Kinematics of a Particle: Impulse and Momentum

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Chapter 15

Kinematics of a Particle: Impulse and Momentum

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

Reading Quiz

Applications

Angular Momentum

Angular Impulse and Momentum Principle

Conservation of Angular Momentum

Concept Quiz

Group Problem Solving

Attention Quiz

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READING QUIZ

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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The box has a mass m and is traveling down the smooth circular ramp such that when it is at the angle θ it is a speed v. Determine its angular

momentum about point O at this instant and the rate of increase in its speed, i.e., a

t

.

Example 15.12

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Solution

Since v is tangent to the path, the angular momentum is

From the FBD, the weight W = mg contributes a moment about O

Since r and m are constant,

Example 15.12

H

O

rmv

; ( sin ) ( )

O O

M H m g r d rm v

dt

  

sin dv dv sin

mgr rm g

dt dt

    

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The 0.4 kg ball B is attached to a cord which passes through a hole at A in a smooth table. When the ball is r

1

= 0.5 m from the hole, it is rotating around in a circle such that its speed is v

1

= 1.2 m/s. By applying a force F the cord is pulled downward through the hole with a constant speed v

c

= 2 m/s.

Example 15.14

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Determine (a) the speed of the ball at the instant it is r

2

= 0.2 m from the hole, and (b) the amount of work done by F in shortening the radial distance from r

1

to r

2

.

Example 15.14

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Solution

Part (a) Free-Body Diagram

As the ball the cord force F on the ball passes through the z axis.

Weight and N

B

are parallel and the conservation of angular momentum applies about the z axis.

Conservation of Angular Momentum

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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Solution

Conservation of Angular Momentum The speed of the ball is thus

Part (b)

The only force that does work on the ball is F.

Example 15.14

2 2

2

(3.0) (2) 3.606m/s

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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Chapter 15

Kinematics of a Particle: Impulse and Momentum

Lecture Notes for Section 15-8~9

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15.8 Steady Flow of a Fluid Stream

• Consider the diversion of a steady stream of fluid by a fixed pipe

• The impulse and momentum diagrams for the

fluid are as shown below

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15.8 Steady Flow of a Fluid Stream

The force ∑F represents the resultant of all the external forces acting on the fluid stream

Since flow is steady, ∑F will be constant during the time interval dt

• Applying the principle of linear impulse and momentum to the fluid stream,

A B

dmvmv

Fdt dmvmv

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15.8 Steady Flow of a Fluid Stream

Principle of Impulse and Momentum

• Solving for the resultant force yields

• It is convenient to express vector equation in the form of two scalar component equations

( 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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15.8 Steady Flow of a Fluid Stream

Principle of Impulse and Momentum

Since flow is steady in the x-y plane, hence we have

• Once the velocity of the fluid flowing onto the device is determined, the mass flow is calculated using

( )

O OB B OA A

M dm d v d v

dt

+

A A A B B B A A B B

dm v A v A Q Q

dt        

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15.8 Steady Flow of a Fluid Stream

Procedure of Analysis Free-Body Diagram

• Draw a FBD of the device which is directing the fluid

Equations of Steady Flow

• Apply the equations of steady flow,

( )

O OB B OA A

M dm d v d v

dt

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Determine the components of reaction which the fixed pipe joint at A exerts on the elbow. If water flowing through the pipe is subjected to a static

gauge pressure of 100 kPa at A. The discharge at B is QB = 0.2 m

3

/s. Water has a density ρw = 1000kg/m

3

, and the water-filled elbow has a mass of 20 kg and center of mass at G.

Example 15.6

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Solution

Since the density of water is constant, Q

B

= Q

A

= Q,

Free-Body Diagram

Since 1 kPa = 1000 N/m

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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Solution

Equations of Steady Flow

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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Solution

Equations of Steady Flow

If moments are summed about point O, then F

x

, F

y

, and static pressure F

A

are eliminated, as well as

moment of momentum of water entering at A,

Example 15.16

( )

20(9.81)(0.125) 200[(0.3)(25.46) 0]

1.50

O OB B OA A

O

O

M dm d v d v dt

M

M kN m

 

  

 

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15.9 Propulsion with Variable Mass

A Control Volume That Loses Mass

• Consider a device such as a rocket which at an instant of time has a mass m and is moving

forward with a velocity v

The “closed system” included both the mass m of

the device and the expelled mass m

e

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15.9 Propulsion with Variable Mass

A Control Volume That Loses Mass

• Applying the principle of impulse and momentum to the system,

• The relative velocity of the device as seen by the observer moving with the particles of the ejected mass is v

D/e

= (v + v

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

dv dm

F m v

dt dt

 

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15.9 Propulsion with Variable Mass

A Control Volume That Gain Mass

• A device such as a scoop or a shovel may gain mass as it moves forward

The device shown has a mass m and is moving

forward with a velocity v

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15.9 Propulsion with Variable Mass

A Control Volume That Gain Mass

• Applying the principle of impulse and momentum to the system,

• 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

dv dm

F m v v

dt dt

  

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The initial combined mass of a rocket and its fuel is m

0

. A total mass mf of fuel is consumed as a constant rate of dm

e

/dt = c and expelled at a constant speed of u relative to the rocket. Determine the max velocity of the rocket i.e., at the instant the fuel runs out.

Example 15.18

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Solution

The rocket is losing mass as it moves upward. The only external force acting on the system consisting of the rocket and a portion of the expelled mass is the weight W

Since W = mg,

Example 15.18

/ e

;

s D e

dv dm dv

F m v W m uc

dt dt dt

       

0 0

( ) ( ) dv

m ct g m ct uc

    dt

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Solution

Separating the variables and integrating, realizing that v = 0 at t = 0, we have

The time t’ needed to consume all the fuel is given by

By sub, we get

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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