DYNAMICS Ferdinand P. Beer Systems of Particles E. Russell - - PowerPoint PPT Presentation

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Seventh Edition Ferdinand P. Beer

• E. Russell Johnston, Jr.

Lecture Notes:

• J. Walt Oler

Texas Tech University CHAPTER

• Systems of Particles

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 2

Contents

Introduction Application of Newton’s Laws: Effective Forces Linear and Angular Momentum Motion of Mass Center of System of Particles Angular Momentum About Mass Center Conservation of Momentum Sample Problem 14.2 Kinetic Energy Work-Energy Principle. Conservation

• f Energy

Principle of Impulse and Momentum Sample Problem 14.4 Sample Problem 14.5 Variable Systems of Particles Steady Stream of Particles Steady Stream of Particles. Applications Streams Gaining or Losing Mass Sample Problem 14.6

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 3

Introduction

• In the current chapter, you will study the motion of systems of

particles.

• The effective force of a particle is defined as the product of it

mass and acceleration. It will be shown that the system of external forces acting on a system of particles is equipollent with the system of effective forces of the system.

• The mass center of a system of particles will be defined and

its motion described.

• Application of the work-energy principle and the impulse-

momentum principle to a system of particles will be

• described. Result obtained are also applicable to a system of

rigidly connected particles, i.e., a rigid body.

• Analysis methods will be presented for variable systems of

particles, i.e., systems in which the particles included in the system change.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 4

Application of Newton’s Laws. Effective Forces

• Newton’s second law for each particle Pi in a

system of n particles,

( )

force effective forces internal force external

1 1

= = = × = × + × = +

• =

= i i ij i i i i n j ij i i i i i n j ij i

a m f F a m r f r F r a m f F

• The system of external and internal forces on a

particle is equivalent to the effective force of the particle.

• The system of external and internal forces

acting on the entire system of particles is equivalent to the system of effective forces.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 5

Application of Newton’s Laws. Effective Forces

• Summing over all the elements,

( )

( )

( )

• =

= = = = = = =

× = × + × = +

n i i i i n i n j ij i n i i i n i i i n i n j ij n i i

a m r f r F r a m f F

1 1 1 1 1 1 1 1

• Since the internal forces occur in equal and
• pposite collinear pairs, the resultant force

and couple due to the internal forces are zero,

( )

( )

• ×

= × =

i i i i i i i i

a m r F r a m F

• The system of external forces and the

system of effective forces are equipollent by not equivalent.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 6

Linear & Angular Momentum

• Linear momentum of the system of

particles,

• =

= =

= = =

n i i i n i i i n i i i

a m v m L v m L

1 1 1

• Angular momentum about fixed point O of

system of particles,

( )

( ) ( )

( )

• =

= = =

× = × + × = × =

n i i i i n i i i i n i i i i O n i i i i O

a m r v m r v m r H v m r H

1 1 1 1

• Resultant of the external forces is

equal to rate of change of linear momentum of the system of particles, L F

• =
• O

O

H M

• =
• Moment resultant about fixed point O of

the external forces is equal to the rate of change of angular momentum of the system

• f particles,

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 7

Motion of the Mass Center of a System of Particles

• Mass center G of system of particles is defined by

position vector which satisfies

G

r

• =

=

n i i i G

r m r m

1

• Differentiating twice,
• =

= = = =

= =

F L a m L v m v m r m r m

G n i i i G n i i i G

• 1

1

• The mass center moves as if the entire mass and all
• f the external forces were concentrated at that

point.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 8

Angular Momentum About the Mass Center

( ) ( ) ( ) ( ) ( ) ( )

( )

• =

× ′ = × ′ = ×

• ′

− × ′ = − × ′ = ′ × ′ = ′ ′ × ′ = ′

= = = = = = = G n i i i n i i i i G n i i n i i i i n i G i i i n i i i i G n i i i i G

M F r a m r a r m a m r a a m r a m r H v m r H

• 1

1 1 1 1 1 1

• The angular momentum of the system of particles

about the mass center,

• The moment resultant about G of the external forces

is equal to the rate of change of angular momentum about G of the system of particles.

• The centroidal frame is not, in

general, a Newtonian frame.

• Consider the centroidal frame
• f reference Gx’y’z’, which

translates with respect to the Newtonian frame Oxyz.

i G i

a a a ′ + =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 9

Angular Momentum About the Mass Center

• Angular momentum about G of particles in their

absolute motion relative to the Newtonian Oxyz frame of reference.

( ) ( ) ( ) ( )

• =

′ = × ′ + ×

• ′

= ′ + × ′ = × ′ =

= = = = G G G n i i i i G n i i i n i i G i i n i i i i G

M H H v m r v r m v v m r v m r H

• 1

1 1 1

• Angular momentum about G of the

particles in their motion relative to the centroidal Gx’y’z’ frame of reference,

( )

• =

′ × ′ = ′

n i i i i G

v m r H

1

• G

G i

v v v ′ + =

• Angular momentum about G of the particle

momenta can be calculated with respect to either the Newtonian or centroidal frames of reference.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 10

Conservation of Momentum

• If no external forces act on the

particles of a system, then the linear momentum and angular momentum about the fixed point O are conserved. constant constant = = = = = =

• O

O O

H L M H F L

• In some applications, such as problems

involving central forces, constant constant = ≠ = = ≠ =

• O

O O

H L M H F L

• Concept of conservation of momentum also

applies to the analysis of the mass center motion, constant constant constant = = = = = = = =

• G

G G G G

H v v m L M H F L

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 11

Sample Problem 14.2

A 10 kg projectile is moving with a velocity

• f 30 m/s when it explodes into 2.5 and 7.5

kg fragments. Immediately after the explosion, the fragments travel in the directions θA = 45o and θB = 30o. Determine the velocity of each fragment. SOLUTION:

• Since there are no external forces, the

linear momentum of the system is conserved.

• Write separate component equations

for the conservation of linear momentum.

• Solve the equations simultaneously for

the fragment velocities.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 12

Sample Problem 14.2

SOLUTION:

• Since there are no external forces, the

linear momentum of the system is conserved. x y

• Write separate component equations for the

conservation of linear momentum.

( ) ( ) ( ) 0

10 5 . 7 5 . 2 v v v v m v m v m

B A B B A A

• =

+ = +

x components:

( )

30 10 30 cos 5 . 7 45 cos 5 . 2 = ° + °

B A

v v

y components:

30 sin 5 . 7 45 sin 5 . 2 = ° − °

B A

v v

• Solve the equations simultaneously for the

fragment velocities.

s m 3 . 29 s / m 1 . 62 = =

B A

v v

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 13

Kinetic Energy

• Kinetic energy of a system of particles,

( )

• =

=

=

• =

n i i i n i i i i

v m v v m T

1 2 2 1 1 2 1

• i

G i

v v v ′ + =

• Expressing the velocity in terms of the

centroidal reference frame,

( ) ( ) [ ]

• =

= = = =

′ + = ′ + ′

• +
• =

′ +

• ′

+ =

n i i i G n i i i n i i i G G n i i n i i G i G i

v m v m v m v m v v m v v v v m T

1 2 2 1 2 2 1 1 2 2 1 1 2 1 2 1 1 2 1

• Kinetic energy is equal to kinetic energy of

mass center plus kinetic energy relative to the centroidal frame.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 14

Work-Energy Principle. Conservation of Energy

• Principle of work and energy can be applied to each particle Pi ,

2 2 1 1

T U T = +

→

where represents the work done by the internal forces and the resultant external force acting on Pi .

ij

f

• i

F

• 2

1→

U

• Principle of work and energy can be applied to the entire system by

adding the kinetic energies of all particles and considering the work done by all external and internal forces.

• Although are equal and opposite, the work of these forces

will not, in general, cancel out.

ji ij

f f

• and
• If the forces acting on the particles are conservative, the work is equal

to the change in potential energy and

2 2 1 1

V T V T + = + which expresses the principle of conservation of energy for the system of particles.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 15

Principle of Impulse and Momentum

2 1 1 2

2 1 2 1

L dt F L L L dt F L F

t t t t

• =

+ − = =

• 2

1 1 2

2 1 2 1

H dt M H H H dt M H M

t t O t t O O O

• =

+ − = =

• The momenta of the particles at time t1 and the impulse of the forces from

t1 to t2 form a system of vectors equipollent to the system of momenta of the particles at time t2 .

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 16

Sample Problem 14.4

Ball B, of mass mB,is suspended from a cord, of length l, attached to cart A, of mass mA, which can roll freely on a frictionless horizontal tract. While the cart is at rest, the ball is given an initial velocity Determine (a) the velocity of B as it reaches it maximum elevation, and (b) the maximum vertical distance h through which B will rise. . 2 gl v = SOLUTION:

• With no external horizontal forces, it

follows from the impulse-momentum principle that the horizontal component of momentum is conserved. This relation can be solved for the velocity of B at its maximum elevation.

• The conservation of energy principle can

be applied to relate the initial kinetic energy to the maximum potential energy. The maximum vertical distance is determined from this relation.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 17

Sample Problem 14.4

SOLUTION:

• With no external horizontal forces, it follows from the

impulse-momentum principle that the horizontal component of momentum is conserved. This relation can be solved for the velocity of B at its maximum elevation.

2 1

2 1

L dt F L

t t

• =

+ (velocity of B relative to A is zero at position 2)

2 , 2 , 2 , 2 , 1 , 1 , A A B A B B A

v v v v v v v = + = = = Velocities at positions 1 and 2 are

( )

2 , A B A B

v m m v m + =

2 , 2 ,

v m m m v v

B A B B A

+ = = x component equation:

2 , 2 , 1 , 1 , B B A A B B A A

v m v m v m v m + = + x y

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 18

Sample Problem 14.4

• The conservation of energy principle can be applied to relate the

initial kinetic energy to the maximum potential energy.

2 2 1 1

V T V T + = + Position 1 - Potential Energy: Kinetic Energy: Position 2 - Potential Energy: Kinetic Energy: gl m V

A

=

1 2 2 1 1

v m T

B

= gh m gl m V

B A

+ =

2

( ) 2

2 , 2 1 2 A B A

v m m T + =

( )

gh m gl m v m m gl m v m

B A A B A A B

+ + + = +

2 2 , 2 1 2 2 1

g v m m m h

B A A

2

2

+ =

2 2 2 2 , 2

2 2 2 2

• +

+ − = + − = v m m m m g m m g v g v m m m g v h

B A B B B A A B B A

g v m m m g v h

B A B

2 2

2 2

+ − =

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 19

Sample Problem 14.5

Ball A has initial velocity v0 = 3 m/s parallel to the axis of the table. It hits ball B and then ball C which are both at rest. Balls A and C hit the sides of the table squarely at A’ and C’ and ball B hits

• bliquely at B’.

Assuming perfectly elastic collisions, determine velocities vA, vB, and vC with which the balls hit the sides of the table. SOLUTION:

• There are four unknowns: vA, vB,x, vB,y, and

vC.

• Write the conservation equations in terms
• f the unknown velocities and solve

simultaneously.

• Solution requires four equations:

conservation principles for linear momentum (two component equations), angular momentum, and energy.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 20

Sample Problem 14.5

x y i v v j v i v v j v v

C C y B x B B A A

• =

+ = =

, ,

SOLUTION:

• There are four unknowns: vA,

vB,x, vB,y, and vC.

• The conservation of momentum and energy equations,

y B A C x B

mv mv mv mv mv L dt F L

, , 2 1

− = + = = +

• (

)

2 2 1 2 , 2 , 2 1 2 2 1 2 2 1 2 2 1 1 C y B x B A

mv v v m mv mv V T V T + + + = + = +

( ) ( ) ( ) ( )

C y B A O O O

mv mv mv mv H dt M H m 9 . m 1 . 2 m 4 . 2 m 6 .

, 2 , 1 ,

− − = − = +

• Solving the first three equations in terms of vC,

C x B C y B A

v v v v v − = − = = 3 6 3

, ,

Substituting into the energy equation,

( ) ( )

72 78 20 9 3 6 3 2

2 2 2 2

= + − = + − + −

C C C C C

v v v v v

( )

s m 34 . 1 s m 4 2 s m 4 . 2 s m 2 . 1 = − = = =

B B C A

v j i v v v

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 21

Variable Systems of Particles

• Kinetics principles established so far were derived for

constant systems of particles, i.e., systems which neither gain nor lose particles.

• A large number of engineering applications require the

consideration of variable systems of particles, e.g., hydraulic turbine, rocket engine, etc.

• For analyses, consider auxiliary systems which consist of

the particles instantaneously within the system plus the particles that enter or leave the system during a short time

• interval. The auxiliary systems, thus defined, are constant

systems of particles.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 22

Steady Stream of Particles

• System consists of a steady stream of particles

against a vane or through a duct. ( ) [ ] ( ) [ ]

B i i A i i t t

v m v m t F v m v m L dt F L

• ∆

+ = ∆ + ∆ + = +

• 2

1

2 1

• The auxiliary system is a constant system of

particles over ∆t.

• Define auxiliary system which includes

particles which flow in and out over ∆t. ( )

A B

v v dt dm F

• −

=

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 23

Steady Stream of Particles. Applications

• Fluid Flowing Through a Pipe
• Jet Engine
• Fan
• Fluid Stream Diverted by Vane or

Duct

• Helicopter

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 24

Streams Gaining or Losing Mass

• Define auxiliary system to include particles of

mass m within system at time t plus the particles of mass ∆m which enter the system

• ver time interval ∆t.

2 1

2 1

L dt F L

t t

• =

+

• The auxiliary system is a constant system of

particles.

u dt dm F a m u dt dm dt v d m F

• −

= + =

• (

) [ ] ( )( ) ( ) ( ) v

m v v m v m t F v v m m t F v m v m

a a

• ∆

∆ + − ∆ + ∆ = ∆ ∆ + ∆ + = ∆ + ∆ +

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 25

Sample Problem 14.6

Grain falls onto a chute at the rate of 120 kg/s. It hits the chute with a velocity of 6 m/s and leaves with a velocity of 4.5 m/s. The combined weight of the chute and the grain it carries is 3 kN with the center of gravity at G. Determine the reactions at C and B. SOLUTION:

• Define a system consisting of the mass of

grain on the chute plus the mass that is added and removed during the time interval ∆t.

• Apply the principles of conservation of

linear and angular momentum for three equations for the three unknown reactions.

Vector Mechanics for Engineers: Dynamics

Seventh Edition 14 - 26

Sample Problem 14.6

SOLUTION:

• Define a system consisting of the

mass of grain on the chute plus the mass that is added and removed during the time interval ∆t.

• Apply the principles of conservation of

linear and angular momentum for three equations for the three unknown reactions.

( ) ( )

( )

( )

° ∆ − = ∆ + − + ∆ − ° ∆ = ∆ = + 10 sin 10 cos

2 1 B y A B x

v m t B W C v m v m t C L dt F L

• (

) ( ) ( ) ( )

° ∆ − ° ∆ = ∆ + − + ∆ − = + 10 sin 6 . 3 10 cos 8 . 1 6 . 3 1 . 2 9 .

2 , 1 , B B A C C C

v m v m t B W v m H dt M H

• Solve for Cx, Cy, and B with

s slug 45 . 7 s ft 32.2 s kg 120

2 =

= ∆ ∆ t m

( )N

307 1 . 110 N 2102 j i C B

• +

= =