to the units that may be used in this equation. We repeat that, for this equation, we must use radian measure. Also, any equation that is derived from s = Rθ will have this same restriction.
Suppose that a body moves a small distance along the path and sweeps out a small central angle.
The usual mathematical notation for a very small quantity is the use of the Greek letter Delta (Γ).
Γs = R Γθ
Let us divide both sides of this equation by the time, (Γt) during which the motion occurred.
= R
We can write:
ν= Rω
If this velocity in the path is changing, there is also a change in the angular velocity. Assume that this change occurs in the small time interval (At).
We can write:
Γv = R Γω Next we divide left and right members by Γt.
=R
The tangential acceleration (a) in the left side is the rate at which a body moving in a circular path is picking up speed in the path. It is equal to the radius times the angular acceleration (a).
We can write:
a = Rα
Let us summarize the three important equations we have derived:
s = Rθ ν = Rω
a = Rα
All three of these equations require the use of radian measure. This means that:
θ must be in radians ω must be in rad/min. or rad./sec.
α must be in rad./min2or rad./sec2
Note that the radian is called a "dimensionless" unit. We put it in or take it out for clarity.
EXAMPLES:
A car is moving on a circular racetrack of radius 150 ft. It sweeps out an angle of 2000. How far has it travelled?
We note that:
θ = 200° x = 3.49 rads.
s = Rθ
s= (150 ft.) (3.49 rad.) s = 523 ft.
3.36 rev./min
EXAMPLE:
A race car is travelling at a speed of 176 ft./sec. (120 MPH) around a circular racetrack of radius 500 ft.
What is the angular velocity of this car in rev./min.?
Use the equation:
v = Rω or ω= =
ω = 0.352 rad./sec.
Note that we knew that the unit of our answer is rad./sec. and not rev./sec. since the equation we used always is in radian measure. The units in the right side of the second equation above actually come out as "nothing'Vsec. We put in the radian unit in the numerator for clarity.
in order to find our answer in rev./min. we use the proper conversion factors.
ω=
3.36 rev/min.
Problems
1. A propeller starts from rest and accelerates at 120 rev/sec2 for 4 seconds. What is its final angular velocity in rev/sec? In rev/min?
2. A rotating turntable starts from rest and accelerates at 5 rev/min2for 3 min. Through how many revolutions has it turned?
3. A helicopter main rotor starts from an initial angular velocity of 2 rev/min and accelerates at 60 rev/min2 while turning through 400 revolutions. What is its final angular velocity?
4. A plane is circling O'Hare in a circular pattern of radius 15,000 ft. It sweeps out an angle of 340°? How far has it travelled?
5. A plane is circling an airport in a circle of radius 5,000 ft. How far has it travelled after 4 revolutions?
6. A race car is moving on a circular track of radius 600 ft. It is travelling at a speed of 100 ft/s.
What is its angular velocity in rev/min?
EXAMPLE:
A race car is moving on a circular racetrack of radius 4,000 ft.. It is increasing its speed at a rate of 15 ft./sec.2 What is its angular acceleration rev./sec.2?
We use the equation:
a = Rα α= =
α = 0.00375 rad./sec.2
We note that the unit is rad./sec.2 because the equation that we have used requires radian measure.
To obtain a in rev./sec.2, we must use the standard conversion factor.
α=
a - 0.000597 rev ./sec.2
7. A race car is moving on a circular racetrack of radius 800 ft. It is accelerating at a rate of 10 ft/sec2 What is its angular acceleration in rev/sec2?
8. A helicopter tail rotor starts with an initial angular velocity of 15 rev/sec and decelerates at a rate of 2.00 rev/sec2 until it comes to rest. Through how many revolutions has the rotor turned while it comes to rest?
Answers
1. 480 rev/sec. 28,800 rev/min.
2. 22.5 rev.
3. 219 rev/min.
4. 89,000 ft.
5. 23.8 miles
6. 1/6 rad/s, 5ln rev/min
7. 1/16071 rev/sec2 8. 56.3 rev.