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# Kinematics of a Particle: Impulse and Momentum

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### Lecture Notes for Section 15-5~7

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ANGULAR MOMENTUM, MOMENT OF A FORCE AND PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM

Today’s Objectives:

Students will be able to:

1. Determine the angular

momentum of a particle and apply the principle of angular impulse & momentum.

2. Use conservation of angular momentum to solve problems.

In-Class Activities:

Check Homework

Applications

Angular Momentum

Angular Impulse and Momentum Principle

Conservation of Angular Momentum

Concept Quiz

Group Problem Solving

Attention Quiz

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1. Select the correct expression for the angular momentum of a particle about a point.

A) r × v B) r × (m v)

C) v × r D) (m v) × r

2. The sum of the moments of all external forces acting on a particle is equal to

A) angular momentum of the particle.

B) linear momentum of the particle.

C) time rate of change of angular momentum.

D) time rate of change of linear momentum.

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APPLICATIONS

Planets and most satellites move in elliptical orbits. This motion is caused by gravitational attraction forces. Since

these forces act in pairs, the sum of the moments of the forces acting on the system will be zero. This means that angular momentum is conserved.

If the angular momentum is constant, does it mean the linear momentum is also constant? Why or why not?

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APPLICATIONS (continued)

The passengers on the amusement-park ride experience conservation of angular momentum about the axis of rotation (the z-axis). As shown on the free body diagram, the line of action of the normal force, N, passes through the z-axis and the weight’s line of action is parallel to it.

Therefore, the sum of moments of these two forces about the z-axis is zero.

If the passenger moves away from the z- axis, will his speed increase or decrease?

Why?

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ANGULAR MOMENTUM (Section 15.5)

The angular momentum of a particle about point O is

defined as the “moment” of the particle’s linear momentum about O.

i j k

Ho = r × mv = rx ry rz mvx mvy mvz

The magnitude of Ho is (Ho)z = mv d

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RELATIONSHIP BETWEEN MOMENT OF A FORCE AND ANGULAR MOMENTUM

(Section 15.6)

The resultant force acting on the particle is equal to the time rate of change of the particle’s linear momentum. Showing the time derivative using the familiar “dot” notation results in the equation

F = L = mv

We can prove that the resultant moment acting on the particle about point O is equal to the time rate of change of the

particle’s angular momentum about point O or

Mo = r × F = H o

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PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM (Section 15.7)

This equation is referred to as the principle of angular impulse and momentum. The second term on the left side,  Mo dt, is called the angular impulse. In cases of 2D motion, it can be applied as a scalar equation using components about the z-axis.

Considering the relationship between moment and time rate of change of angular momentum

Mo = Ho = dHo/dt

By integrating between the time interval t1 to t2

2

1

)1 2 (

) (

t

t

Ho Ho

dt

Mo (Ho )1

###  

2 (Ho )2

1

t

t

dt Mo

+

or

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CONSERVATION OF ANGULAR MOMENTUM

When the sum of angular impulses acting on a particle or a system of particles is zero during the time t1 to t2, the

angular momentum is conserved. Thus, (HO)1 = (HO)2

An example of this condition occurs when a particle is subjected only to a central force. In the figure, the force F is always directed toward point O. Thus, the angular impulse of F about O is always zero, and angular momentum of the particle about O is conserved.

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EXAMPLE

Given:A satellite has an elliptical orbit about earth.

msatellite = 700 kg

mearth = 5.976 × 1024 kg vA = 10 km/s

rA = 15 × 106 m

A = 70°

Find: The speed, vB, of the satellite at its closest distance, rB, from the center of the earth.

Plan: Apply the principles of conservation of energy and conservation of angular momentum to the system.

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EXAMPLE (continued)

Solution:

Conservation of energy: TA + VA = TB + VB becomes ms vA2 = ms vB2

where G = 66.73×10-12 m3/(kg·s2). Dividing through by ms and substituting values yields:

1 G ms me 1 G ms me

2 rA 2 rB

rB

1024)

-12(5.976 × 10

73 × . v 66

5 . 0 106

x 15

1024)

-12(5.976 × 10

73 × .

2 66 ) 000 , 10 ( 5 . 0

2 B

23.4 × 106 = 0.5 (vB)2 – (3.99 × 1014)/rB or

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### EXAMPLE (continued)

Solving the two equations for rB and vB yields rB = 13.8 × 106 m vB = 10.2 km/s

Now use Conservation of Angular Momentum.

(rA ms vA) sin A = rB ms vB

(15 × 106)(10,000) sin 70° = rB vB or rB = (140.95 × 109)/vB Solution:

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CONCEPT QUIZ

1. If a particle moves in the x - y plane, its angular momentum vector is in the

A) x direction. B) y direction.

C) z direction. D) x - y direction.

2. If there are no external impulses acting on a particle 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.

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GROUP PROBLEM SOLVING

Given: The four 5 lb spheres are rigidly attached to the

crossbar frame, which has a negligible weight.

A moment acts on the shaft as shown, M = 0.5t + 0.8 lb·ft).

Find: The velocity of the spheres after 4 seconds, starting from rest.

Plan:

Apply the principle of angular impulse and momentum about the axis of rotation (z-axis).

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### GROUP PROBLEM SOLVING (continued)

Angular momentum: HZ = r × mv reduces to a scalar equation.

(HZ)1 = 0 and (HZ)2 = 4×{(5/32.2) (0.6) v2} = 0.3727 v2

Apply the principle of angular impulse and momentum.

0 + 7.2 = 0.3727 v2  v2 = 19.4 ft/s Solution:

Angular impulse:

M dt = (0.5t + 0.8) dt = [(0.5/2) t

### 

2 2 + 0.8 t] = 7.2 lb·ft·s

1

t

t

### 

2

1

t

t 0

4

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ATTENTION QUIZ

1. A ball is traveling on a smooth surface in a 3 ft radius circle with a speed of 6 ft/s. If the attached cord is pulled down with a constant speed of 2 ft/s, which of the following

principles can be applied to solve for the velocity of the ball when r = 2 ft?

A) Conservation of energy

B) Conservation of angular momentum C) Conservation of linear momentum D) Conservation of mass

2. If a particle moves in the z - y plane, its angular momentum vector is in the

A) x direction. B) y direction.

C) z direction. D) z - y direction.

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t

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O

O O

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1

1

c

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2

1

2

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B

### Example 15.14

1 2

1 1 2 2

2 2

(0.5)(0.4)(1.2) (0.2)(0.4) 3m/s

B B

rm v r m v

v v

 

 

  

H H

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2 2

2

### v   

1 1 2 2

2 2

1 1

(0.4)(1.2) (0.4)(3.606) 2.313

2 F 2 F

T U T

U U J

   

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A B

dmvmv

Fdt dmvmv

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( B A) dm

dt

F v v

( )

( )

x Bx Ax

y By Ay

F dm v v dt

F dm v v dt

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O OB B OA A

### +

A A A B B B A A B B

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O OB B OA A

## 

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3

3

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B

A

2

### Example 15.16

2

200 kg/s

0.2 25.46 m/s ; 6.37 m/s (0.05)

w

B A

B A

dm Q

dt

Q Q

v v

A A

3 2

[100(10 )][ (0.1) ] 3141.6N

A A A

Fp A

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### Example 15.16

( );

3141.6 200(0 6.37) 4.41

( );

20(9.81) 200( 25.46 0) 4.90

x Bx Ax

x

x

y By Ay

y

F dm v v dt

F

F kN

F dm v v dt

Fy

F kN

 

   

   

    

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x

y

A

O OB B OA A

O

O

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e

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D/e

e

### ),

( )( ) ( )

e e CV e e e e

CV e e e e

mv m v F dt m dm v dv m dm v F dt v dm m dv dm dv v dm

      

    

/ e

CV D e

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### • We may write this equation as

( )( ) ( )

i i CV i i i i

mv m v F dt m dm v dv m dm v

 

    

i

CV i

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0

e

(39)

/ e

s D e

0 0

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### Example 15.18

0 0 0

0 0

0 0

ln( ) ln

v t uc t m

dv g dt v u m ct gt u gt

m ct m ct

  

###  

/

f e f

m dm t ct t m c dt

    

     

0 max

0

ln f

f

m gm

v u

m m c

 

    

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### CH015 HW

End section 15-1

_01: 15-2, 15-8, 15-13, 15-20

End section 15-2

_02: 15-31, 15-48, 15-64

End section 15-3

_03: 15-82, 15-88

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