Engineers often need to analyze the dynamics of systems of particles – this is the basis for many fluid dynamics

Chapter 14

Engineering Mechanics

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 Kinetic Energy

Work-Energy Principle.

Conservation of Energy Principle of Impulse and

Momentum

Engineers often need to analyze the dynamics of systems of particles – this is the basis for many fluid dynamics

applications, and will also help establish the principles used

in analyzing rigid bodies

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.

Application of Newton’s Laws. Effective Forces

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.

Application of Newton’s Laws. Effective Forces

       ^{ }





 





 





 

















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 opposite 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 but not equivalent.

Linear & Angular Momentum

14 - 7

• 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 of particles,

Motion of the Mass Center of a System of Particles

14 - 8

• Mass center G of system of particles is defined by position vector which satisfiesr_G



 ⁿ

i

i i

G m r

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 of the external forces were concentrated at that point.

Angular Momentum About the Mass Center

14 - 9

 

     

 

   































 



 











 



 

 

 

















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 of reference Gx’y’z’, which translates with respect to the Newtonian frame Oxyz.

i G

i a a

a    

Angular Momentum About the Mass Center

14 - 10

• 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 r m v

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.

Conservation of Momentum

14 - 11

• 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

0 0













 

O

O O

H L

M H

F

L 

 

 

• In some applications, such as

problems involving central forces,

constant constant

0 0













 

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

0 0

















 

G G

G

G G

H v

v m L

M H

F L

 

 

 

 

Concept Question

2 - 12

Three small identical spheres A, B, and C, which can slide on a horizontal, frictionless surface, are attached to three 200-mm-long

strings, which are tied to a ring G. Initially, each of the spheres rotate clockwise about the ring with a relative velocity of v_rel.

Which of the following is true?

a) The linear momentum of the system is in the positive x direction b) The angular momentum of the system is in the positive y direction c) The angular momentum of the system about G is zero

d) The linear momentum of the system is zero

v_rel

v_rel

v_rel

x

Sample Problem 14.2

14 - 13

A 10-kg projectile is moving with a

velocity of 30 m/s when it explodes into 2.5 and 7.5-kg fragments. Immediately after the explosion, the fragments travel in the directions q_A = 45^o and q_B = 30^o. 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.

Sample Problem 14.2

14 - 14

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

2.5 7.5 10 0

A A B B

A B

m v m v mv

v v v

+ =

+ =

x components:

( )

2.5v_A cos45° +7.5v_B cos30° =10 30

y components:

2.5v_A sin 45° - 7.5v_B sin30° = 0

• Solve the equations simultaneously for the fragment velocities.

62.2m s 45 29.3m s 30

A B

v = a v = c

Kinetic Energy

14 - 18

• Kinetic energy of a system of particles,

  _



 ⁿ

i

i i n

i

i i

i v v m v

m T

1

2 2

1 2 1

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

2 1 2 1

1

2 2

1 1

2 2 1

1 2 1 1















• Kinetic energy is equal to kinetic energy of mass center plus kinetic energy relative to the centroidal frame.

Work-Energy Principle. Conservation of Energy

14 - 19

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

2 2

1

1 U T

T  _ 

where represents the work done by the internal forces and the resultant external force acting on P_i . ^ij

f Fi

12

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.

Principle of Impulse and Momentum

14 - 20

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 .

Sample Problem 14.4

14 - 21

Ball B, of mass m_B,is suspended from a cord, of length l, attached to cart A, of mass m_A, 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.

.

0 2gl 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.

Sample Problem 14.4

14 - 22

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

,

0 1 , 1

, 0

A A

B A

B

B A

v v

v v

v v

v











Velocities at positions 1 and 2 are

 

0 A B A

Bv m m v

m  

0 2

, 2

, v

m m

v m v

B A

B B

A   

x component equation:

2 , 2

, 1

, 1

, B B A A B B

A

Av m v m v m v

m   

x y

Sample Problem 14.4

14 - 23

• 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 V₁  _A

2 2 0

1 1m v T  _B

gh m gl m

V₂  _A  _B

 

2

2 1 mA mB vA

T  





gl m v

m_B  _A  _A  _{B A}²_,2  _A  _B

2 2 1

2 0 1

g v m m

h m

B A

A

2

2

 0



2 0 2

0 2

2 2 ,

0

2 2

2

2 

 







 

 



 v

m m

m m

g m m

g v g

v m

m m

g h v

B A

B B

B A A

B B A

g v m m

m g

h v

B A

B

2 2

2 0 2

0  

