LINEAR AND ANGULAR MOMENTUM, PRINCIPLE

OF IMPULSE AND MOMENTUM

Today’s Objectives:
Students will be able to:
1. Develop formulations for the linear and
In-Class Activities:
angular momentum of a body.
2. Apply the principle of linear and angular • Check Homework
impulse and momentum. • Reading Quiz
• Applications
• Linear and Angular
Momentum
• Principle of Impulse and
Momentum
• Concept Quiz
• Group Problem Solving
• Attention Quiz

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 READING QUIZ

1. The angular momentum of a rotating two-dimensional rigid

body about its center of mass G is ___________.
A) m vG B) IG vG
C) mw D) IG w

2. If a rigid body rotates about a fixed axis passing through its

center of mass, the body’s linear momentum is __________.
A) a constant B) zero
C) m vG D) IG w

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 APPLICATIONS

A swing bridge opens and

closes by turning, using a
motor located at A under the
center of the deck that
applies a torque M to the
bridge.

If the bridge was supported by and rotated about at its end B,

would the same torque open the bridge in the same time, or
would it open slower or faster?
What are the benefits of making the bridge with the variable
depth (thickness) substructure shown?

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 APPLICATIONS (continued)

As the pendulum of the Charpy tester swings downward, its

angular momentum and linear momentum both increase. By
calculating its momenta in the vertical position, we can
calculate the impulse the pendulum exerts when it hits the test
specimen.
As the pendulum rotates about point O, what is its angular
momentum about point O?
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
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 APPLICATIONS (continued)

The space shuttle, now retired from NASA’s fleet, has

several engines that exerted thrust on the shuttle when they
were fired. By firing different engines, the pilot could
control the motion and direction of the shuttle.

If only engine A was fired, about which axis did the shuttle
tend to rotate?

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

R.C. Hibbeler All rights reserved.
LINEAR AND ANGULAR MOMENTUM (Section 19.1)

The linear momentum of a rigid body is defined as

L = m vG
This equation states that the linear momentum vector L has a
magnitude equal to (mvG) and a direction defined by vG.

The angular momentum of a rigid

body is defined as
HG = I G w
Remember that the direction of
HG is perpendicular to the plane
of rotation.

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 LINEAR AND ANGULAR MOMENTUM (continued)

Translation
When a rigid body undergoes
rectilinear or curvilinear
translation, its angular momentum
is zero because w = 0.

Therefore,
L = m vG
and
HG = 0

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 LINEAR AND ANGULAR MOMENTUM (continued)

Rotation about a fixed axis

When a rigid body is rotating
about a fixed axis passing
through point O, the body’s
linear momentum and angular
momentum about G are:
L = mvG
HG = IG w

It is sometimes convenient to compute the angular momentum

of the body about the center of rotation O.
HO = ( rG × mvG) + IG w = IO w
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
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 LINEAR AND ANGULAR MOMENTUM (continued)

General plane motion

When a rigid body is subjected to
general plane motion, both the linear
momentum and the angular momentum
computed about G are required.
L = m vG
HG = IG w
The angular momentum about point A is
HA = IGw + mvG (d)

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 PRINCIPLE OF IMPULSE AND MOMENTUM
(Section 19.2)

As in the case of particle motion, the principle of impulse

and momentum for a rigid body is developed by
combining the equation of motion with kinematics. The
resulting equations allow a direct solution to problems
involving force, velocity, and time.
Linear impulse-linear momentum equation:
t2 t2
L1 + å òF dt = L2 or (mvG)1+ åò F dt = (mvG)2
t1 t1

Angular impulse-angular momentum equation:

t2 t2
(HG)1 + åò MG dt = (HG)2 or IGw1 + åòMG dt = IGw2
t1 t1

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 PRINCIPLE OF IMPULSE AND MOMENTUM
(continued)
The previous relations can be represented graphically by
drawing the impulse-momentum diagram.

To summarize, if motion is occurring in the x-y plane, the

linear impulse-linear momentum relation can be applied to
the x and y directions and the angular momentum-angular
impulse relation is applied about a z-axis passing through
any point (i.e., G). Therefore, the principle yields three
scalar equations describing the planar motion of the body.
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
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 PROCEDURE FOR ANALYSIS

• Establish the x, y, z-inertial frame of reference.

• Draw the impulse-momentum diagrams for the body.

• Compute IG, as necessary.

• Apply the equations of impulse and momentum (one vector

and one scalar or the three scalar equations).

• If more than three unknowns are involved, kinematic equations

relating the velocity of the mass center G and the angular
velocity w should be used to furnish additional equations.

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 EXAMPLE

Given: The 300 kg wheel has a

radius of gyration about its mass
center O of kO = 0.4 m.
The wheel is subjected to a
couple moment of 300 Nm.
A

Find: The angular velocity after 6 seconds if it starts from rest

and no slipping occurs.

Plan: Time as a parameter should make you think Impulse and

Momentum! Since the body rolls without slipping, point
A is the center of rotation. Therefore, applying the
angular impulse and momentum relationships along with
kinematics should solve the problem.
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
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 EXAMPLE (continued)

Solution: y
Impulse-momentum x
diagram: Mt Wt
(m vG)1 (m vG)2
G + r G
A
=
IGw1 Ft IG w2

Kinematics: (vG)2 = r w2 Nt
t2
Impulse & Momentum: (HA)1 + åò MA dt = (HA)2
t1

0 + M t = m(vG)2 r + IG w2 = m r2 w2 + m(kO)2 w2 = m{r2 + (kO)2}w2

Mt 300 (6)
w2 = = = 11.5 rad/s
m {r2 + (kO)2} 300 {0.62 + 0.42}
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
R.C. Hibbeler All rights reserved.
 CONCEPT QUIZ

1. If a slab is rotating about its center of

mass G, its angular momentum about
any arbitrary point P is __________ its
angular momentum computed about G
(i.e., IG w).
A) larger than B) less than
C) the same as D) None of the above.

2. The linear momentum of the slab in question 1 is

__________.
A) constant B) zero
C) increasing linearly D) decreasing linearly
with time
Dynamics, Fourteenth Edition
with time
Copyright ©2016 by Pearson Education, Inc.
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 GROUP PROBLEM SOLVING

Given: A gear set with:

WA = 15 lb
WB = 10 lb
kA = 0.5 ft
kB = 0.35 ft
M = 2(1 – e-0.5t) ft·lb
Find: The angular velocity of gear A after 5 seconds if the
gears start turning from rest.
Plan: Time is a parameter, thus Impulse and Momentum is
recommended. First, relate the angular velocities of the
two gears using kinematics. Then apply angular impulse
and momentum to both gears.
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
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 GROUP PROBLEM SOLVING (continued)
Solution:
Impulse-momentum diagrams: Note that the initial momentum is
zero for both gears.
Gear A: W At y

Axt rA x
A
Ayt =
IA w A
Ft
Gear B: Ft WB t
IB w B

t2 Bx t
= rB

òt M dt By t
1

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 GROUP PROBLEM SOLVING (continued)

Kinematics: rA wA = rB wB

Angular impulse & momentum relation for gear A about point

A yields:

(F t) rA = IA wA or F t = IA wA/rA
t2

For gear B:
òt M dt – (F t) rB = IB wB
1

t2

Combining the two equations: ò M dt = (rB/rA) IA wA + IB wB

t1

Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.

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 GROUP PROBLEM SOLVING (continued)

Substituting from kinematics for wA = (rB/rA) wB, yields

t2

t2 5
òM
t1
dt = wB [(r B/r A ) 2
IA + IB ]

where òM dt = ò2(1 – e-0.5t) dt = 2t + 4e-0.5t

t 0
1
= (10.328) – (4) = 6.328 ft·lb·s

IA = mA (kA)2 = (15/32.2)(0.5)2 = 0.116 slug·ft2

IB = mB (kB)2 = (10/32.2)(0.35)2 = 0.038 slug·ft2
6.328
Therefore, wB = = 75.9 rad/s
0.5 2
[(
) (0.116) + (0.038)]
0.8
and wA = (0.5/0.8)(75.9) = 47.4 rad/s
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
R.C. Hibbeler All rights reserved.
 ATTENTION QUIZ

1. If a slender bar rotates about end A, its A

angular momentum with respect to A is? w

A) (1/12) m l2 w B) (1/6) m G
l2 w l
C) (1/3) m l2 w D) m l2 w

2. As in the principle of work and energy, if a force does no

work, it does not need to be shown on the impulse and
momentum diagram/equation.
A) False B) True
C) Depends on the case. D) No clue!
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
R.C. Hibbeler All rights reserved.
Dynamics, Fourteenth Edition Copyright ©2016 by Pearson Education, Inc.
R.C. Hibbeler All rights reserved.