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Note 10 Rotational Motion I
Sections Covered in the Text: Chapter 13
Motion is classified as being of one of three types:
translational, rotational or vibrational. Translational
motion is the motion executed by the center of mass of
an object modelled as a particle (Notes 03 and 06). The
motion of a baseball hit in a line drive is largely trans-
lational. But a baseball can also rotate, so the motion
of a baseball can in general possess both translational
and rotational components. A baseball does not
usually vibrate. We shall study vibrational motion in
Note 12.
At first thought the motion of a baseball might seem
impossibly complicated to describe mathematically.
Physics shows, however, that the motion of any object
can be separated into its translational and rotational
components and those components solved for
separately. This is possible because translational and Figure 10-1. A particle P in a rigid body is located with
rotational components of the motion of a rigid body respect to O by the polar coordinates (r,θ).
1
do not interact with one another.
We have already studied a particularly simple form
of rotational motion in Note 05: a particle rotating Any point P in the object can be located relative to the
about a point external to it. This motion is called point O with polar coordinates (r,θ). As the object
circular motion. Here we survey a few aspects of the rotates, P follows a circle of radius r. Every other point
rotational motion of a rigid extended body rotating in the object also follows a circular path, but a path
about a point internal to itself. with a different radius. Let us suppose that in some
We begin by laying down the kinematics of a rota- elapsed time ∆t, P moves from a position on the
ting body as we did for a body in translational positive x-axis to the point where it is shown in the
motion. We derive the kinematic equations of rota- figure. Then the subtended angle θ is called the
tional motion and deduce the relationships between angular displacement of P measured relative to the
rotational and translational quantities. We introduce positive x-axis. Because the object is rigid, the angular
the construct of torque. displacement of every particle in the object is the same
as the angular displacement of P.
Angular displacement is measured in the dimen-
Angular Speed and Angular Acceleration sionless unit called radian (abbreviated rad) and in the
following way. The arc length of the circle on which P
For convenience we consider a rigid extended object moves (the distance travelled by P) is related to the
whose mass is confined to a plane and whose trans- radius r and angle θ by
lational motion is zero (Figure 10-1). We suppose this
object is rotating in a counterclockwise direction s = rθ , …[10-1]
about an axis perpendicular to its plane passing
through a point O. In the coordinate system of the s
figure, this axis can be thought of as the z-axis. and therefore θ = . …[10-2]
We assume that the object is a rigid, extended body. r
By this we mean it cannot be modelled as a single
particle. It can, however, be modelled (approximately) (rad). If s = r then θ = 1 radian. If the object executes
as a collection of particles whose positions are fixed one complete revolution, then the angular displace-
relative to one another. A CD is a rigid body whereas ment of P is 2π radians. 2π radians is equivalent to
a soap bubble is not. 360˚. If the object executes two revolutions, then its
angular displacement is 4π radians. By its nature,
angular displacement is a cumulative quantity.
1 But the body must be rigid. Examples of non-rigid bodies
whose modes of translation and rotation interact are typically
studied in a higher-level course in classical mechanics.
10-1
Note 10
We now have the tools we need to define the rota- Δθ dθ
tional equivalents of the translational quantities we ω≡lim = . …[10-5]
Δt→0 Δt dt
defined in Note 03. We begin by making Figure 10-1
more general (Figure 10-2) by supposing that in some We now extend the math to allow for changes in the
elapsed time ∆t = tf – ti any arbitrary point P in the 3
instantaneous angular velocity. If the instantaneous
object moves from a position i to a position f. These angular velocity is changing, then the object is by
positions are shown as [A] and [B] in the figure. The definition, undergoing an angular acceleration. Let
corresponding angular positions are θi and θ f the instantaneous angular velocity of the point P at
respectively. positions i and f be ω and ω respectively. The change
i f
in the instantaneous angular velocity divided by the
corresponding elapsed time is defined as the average
angular acceleration:
ωf – ωi Δω
α≡ = . …[10-6]
t – t Δt
f i
–2
Average angular acceleration has dimension T and
–2
units s . The limit of the average angular acceleration
as ∆t → 0 is defined as the instantaneous angular
acceleration:
Δω dω
α≡ lim = . …[10-7]
Δt→0 Δt dt
Figure 10-2. A more general representation of a rotating ω and α are, in fact, vector quantities (pseudo-vectors)
body than that shown in Figure 10-1. In some elapsed time whose complete vector nature is beyond the scope of
∆t a particle P in the body moves counterclockwise between 4
these notes to describe adequately. But you can find
two arbitrary angular positions. the direction of the vectors with the help of the right-
hand rule as was introduced for the vector cross
product (Figure 10-3).
The angular displacement of P in the interval chosen is
defined as the difference between the angular
positions that define the interval:
∆θ = θ – θ …[10-3]
f i
2
(rad). The angular displacement divided by the
corresponding elapsed time is defined as the average
angular velocity:
θf –θi Δθ
ω≡ = . …[10-4]
t –t Δt
f i
Average angular velocity has dimension T–1 and units
–1 –1
rad.s (or just s since rad has no dimension). The
limit of the average angular velocity as ∆t → 0 is Figure 10-3. How to use the right hand rule to find the
defined as the instantaneous angular velocity: direction of the vector ω of a rotating body.
3 Of course, in order for the object’s angular velocity to change,
2 The alert reader will notice the word displacement here the object must be subject to a force applied to it in a special way.
implying a vector quantity. Angular displacement is, in fact, a For the moment we ignore this force, as we did in Notes 02 and 03,
vector quantity, or more correctly, a pseudo-vector. This aspect of and stay within the area of kinematics.
rotational motion is somewhat advanced and best left to a second 4 We leave this description to a higher-level course in classical
year course in classical mechanics. mechanics.
10-2
Note 10
–1 –1
To find the direction of the ω vector, extend your right The final angular speed is 10.0 s or 10.0 rad.s .
hand, curl your fingers as if you are to grip something (b) Using the second equation in Table 10-1 we have
and extend your thumb. Now curl your fingers in the for the angular displacement
direction of the angular displacement of the object
(the direction the object is rotating). Then your thumb θ =θ +1(ω +ω )t
points in the direction of the ω vector. By convention, f i 2 i f
the ω vector is placed on a diagram along the body’s
axis of rotation. Later in this note we shall see other = 0 + 1(0 +10.0rad.s–1)(5.00s)
uses of the right hand rule. 2
= 25.0 radians.
Rotational Kinematics
As implied in the previous section, a set of kinematic The number n of revolutions is this number divided
equations exist for rotational motion just as they do by the number of radians per revolution (i.e., 2π):
for translational motion. They have a similar form and
are derived in a similar fashion. We shall therefore n = 25.0(rad) = 3.98 rev.
just list them (Table 10-1).
2π rad
rev
Table 10-1. Comparison of translational and rotational kine-
matic equations. Thus nearly 4 revolutions are required for the wheel
Translational Motion Rotational Motion to accelerate to the final angular speed of 10.0 s–1.
vf = vi + at ωf = ωi +αt €
1 2 1 2
xf = xi + vit + 2 at θf =θi +ωit + 2αt Relations exist between the angular and tangential
x = x + 1(v +v )t θ =θ +1(ω +ω )t speeds of a particle in a rotating rigid body, and bet-
f i 2 i f f i 2 i f ween the angular and tangential accelerations. Since
v2 = v2 + 2a(x – x ) ω2 = ω2 + 2α(θ –θ ) these relations are useful in solving rotational motion
f i f i f i f i problems we consider them next.
Problems in rotational kinematics can be solved much
like problems in translational kinematics. Assuming Relations Between Rotational and
you have memorized the translational equations and
know the rotational equivalents, you can easily Translational Variables
reconstruct the rotational equations. Let us consider Suppose that a rigid extended body rotates about an
an example. axis that passes through an internal point O as shown
in Figure 10-4. Consider a point P in this body.
Example Problem 10-1
A Problem in Rotational Kinematics
Starting from rest a wheel is rotated with a constant
angular acceleration of 2.00 rad.s–2 for 5.00 s. (a) What
is the final angular speed of the wheel? (b) How many
complete revolutions does the wheel execute in the
elapsed time of 5.00 s?
Solution:
(a) Using the first equation in Table 10-1 we have for
the final angular speed
ω =ω +αt=0+(2.00s–2)(5.00s)
f i
= 10.0 s–1.
Figure 10-4. A particle P in a rotating rigid body.
10-3
Note 10
The magnitude of the tangential velocity of P is, from Table 10-2. Relationships between the magnitudes of trans-
eq[10-1], lational and rotational variables.
ds dθ Translational Rotational Relationship
v = = r , x θ x =θr
dt dt
v ω v =ωr
since r is constant. Thus using the definition of ω in a α a =αr =ω2r
eq[10-5]
v =rω . …[10-8] Let us consider an example.
The tangential acceleration of P is, using eq[10-8],
dv dω Example Problem 10-2
at = =r , Tangential and Angular Speeds
dt dt
again since r is constant, or using the definition of α in A bicycle wheel of diameter 1.00 m spins freely on its
–1
eq[10-7], axis at an angular speed of 2.00 rad.s . (a) What is the
at = rα. …[10-9] tangential speed of a point on the rim of the wheel?
(b) What is the tangential speed of a point halfway
Now since P is moving in a circle it is undergoing a between the axis and the rim?
centripetal acceleration. The magnitude of this
centripetal or radial component of the acceleration is Solution:
(a) Using eq[10-8] the tangential speed of a point on
v2 the rim of the wheel is
a = =rω2.
c r z
1.00m
–1 –1
using eq[10-8]. v =rω = (2.00rad.s ) =1.00 m.s .
2
The relationship between tangential and radial com-
ponents of the acceleration of P can be seen with the
help of Figure 10-5. The total acceleration of P is the (b) A point halfway between axis and rim will have a
sum of the a and a vectors. The complete vector tangential speed one half of this value, or
t r
nature of a and a is beyond the scope of these notes
t c v = 0.50 m.s–1.
to describe. The relationships between the magnitudes
of these quantities are summarized in Table 10-2.
Clearly, the further a point on the wheel is from the
axis of rotation the greater is its tangential speed.
Though the two points have different tangential
speeds they have the same angular speed.
We stated in Note 07 without proof that the centre of
mass of a rigid body can be taken to be the body’s
geometric centre. We are now ready to extend the idea
of the centre of mass to a system of discrete bodies,
and to define the centre of mass in a proper
mathematical fashion.
Figure 10-5. The resultant acceleration of any particle P in a
rotating rigid body is the vector sum of the tangential and
radial acceleration vectors.
10-4
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