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Outline
CONSERVATION OF
MOMENTUM
Ozkaya and Nordin
p. 317-335
Newton’s second law of motion
Linear motion: If a non-zero force F acts on a rigid body of
mass m, the body’s centre of gravity will have an
acceleration aCG, according to the law F=maCG.
Angular motion: If there exists a non-zero moment MCG
about a body’s centre of gravity, the body will have an
angular acceleration !, according to the law MCG=ICG!,
where ICG is the moment of inertia of the body about its
centre of gravity.
ma
! Fx = m(aCG ) x
! Fy = m(aCG ) y
! MCG = ICG"
F2
CG
F1
• Newton’s equations of motion (review)
• impulse-momentum
• conservation of linear momentum
• conservation of angular momentum
• moment of inertia
Newton’s second law is the law of
conservation of momentum
conservation of linear and angular momentum
d
(mvx )
dt
d
Fy = ( mv y )
dt
Fx =
MCG =
law of impulse-momentum
t2
(
!t
t
!t Fy dt = m(vy
1
Fx dt = m vx 1 " vx 2
2
ICG!
F3
F4
d
(I ! )
dt CG
1
1
" vy 2
t2
) !t
)
1
Mdt = ICG (# 1 " # 2 )
Ground reaction forces mirror COG
acceleration
vertical GRF >BW: upward
acceleration or downward
deceleration (of magnitude
((Fy/(body mass)-g))
vertical GRF >BW:
downward acceleration or
upward deceleration
3
COG
vertical
position (m)
Walk
Run
Vertical Force Fy
(BW)
Ground reaction forces
(measured from a force
plate) mirror acceleration of
the whole-body centre of
gravity.
Ground reaction forces during walking
COG
vertical
acceleration
(m/s2)
2
RHS
RHS
0
0.0
1 upward acceleration
2 upward deceleration
200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time (seconds)
Activity
Walking
Running
VB Spike Landing
Somersault Landing
Peak Vertical GRF
1.1 BW
2.8 BW
4.8 BW
13.6 BW
Ground reaction forces during running
high-frequency
“heel strike
transient” (occurs
during running
but not walking)
LHS RTO
1
3 downward acceleration
1
Foot Contact
150
Forces
(%Body Weight)
100
Vertical
Med (+) - Lat (-)
50
Post (+) - Ant (-)
2
4
3
4 downward deceleration
5 forward braking
6 forward propulsion
5
0
6
-50
0
25
50
75
100
% Gait Cycle
Law of impulse-momentum
Impulse applied
to the system
between instants
1 and 2
t2
=
(
t
!t Fy dt = m(vy
!t
1
Change in
momentum of
the system
Fx dt = m vx 1 " vx 2
2
1
t2
!t
1
1
" vy 2
)
)
Mdt = ICG (# 1 " # 2 )
Computing the impulse (a)
Computing the impulse (b)
force (N)
force (N)
3
3
2
2
1
1
1
2
3
4
5
6
7
time (s)
impulse =
1
2
3
4
5
6
7
time (s)
impulse =
Ground reaction forces during vertical
jumping
Computing the impulse (c)
vert. GRF
800 N
time
Each box:
100 N
represents (100 N)(0.05 s) = 5 Ns
0.05 s
“squat” jump
impulse =
“countermovement”
jump
Conservation of linear momentum
If the net force applied to a system
is zero, then the second law states :
d
d
(mvx ) = 0
(mvy ) = 0
dt
dt
and the momentum in the x and y
directions is constant (conserved) :
(mvx ) = constant = px
(mvy ) = constant = py
Newton says: if
there’s no force,
there’s no
change in
momentum.
Therefore, if m
increases, v must
decrease.
This is the law of conservation of
linear momentum :
m1v x1 = m 2v x2
Linear momentum has
units of [kg][m][s-1]
m1v y1 = m 2v y2
Conservation of angular momentum
If the net moment applied to a system
is zero, then the second law states
d
(I ! ) = 0
dt CG
and the angular momentum is
constant (conserved) :
( ICG! ) = constant = H
This is the law of conservation of
angular momentum :
( ICG )1! 1 = ( ICG ) 2! 2
If I increases, ! must decrease, and vice - versa.
v
-v
2v
m
m
m
0
m
v
v=0
m
m
m
v
-v/2
m
2m
v=0
3m
Moment of Inertia
The moment of inertia I of an
object about a specific axis is
2
I = ! mi ( ri ) = " r 2 dm
i
where r is the distance from the
mass mi (or mass element dm)
to the axis of interest. For the
Y
object shown at left :
2
I X = ! mi ( yi ) = " y 2dm
i
2
IY = ! mi ( xi ) = " x 2 dm
i
m
3v/2
0
m
2m
v/2
3m
The moment of inertia reflects
that a body has an inertia for
turning that depends not only
on the masses, but also on
how far away they are from
the axis of rotation.
X
Clearly, Ix > Iy
in this example
Moment of inertia has
units of [kg][m2]
Example: computing moment of inertia
Calculate the moment of inertia of the object below
(a) about the axis y1 (which passes through the CG
of the system), and (b) about the y2 axis.
y2
y1
20 kg
10 kg
0.5 m
"miri2
1.0 m
Iy1 =
= (20 kg)(0.5
= 15.0 kg·m 2
m)2
+ (10 kg)(1.0
m)2
Iy2 = "miri2 = (20 kg)(1.5 m)2 + (10 kg)(0 m)2
= 45.0 kg·m 2
Example: Parallel Axis Theorem
Same as last problem. However, calculate the
moment of inertia about the y2 axis using the
parallel axis theorem.
y2
y1
20 kg
10 kg
0.5 m
1.0 m
Iy1 = ICG = "miri2 = (20 kg)(0.5 m)2 + (10 kg)(1.0 m)2
= 15.0 kg·m2
Iy2 = I = ICG + mTOTAL(rCG)2 = 15.0 kg·m2 + (30 kg)(1.0 m)2
= 45.0 kg·m 2
The Parallel Axis Theorem
If the moment of inertia about an axis passing
through the center of gravity is known, the
moment of inertia about a second, parallel axis is
given by:
I = ICG + mTOTAL(rCG)2
where
ICG = MOI about axis passing through the CG
mTOTAL = total mass of the object
rCG = perpendicular distance between axes
Conservation of angular momentum
during free-flight
During free-flight, the
net moment about the
body’s COG is zero.
Therefore, while ICG
and " vary
considerably, the
product H= ICG " must
remain constant.
Conservation of angular momentum
during a long jump
avoiding
forward
rotation
of the
trunk
after
take-off
late
stage
preparation
for landing
Conservation of angular momentum
during a cat flip
The cat appears to defy
the law of conservation
of angular momentum,
by rotating forward and
hind segments in
different directions, and
appropriately drawing
the legs in an out to
modify the MOI of each
segment.
GRF
Angular momentum is conserved, but
energy is not.
A classic example of
conservation of angular
momentum is the figure skater
pulling her arms in (and
decreasing her ICG) to increase
her ". Angular momentum is
conserved, but energy increases.
Where does the energy come
from? (HINT: think about the
centrifugal force acting to pull
her hands outward)
Momentum is a vector, and must be
conserved about a given axis
Another classic example of
conservation of angular
momentum. When the
spinning wheel is moved so its
axis of rotation is vertical, the
frictionless turntable spins in a
direction opposite to the wheel
(conserving angular
momentum about the vertical
axis).
Review Questions
Under what conditions is a body’s linear momentum
constant? Angular momentum?
! Can you compute a linear impulse? An angular impulse?
! When angular momentum is constant, what is the effect of
changing moment of inertia?
! What influences the moment of inertia of the human body?
! Can you compute moment of inertia?
! Can you apply the parallel axis theorem?
! Can you apply the principles of conservation of linear
momentum/ conservation of angular momentum?
! What is the “cat twist” technique?
! How does a (long jumper, gymnast, volleyball player) control
movement while in flight?
!
The gymnast shown in the figure
at right loses balance while
performing on the beam, and
begins to rotate counterclockwise
(CCW). (a) Mark on the diagram
(with arrows) the direction she
should move her arms to help her
regain balance. (b) Why does
this work? Explain the physics
underlying this phenomenon.
Example problem: cycling around a
turn
Example problem: moment of
inertia
Show that a narrow rod with
uniformly - distributed mass
(often used to model body
segments) has (a) a moment
of inertia about its center of
1
gravity equal to I CG = ml 2 ,
12
and (b) a moment of inertia
1
about its end equal to I0 = ml 2 .
3
Example problem: conservation
of angular momentum
0
CG
l
Consider a cyclist of body mass m going
around a turn of radius r, with constant
angular velocity ". (a) Identify the
magnitude and direction of the cyclist’s
velocity and acceleration. (b) Mark on a
free body diagram the centrifugal force
applied to the cyclist’s centre of gravity,
and the associated centripetal force applied
to the tires of the bike. (c) Explain, using a
free body diagram, why cyclists must lean
into a corner to keep their balance. (d)
Explain, again with a free body diagram,
why one should be able to cycle faster
around banked corners.
Conservation of angular momentum
during free-flight (cont)
Estimate the change in angular
velocity between (b) and (d) below,
given the ICG values shown at left.
ICG = 3.5 kg·m2
ICG = 6.5 kg·m2
ICG = 15.0 kg·m2
IHBAR = 85.0 kg·m2