CEE 271: Applied Mechanics II, Dynamics

– Lecture 33: Ch.19, Sec.1–3 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Date: __________________

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LINEAR AND ANGULAR MOMENTUM,

PRINCIPLE OF IMPULSE AND MOMENTUM

Today’s objectives: Students

will be able to

1 Develop formulations for

the linear and angular

momentum of a body.

2 Apply the principle of linear

and angular impulse and

momentum.

In-class activities:

• Reading Quiz

• Applications

• Linear and Angular

Momentum

• Principle of Impulse and

Momentum

• Concept Quiz

• Group Problem Solving

• Attention Quiz

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READING QUIZ

1 The angular momentum of a rotating two-dimensional rigidbody about its center of mass G is .

(a) mvG(b) IGvG(c) mω(d) IGω

ANS: (d)

2 If a rigid body rotates about a fixed axis passing through itscenter of mass, the body’s linear momentum is .

(a) a constant

(b) zero

(c) mvG(d) IGω

ANS: (b)

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APPLICATIONS

• The swing bridge opens and

closes by turning using a

motor located under the

center of the deck at A that

applies a torque M to the

bridge.

• If the bridge was supported 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 as shown?

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PROCEDURE FOR ANALYSIS

1 Establish the x, y, z inertial frame of reference.

2 Draw the impulse-momentum diagrams for the body.

3 Compute IG, as necessary.

4 Apply the equations of impulse and momentum (one vector

and one scalar or the three scalar equations).

5 If more than three unknowns are involved, kinematic

equations relating the velocity of the mass center G and

the angular velocity ω should be used to furnish additional

equations.

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EXAMPLE

• Given: The 300 kg wheel has a

radius of gyration about its mass

center O of kO = 0.4meter. The

wheel is subjected to a couple

moment of M = 300 N ·m.

• 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.

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EXAMPLE (Solution)

• Impulse-momentum diagram:

+ =

Wt

Ft

IGω2

mvG2

Nt

Mt

mvG1

IGω1

• Impulse and Momentum using kinematics: (vG)2 = rω2:

(HA)1 +Σ

∫

t2

t1

MAdt = (HA)2

0 +Mt = m(vG)2r + IGω2

= mr2ω2 +m(kO)2ω2

= m[

r2 + (kO)2]

ω2

ω2 (t = 6 s) =300(6)

300(0.62 + 0.42)= 11.5 rad/s

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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., IGω).

(a) larger than

(b) less than

(c) the same as

(d) None of the above

ANS: (c)

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

(a) constant

(b) zero

(c) increasing linearly with time

(d) decreasing linearly with time

ANS: (b)

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ATTENTION QUIZ

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

momentum with respect to A is?

������������������

������������������A

ω

l

G

(a) 1

12ml2ω

(b) 1

6ml2ω

(c) 1

3ml2ω

(d) ml2ω

ANS: (c)

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ATTENTION QUIZ (Continued)

2. As in the principle of work and energy, if a force does nowork, it does not need to be shown on the impulse andmomentum diagram/equation.

(a) False

(b) True

(c) Depends on the case

(d) No clue!

ANS: (a)

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CONSERVATION OF MOMENTUM

Today’s objectives: Students

will be able to

1 Understand the conditions

for conservation of linear

and angular momentum.

2 Use the condition of

conservation of linear/

angular momentum.

In-class activities:

• Reading Quiz

• Applications

• Conservation of Linear and

Angular Momentum

• Concept Quiz

• Group Problem Solving

• Attention Quiz

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READING QUIZ

1 If there are no external impulses acting on a body .

(a) only linear momentum is conserved

(b) only angular momentum is conserved

(c) both linear momentum and angular momentum are

conserved

(d) neither linear momentum nor angular momentum are

conserved

ANS: (c)

2 If a rigid body rotates about a fixed axis passing through itscenter of mass, the body’s linear momentum is .

(a) a constant

(b) zero

(c) mvG(d) IGω

ANS: (b)

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APPLICATIONS

• A skater spends a lot of time either spinning on the ice or

rotating through the air. To spin fast, or for a long time, the

skater must develop a large amount of angular momentum.

• If the skater’s angular momentum is constant, can the skater

vary her rotational speed? How?

• The skater spins faster when the arms are drawn in and

slower when the arms are extended. Why?

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APPLICATIONS(continued)

• Conservation of angular momentum allows cats to land on

their feet and divers to flip, twist, spiral and turn. It also

helps teachers make their heads spin!

• Conservation of angular momentum makes

water circle the drain faster as it gets closer to the drain.

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CONSERVATION OF LINEAR MOMENTUM (19.3)

• Recall that the linear impulse and momentum relationship

is

L1 +∑

∫

t2

t1

F dt = L2

where L1 = (mvG)1 and L2 = (mvG)2.

• If the sum of all the linear impulses acting on the rigid body

(or a system of rigid bodies) is zero, all the impulse terms

are zero. Thus, the linear momentum for a rigid body (or

system) is constant, or conserved. So L1 = L2.

• This equation is referred to as the

conservation of linear momentum. The conservation of

linear momentum equation can be used if the linear

impulses are small or non-impulsive.

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CONSERVATION OF ANGULAR MOMENTUM

• Similarly, if the sum of all the angular impulses due to

external forces acting on the rigid body (or a system of

rigid bodies) is zero, all the impulse terms are zero. Thus,

angular momentum is conserved

HG1 +∑

∫

t2

t1

MGdt = HG2

where HG1 = IGω1 and HG2 = IGω2.

• The resulting equation is referred to as the conservation of

angular momentum or HG1 = HG2.

• If the initial condition of the rigid body (or system) is known,

conservation of momentum is often used to determine the

final linear or angular velocity of a body just after an event

occurs.

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GROUP PROBLEM SOLVING

• Given: Two children

(mA = mB = 30 kg) sit at the

edge of the merry-go-round,

which has a mass of 180 kg and

a radius of gyration of kz = 0.6 m.

• Find: The angular velocity of the merry-go-round if A jumps

off horizontally in the +t direction with a speed of 2 m/s,

measured relative to the merry-go-round.

• Plan: Draw an impulse-momentum diagram. The

conservation of angular momentum can be used to find the

angular velocity.

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GROUP PROBLEM SOLVING(Solution)

M

B

B

A

vA/M

AM

Figure: Impulse-

• Apply the conservation of angular

momentum equation:

∑

H1 =∑

H2

180(0.6)2(2) + 2× [(30)2(0.75)2]

= 180(0.6)2ω + (30)ω(0.75)2

+(30)(0.75ω + 2)(0.75)

Now we solve

197.1 = 98.55ω + 45

ω = 1.54 rad/s

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ATTENTION QUIZ

1. Using conservation of linear and angular momentumrequires that

(a) all linear impulses sum to zero

(b) all angular impulses sum to zero

(c) both linear and angular impulses sum to zero

(d) None of the above

ANS: (c)

2. The angular momentum of a body about a point A that isthe fixed axis of rotation but not the mass center (G) is

(a) IAω(b) IGω(c) rG(mvG) + IGω(d) Both (a) and (c)

ANS: (d)

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Note

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