Consider the general curvilinear motion in space
 of a particle of mass m, where
the particle is located by its position vector r measured from a fixed origin O.
 
v
The velocity of the particle is  r  is tangent
to its path. The resultant force  F of all
forces on m isin the
 direction of its
acceleration a  v .
We may write the basic equation of motion for
the particle, as




F  ma  mv
or




d 
 d

F  mv  mv  
G G
dt
dt
Where the product
of the mass and velocity is defined as the linear

momentum G  mv of the particle. This equation states that the resultant of
all forces acting on a particle equals its time rate of change of linear
momentum.
In SI, the units of linear momentum

mv
are seen to be kg.m/s, which also
equals N.s.
Linear momentum equation is one of the most useful and important
relationships in dynamics, and it is valid as long as mass m of the particle is
not changing with time.
We now write the three scalar components of linear momentum equation as

Fx  G x

Fy  G y

Fz  G z
These equations may be applied independently of one another.
The Linear Impulse-Momentum Principle
All that we have done so far is to rewrite Newton’s second law in an
alternative form in terms of momentum. But we may describe the effect of
the resultant force on the linear momentum of the particle over a finite
period of time simply by integrating the linear momentum equation with
respect to time t. Multiplying the equation by dt gives
we integrate from time t1 to time t2 to obtain


 dG
F
dt
t2



Fdt 
t1

G2


G1



Fdt  dG , which
 


dG  G2  G1  G




linear impulse
change in linear momentum
Here the linear momentum at time t2 is G2=mv2 and the linear momentum at
time t1 is G1=mv1. The product of force and time is defined as the linear
impulse of the force, and this equation states that the total linear impulse
on m equals the corresponding change in linear momentum of m.
Alternatively, we may write

G1 


Fdt  G2






I
which says that the initial linear momentum of the body plus the linear impulse
applied to it equals its final linear momentum.
m


G2  mv2
v1


G1  mv1
=
+


Fdt
The impulse integral is a vector which, in general, we may involve changes in
both magnitude and direction during the time interval. Under these


F and G in component form
conditions, it will be necessary to express

and then combine the integrated components. The components become the
scalar equations, which are independent of one another.
t2
 F dt  mv
x
  mv x 1  G x
x 2
2
 G x1  G x
y2
 G y1  G y
t1
t2
 F dt  mv   mv   G
y
y 2
y 1
t1
t2
 F dt  mv
z
t1
  mv z 1  Gz
z 2
2
 G z1  G z
There are cases where a force acting on a particle changes with the time in a
manner determined by experimental measurements or by other approximate
means. In this case, a graphical or numerical integration must be performed.
If, for example, a force acting on a particle in a given direction changes with
t2
the time as indicated in the figure, the impulse,
t1 to t2 is the shaded area under the curve.

t1

F dt , of this force from
Conservation of Linear Momentum
If the resultant force on a particle is zero during an interval of time, its
linear momentum G remains constant. In this case, the linear momentum of
the particle is said to be conserved. Linear momentum may be conserved in
one direction, such as x, but not necessarily in the y- or z- direction.



G  0  G1  G2


mv1  mv2
This equation expresses the principle of conservation of linear momentum.
PROBLEMS
1. The 200-kg lunar lander is descending onto the moon’s surface with a
velocity of 6 m/s when its retro-engine is fired. If the engine produces a
thrust T for 4 s which varies with the time as shown and then cuts off,
calculate the velocity of the lander when t=5 s, assuming that it has not yet
landed. Gravitational acceleration at the moon’s surface is 1.62 m/s2.
SOLUTION
m  200 kg ,
v1  6 m / s ,
g  1.62 m / s 2
,
Fdt  mv 2  mv1
v2  ?
motion
1
mg (5)  (800 ) 2  (800 )2  200 v2  6 
2
1620  800  1600  200 v2  6 
v2  6  3.9
v 2  2 .1 m / s
mg
t  5 s,
+
T
PROBLEMS
2. The 9-kg block is moving to the right with a velocity of 0.6 m/s on a
horizontal surface when a force P is applied to it at time t=0. Calculate the
velocity v of the block when t=0.4 s. The kinetic coefficient of friction is
mk=0.3.
y
SOLUTION
motion
x
W=mg
P
F y  0

N
N  mg  0
N  9(9.81)  88 .3 N

Ff=mkN
F f  m k N  0.3(88 .3)
in x direction
t
 Fdt  mv  mv
 72 dt   36 dt 
0
t1  0.2
0
2
1
t 2  0.4
t 2  0.4
t1  0.2
0
0.3(88 .3)dt  9(v2  0.6)
72 (0.2)  36 (0.2)  26 .49 (0.4)  9v2  5.4

v2  1.823 m / s
PROBLEMS
3. A tennis player strikes the tennis ball with her racket while the ball is still
rising. The ball speed before impact with the racket is v1=15 m/s and after
impact its speed is v2=22 m/s, with directions as shown in the figure. If the
60-g ball is in contact with the racket for 0.05 s, determine the magnitude of
the average force R exerted by the racket on the ball. Find the angle b made
by R with the horizontal.
SOLUTION

v2 y

v2
y
in x direction

t
0
Fx dt  mv 2 x  mv1 x
0.05
Rx t 0
 0.06 22 cos 20   0.06  15 cos10 
0.05 R x  2.127
20°

10° v

 1x
v1 y v1
 F dt  mv
y
0
Ryt
0.05
0

2 y
0.05 R y  0.325
R  43 .02 N
Rx
 mv1  y
0.05
 0.06 (9.81)t 0
 0.06 22 sin 20   0.06 15 sin 10 
R
R y  6.49 N
tan b 
x
W=mg
R x  42 .53 N
in y direction
t

v2 x
Ry
Rx

b  8.68 
Ry
Ry
b
Rx
R
PROBLEMS
4. The 40-kg boy has taken a running jump from the upper surface and lands
on his 5-kg skateboard with a velocity of 5 m/s in the plane of the figure as
shown. If his impact with the skateboard has a time duration of 0.05 s,
determine the final speed v along the horizontal surface and the total normal
force N exerted by the surface on the skateboard wheels during the impact.
PROBLEMS
(mB+mS)g
y
x
N
Linear momentum is conserved in x-direction;
mB vBx  mS vSx  mB  mS v
40  5 cos 30  0  40  5  v 
in y direction
m B v By  mS v Sy 
0.05
 N  m
B
v  3.85 m / s
 mS g dt  0
0
 40  5 sin 30   0  N 0.05   45 9.810.05   0

N  2440 N or N  2.44 kN
In addition to the equations of linear impulse and linear momentum, there exists
a parallel set of equations for angular impulse and angular momentum. First, we
define the term angular momentum. Figure shows a particle P of mass m moving

along a curve in space. The particle is located by its position vector r
respect to a convenient origin O of fixed coordinates x-y-z.
y
with

 

The velocity of the particle is v  r, and its linear momentum is G  mv . The

moment of the linear momentum vector mv about the origin O is defined as

the angular momentum H O of P about O and is given by the cross-product
relation for the moment of a vector


  
H o  r  mv  r  G
The angular momentum is a vector perpendicular to the plane A defined by
and
 . The sense of
v
products.

H O is clearly defined by the right-hand rule for cross

r
The scalar components of angular momentum may be obtained from the
expansion

i

k

j



z  m yv z  zv y i  mzv x  xv z  j  m xv y  yv x k

Ho  m x
y
vx
vy
so that

H ox  m yvz  zv y



H o  r  mv





vz
H oy  mzv x  xvz 
In SI units, angular momentum has the units
kg.m2/s =N.m.s.

H oz  m xv y  yvx

If


F represents the resultant of all forces acting on the particle P, the

moment M o about the origin O is the vector cross product

We now differentiate


Mo  r 

 
 

F  r  mv



H o  r  mv
with time, using the rule for the
differentiation of a cross product and obtain

 
 

d 
H o  r  mv   rm
v  r m
v

dt
 
a
r mr  0 m


Mo


The term v  mv is zero since the cross product of parallel vectors is zero.
Substitution into the expression for moment about O gives



M o  Ho
The scalar components of this equation is
M
ox
 H ox

M oy  H oy
M
oz
 H oz
The Angular Impulse-Momentum Principle
To obtain the effect of the moment on the angular momentum of the particle


M o  H o from time t1 to t2.
over a finite period of time, we integrate
t2

or

M o dt 
t1
t2


M o dt

H o 2 

   

dH o  H o
 1

Ho

2  Ho
1

 H o





 r2  mv2   r1  mv1   H o

t1


change in angular momentum
total angular impulse
The total angular impulse on m about the fixed point O equals the
corresponding change in angular momentum of m about O.
Alternatively, we may write
 
 2
t2



H o 1    M o dt  H o
t1
Plane-Motion Application
Most of the applications can be analyzed as plane-motion problems where
moments are taken about a single axis normal to the plane motion. In this case,
the angular momentum may change magnitude and sense, but the direction of
the vector remains unaltered.
t2
M
o dt
H o 2  H o 1
t1
t2
  Fr sindt  mv d
2 2
t1
 mv1d1
Conservation of Angular Momentum
If the resultant moment about a fixed point O of all forces acting on a

particle is zero during an interval of time, its angular momentum H O remains
constant. In this case, the angular momentum of the particle is said to be
conserved. Angular momentum may be conserved about one axis but not about
another axis.



H o  0  H O1  H O2
This equation expresses the principle of conservation of angular momentum.
PROBLEMS
1. The assembly starts from rest and reaches an angular speed of 150
rev/min under the action of a 20 N force T applied to the string for t
seconds. Determine t. Neglect friction and all masses except those of the
four 3-kg spheres, which may be treated as particles.
SOLUTION

t2
t1

v



M z dt  H z2  H z1
  2  
20
0.1 t  4 
3 
0.4150   0.4
 
60  


T r
m
r
sphere link 
pulley
vsphere
t  15 .08 s

v

v

v
z
PROBLEMS
2. A pendulum consists of two 3.2 kg concentrated masses positioned as
shown on a light but rigid bar. The pendulum is swinging through the vertical
position with a clockwise angular velocity w=6 rad/s when a 50-g bullet
traveling with velocity v=300 m/s in the direction shown strikes the lower
mass and becomes embedded in it. Calculate the angular velocity w which the
pendulum has immediately after impact and find the maximum deflection  of
the pendulum.
SOLUTION
Angular momentum is conserved during impact;

t
0
(2)





M O dt  H O2  H O1  0 , H O1  H O2

MO  0 
r  mv 1  r  mv 2
(1)
0.050 300 0.4 cos 20   3.20.22 6  3.20.42 6  0.050  3.20.42 w   3.20.22 w 
w   2.77 rad / s (ccw)

v1´
1
2

v2´
O

w´
v2
w
2
1
v1

SOLUTION

v1´
1
2

v2´
O

w´
v2
w
2

1
v1
Energy considerations after impact;
T1  Vg1  T2  Vg 2
(Datum at O)
1
0.05  3.20.4  2.77 2  1 3.20.2  2.77 2  3.20.29.81  3.2  0.05 0.49.81
2
2
 0  3.20.29.81cos  3.2  0.05 0.49.81 cos
  52 .1o