Center of Mass
P
m1v1 + m2 v2 P
vcm =
=
m1 + m2
M
M
In vector form,




m1v1 + m2 v 2 P
v cm =
=
m1 + m2
M
In an isolated system, the total linear momentum does not change,
therefore the velocity of the center of mass does not change.
(

∑F
EXT
)
 

Δt = Pf − P0 = ΔP
Divide both sides by Δt


ΔP
F
=
∑ EXT Δt


P = Mv CM



Δ
M
v

ΔP
Δv CM
(
CM )
=
=M
= MaCM
Δt
Δt
Δt


∴ ∑ FEXT =MaCM
Newton’s 2nd Law for a system of particles
which act like a single particle of mass M
concentrated at the center of mass
The Principle of Conservation of Linear Momentum
Example: Ice Skaters
Starting from rest, two skaters
push off against each other on
ice where friction is negligible.
One is a 54 kg woman and
one is a 88 kg man. The woman
moves away with a velocity of
+2.5 m/s. Find the recoil velocity
of the man.
The Principle of Conservation of Linear Momentum


Pf = Po
m1v f 1 + m2 v f 2 = 0
vf 2 = −
vf 2
m1v f 1
m2
(
54 kg )(+ 2.5 m s )
=−
= −1.5 m s
88 kg
Center of Mass
Consider the “Ice Skater” example which was an
isolated system:
BEFORE
vcm
m1v1 + m2 v2
=
=0
m1 + m2
AFTER
vcm
(54
kg)(+2.5
m/s)
kg)(-1.5
m/s)
(
)(
)(
88 kg
− 1.5 m
s )++((88
54 kg
+ 2.5 m
s)
=
= 00.02
.002 ≈ 0
88kg
kg++88
54 kg
kg
54
Chapter 8
Rotational Motion
Rotational Motion and Angular Displacement
In the simplest kind of rotation,
points on a rigid object move on
circular paths around an axis of
rotation.
Rotational Motion and Angular Displacement
The angle through which the
object rotates is called the
angular displacement.
Δθ = θ − θ o
Rotational Motion and Angular Displacement
DEFINITION OF ANGULAR DISPLACEMENT
When a rigid body rotates about a fixed axis, the angular
displacement is the angle swept out by a line passing through
any point on the body and intersecting the axis of
rotation perpendicularly.
By convention, the angular displacement
is positive if it is counterclockwise and
negative if it is clockwise.
SI Unit of Angular Displacement: radian (rad)
Rotational Motion and Angular Displacement
Arc length s
θ (in radians) =
=
Radius
r
For a full revolution:
2π r
θ=
= 2π rad
r
2π rad = 360
Rotational Motion and Angular Displacement
Example: Adjacent Synchronous Satellites
Synchronous satellites are put into
an orbit whose radius is 4.23×107m.
If the angular separation of the two
satellites is 2.00 degrees, find the
arc length that separates them.
Rotational Motion and Angular Displacement
Arc length s
θ (in radians) =
=
Radius
r
! 2π rad $
2.00 deg #
& = 0.0349 rad
" 360 deg %
s = rθ = ( 4.23×10 7 m ) ( 0.0349 rad )
= 1.48 ×10 6 m (920 miles)
Rotational Motion and Angular Displacement
Conceptual Example: A Total Eclipse of the Sun
The diameter of the sun is about 400 times greater than
that of the moon. By coincidence, the sun is also about
400 times farther from the earth than is the moon.
For an observer on the earth, compare the angle subtended
by the moon to the angle subtended by the sun and explain
why this result leads to a total
solar eclipse.
Rotational Motion and Angular Displacement
Arc length s
θ (in radians) =
=
Radius
r
Angular Velocity and Angular Acceleration
Δθ = θ − θ o
How do we describe the rate
at which the angular displacement
is changing?
Angular Velocity and Angular Acceleration
DEFINITION OF AVERAGE ANGULAR VELOCITY
Angular displacement
Average angular velocity =
Elapsed time
θ − θ o Δθ
ω=
=
t − to
Δt
SI Unit of Angular Velocity: radian per second (rad/s)
Angular Velocity and Angular Acceleration
Example: Gymnast on a High Bar
A gymnast on a high bar swings through
two revolutions in a time of 1.90 s.
Find the average angular velocity
of the gymnast.
Angular Velocity and Angular Acceleration
# 2π rad &
Δθ = −2.00 rev %
( = −12.6 rad
$ 1 rev '
− 12.6 rad
ω=
= −6.63 rad s
1.90 s
Angular Velocity and Angular Acceleration
INSTANTANEOUS ANGULAR VELOCITY
Δθ
ω = lim ω = lim
Δt →0
Δt →0 Δt
Angular Velocity and Angular Acceleration
Changing angular velocity means that an angular
acceleration is occurring.
DEFINITION OF AVERAGE ANGULAR ACCELERATION
Change in angular velocity
Average angular acceleration =
Elapsed time
ω − ωo Δω
α =
=
t − to
Δt
SI Unit of Angular acceleration: radian per second squared (rad/s2)
Angular Velocity and Angular Acceleration
Example: A Jet Revving Its Engines
As seen from the front of the
engine, the fan blades are
rotating with an angular
speed of -110 rad/s. As the
plane takes off, the angular
velocity of the blades reaches
-330 rad/s in a time of 14 s.
Find the angular acceleration, assuming it to
be constant.
Angular Velocity and Angular Acceleration
α
(
− 330 rad s )− (− 110 rad s )
=
= −16 rad
14 s
s
2
The Equations of Rotational Kinematics
Recall the equations of kinematics for constant acceleration.
Five kinematic variables:
v = vo + at
1. displacement, x
2. acceleration (constant), a
3. final velocity (at time t), v
x=
2
1
2
(vo + v )t
2
o
v = v + 2ax
4. initial velocity, vo
5. elapsed time, t
x = vot + at
1
2
2
The Equations of Rotational Kinematics
The equations of rotational kinematics for constant
angular acceleration:
ANGULAR ACCELERATION
ANGULAR VELOCITY
ω = ωo + α t
θ=
ANGULAR DISPLACEMENT
2
1
2
(ωo + ω )t
2
o
ω = ω + 2αθ
θ = ωo t + α t
1
2
2
TIME
The Equations of Rotational Kinematics
The Equations of Rotational Kinematics
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables.
4. Verify that the information contains values for at least three
of the five kinematic variables. Select the appropriate equation.
5. When the motion is divided into segments, remember that
the final angular velocity of one segment is the initial velocity
for the next.
6. Keep in mind that there may be two possible answers to a
kinematics problem.
The Equations of Rotational Kinematics
Example: Blending with a Blender
The blades are whirling with an
angular velocity of +375 rad/s when
the “puree” button is pushed in.
When the “blend” button is pushed,
the blades accelerate and reach a
greater angular velocity after the
blades have rotated through an
angular displacement of +44.0 rad.
The angular acceleration has a
constant value of +1740 rad/s2.
Find the final angular velocity of the blades.
The Equations of Rotational Kinematics
θ
α
ω
ωo
+44.0 rad
+1740 rad/s2
?
+375 rad/s
t
ω 2 = ωo2 + 2αθ
2
o
ω = ω + 2αθ
=
2
2
375rad
s
+
2
1740
rad
s
(
) (
) ( 44.0rad) = +542 rad s
Angular Variables and Tangential Variables
The relationship between the (tangential) arc length, s,
at some radius, r, and the angular displacement, θ,
has been shown to be
s = rθ (θ in radians)
Let’s find other relationships between the angular
and tangential variables.
Consider skaters “cracking-the-whip” :

v T = tangential velocity
vT = tangential speed
Angular Variables and Tangential Variables
!θ $
s rθ
vT = = = r # &
"t%
t t
vT = rω (ω in rad/s)
ω=
θ
t
Angular Variables and Tangential Variables
The tangential acceleration, aT, is the change
of the tangential velocity per time:
vT − vTo (rω )− (rωo )
ω − ωo
aT =
=
=r
t
t
t
aT = rα
2
(α in rad/s )
ω − ωo
α=
t
Angular Variables and Tangential Variables
Example: A Helicopter Blade
A helicopter blade has an angular speed of 6.50 rev/s and an
angular acceleration of 1.30 rev/s2.
For point 1 on the blade, find
the magnitude of (a) the
tangential speed and (b) the
tangential acceleration.
Angular Variables and Tangential Variables
rev #& 2π rad #
&
ω = $ 6.50
!$
! = 40.8 rad s
s "% 1 rev "
%
vT = rω = (3.00 m ) ( 40.8rad s) = 122 m s
Angular Variables and Tangential Variables
rev #& 2π rad #
&
2
α = $1.30 2 !$
! = 8.17 rad s
s "% 1 rev "
%
aT = rα = (3.00 m ) (8.17rad s2 ) = 24.5m s2
Centripetal Acceleration and Tangential Acceleration
In uniform circular motion, the only acceleration
present is the centripetal acceleration.
2
rω )
v
(
ac = =
= rω 2
r
r
2
T
(ω in rad/s)
In nonuniform circular motion, there are both a
centripetal and a tangential acceleration.
Centripetal Acceleration and Tangential Acceleration
Example: A Discus Thrower
Starting from rest, the thrower
accelerates the discus to a final
angular speed of +15.0 rad/s in
a time of 0.270 s before releasing it.
During the acceleration, the discus
moves in a circular arc of radius
0.810 m.
Find the magnitude of the total
acceleration of the discus just
before the discus is released.
Centripetal Acceleration and Tangential Acceleration
2
ac = rω = ( 0.810 m ) (15.0 rad s)
2
= 182 m s2
" 15.0 rad s %
ω − ω0
aT = rα = r
= ( 0.810 m ) $
'
# 0.270 s &
t
= 45.0 m s2
a = aT2 + ac2 =
2
2
2
182
m
s
+
45.0
m
s
=
187m
s
(
) (
)
φ = tan-1 aT/ac = tan-1 (45.0)/(182) = 13.9o
Rolling Motion
Consider a car moving with
a linear velocity, v.
The tangential speed of a
point on the outer edge of
the tire is equal to the speed
of the car over the ground.
v = vT = rω
Also, the tangential acceleration
of a point on the outer edge of
the tire is equal to the acceleration
of the car over the ground.
a = aT = r α
Rolling Motion
Example: An Accelerating Car
Starting from rest, the car accelerates
for 20.0 s with a constant linear
acceleration of 0.800 m/s2. The
radius of the tires is 0.330 m.
What is the angle through which
each wheel has rotated?
Rolling Motion
a 0.800 m s 2
α= =
= 2.42 rad s 2
r
0.330 m
θ
α
?
-2.42 rad/s2
θ = ωo t + α t
1
2
θ=
1
2
ω
è
= -2.42 rad/s2
since the wheels go clockwise
ωo
t
0 rad/s
20.0 s
2
(−2.42 rad s )(20.0 s)
2
2
α
= −484 rad