### Chapter 15

### Kinematics of a Particle: Impulse and Momentum

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

**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

**READING QUIZ**

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

A) * r* ×

*B)*

**v***× (m v)*

**r**C) * v* ×

*D) (m v) ×*

**r**

**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.

**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?

**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?

**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**

**H**_{o} = r × mv = r_{x} r_{y} r_{z}
mv_{x} mv_{y} mv_{z}

The magnitude of H_{o} is (H_{o})_{z} = mv d

**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

M_{o} = r × * F* = H

^{}

_{o}

**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, M_{o} 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

M_{o} = H^{}_{o} = dH_{o}/dt

By integrating between the time interval t_{1} to t_{2}

###

^{2}

^{}

^{}

1

)1 2 (

) (

t

t

* H*o

*o*

**H**dt

**M**_{o} (^{H}^{o} )_{1}

###

^{2}(

^{H}^{o})

_{2}

1

t

t

dt
**M**_{o}

**+**

or

**CONSERVATION OF ANGULAR MOMENTUM**

When the sum of angular impulses acting on a particle or a
system of particles is zero during the time t_{1} to t_{2}, the

angular momentum is conserved. Thus,
(H_{O})_{1} = (H_{O})_{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.

**EXAMPLE **

**Given:**A satellite has an elliptical
orbit about earth.

m_{satellite} = 700 kg

m_{earth} = 5.976 × 10^{24} kg
v_{A} = 10 km/s

r_{A} = 15 × 10^{6} m

_{A} = 70°

**Find:** The speed, v_{B}, of the satellite at its closest distance,
r_{B}, from the center of the earth.

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

**EXAMPLE **
(continued)

**Solution:**

Conservation of energy: T_{A} + V_{A} = T_{B} + V_{B }becomes
m_{s} v_{A}^{2} – = m_{s} v_{B}^{2} –

where G = 66.73^{×}10^{-12} m^{3}/(kg·s^{2}). Dividing through by m_{s }and
substituting values yields:

1 G m_{s} m_{e} 1 G m_{s} m_{e}

2 r_{A} 2 r_{B}

r_{B}

10^{24})

-12(5.976 × 10

73 × . v 66

5
.
0
10^{6}

x 15

10^{24})

-12(5.976 × 10

73 × .

2 66 ) 000 , 10 ( 5 . 0

2 B

23.4 ^{×} 10^{6} = 0.5 (v_{B})^{2} – (3.99 ^{×} 10^{14})/r_{B}
or

**EXAMPLE ** (continued)

Solving the two equations for r_{B} and v_{B} yields
r_{B} = 13.8 × 10^{6} m v_{B} = 10.2 km/s

Now use Conservation of Angular Momentum.

(r_{A} m_{s} v_{A}) sin _{A} = r_{B} m_{s} v_{B}

(15 × 10^{6})(10,000) sin 70° = r_{B} v_{B} or r_{B} = (140.95 × 10^{9})/v_{B}
**Solution:**

**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.

**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).

**GROUP PROBLEM SOLVING ** (continued)

Angular momentum: **H**_{Z} = r × mv reduces to a scalar equation.

(H_{Z})_{1} = 0 and (H_{Z})_{2} = 4×{(5/32.2) (0.6) v_{2}} = 0.3727 v_{2}

Apply the principle of angular impulse and momentum.

0 + 7.2 = 0.3727 v_{2 } v_{2} = 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

**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.

### 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

### 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*

###

*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

### 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**

**done by F in shortening the radial distance from r**

_{1}

### to *r*

_{2}

### .

### Example 15.14

### Solution

**Part (a) Free-Body Diagram**

**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**

**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**

### Solution

**Conservation of Angular Momentum** The speed of the ball is thus

**Conservation of Angular Momentum**

**Part (b)**

**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*

###

### Chapter 15

### Kinematics of a Particle: Impulse and Momentum

### Lecture Notes for Section 15-8~9

### 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

### 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*

**Since flow is steady, ∑F will be constant during**

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

*A* *B*

*dm*** ^{v}**

*m*

****

^{v}###

^{F}*dt dm*

****

^{v}*m*

^{v}### 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*

###

###

### 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*

### 15.8 Steady Flow of a Fluid Stream

### Procedure of Analysis **Free-Body Diagram**

**Free-Body Diagram**

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

**Equations of Steady Flow**

**Equations of Steady Flow**

### • Apply the equations of steady flow,

### ( )

*O* *OB B* *OA A*

*M* *dm* *d v* *d v*

### *dt*

##

### 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

### Solution

*Since the density of water is constant, Q*

_{B}*= Q*

_{A}*= Q,*

**Free-Body Diagram**

**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*

*F* *p A*

###

### Solution

**Equations of Steady Flow**

**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*

###

###

### Solution

**Equations of Steady Flow**

**Equations of Steady Flow**

**If moments are summed about point O, then F**

**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*

###

###

###

###

### 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}### 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*

###

###

### 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**

### 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*

###

###

### 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

### 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*

### 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*

### 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