Consider the general curvilinear motion in space of a particle of mass m, where the particle is located by its position vector measured from a fixed origin O.
r
The velocity of the particle is is tangent to its path. The resultant force of all forces on m is in the direction of its acceleration .
rv
va
We may write the basic equation of motion for the particle, as
vmamF
or
GGdt
dvm
dt
dvmF
Where the product of the mass and velocity is defined as the linear momentum 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.
F
vmG
In SI, the units of linear momentum 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
xx GF yy GF zz GF
These equations may be applied independently of one another.
vm
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
, which we integrate from time t1 to time t2 to obtain
momentumlinearinchange
G
G
impulselinear
t
t
GGGGddtF 12
2
1
2
1
GddtF
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.
dt
GdF
Alternatively, we may write
21 GdtFG
I
which says that the initial linear momentum of the body plus the linear
impulse applied to it equals its final linear momentum.
m v1+ =
11 vmG
dtF
22 vmG
The impulse integral is a vector which, in general, we may involve
changes in both magnitude and direction during the time interval.
Under these conditions, it will be necessary to express and in
component form and then combine the integrated components. The
components become the scalar equations, which are independent of
one another.
F
G
xxxxx
t
t
x GGGmvmvdtF 12
2
1
12
yyyyy
t
t
y GGGmvmvdtF 12
2
1
12
zzzzz
t
t
z GGGmvmvdtF 12
2
1
12
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 the time as indicated in the
figure, the impulse, , of this force from t1 to t2 is the shaded
area under the curve.
dtF
t
t2
1
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.
210 GGG
21 vmvm
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
smv
v
v
vmg
mvmvFdt
vsmg
stsmvkgm
/ 1.2
9.36
620016008001620
62002)800(22
1)800()5(
? , / 62.1
, 5 , / 6 , 200
2
2
2
2
12
22
1
mg
motion
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 k=0.3.
SOLUTION
smvv
vdtdtdt
mvmvFdt
directionxin
NFNN
mgNF
tt
t
t
t
kf
y
/823.14.59)4.0(49.26)2.0(36)2.0(72
)6.0(9)3.88(3.03672
)3.88(3.0 3.88)81.9(9
0 0
22
2
4.0
0
4.0
2.0
2.0
0
10
2
22
1
1
motionx
y
P
W=mg
N Ff=kN
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 made by R with the horizontal.
68.8 tan 02.43
49.6 325.005.0
10sin1506.020sin2206.0)81.9(06.0
53.42 127.205.0
10cos1506.020cos2206.0
05.0
0
05.0
0
10
2
05.0
0
10
2
x
y
yy
y
y
t
yy
xx
x
x
t
xx
R
RNR
NRR
ttR
mvmvdtF
NRR
tR
mvmvdtF
SOLUTION
Rx
RyR
W=mg
Rx
RyR
in x direction
in y direction
x
1v
2v
10°20°
xv1
yv1
xv2yv2
y
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
N
y
x
s/m.vvcos
vmmvmvm SBSxSBxB
853540030540
Linear momentum is conserved in x-direction;
kNNorNN
N
dtgmmNvmvm SBSySByB
44.22440
005.081.94505.0030sin540
005.0
0
in y direction
r
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 with respect to a convenient origin O of
fixed coordinates x-y-z.
y
The velocity of the particle is , and its linear momentum is .
The moment of the linear momentum vector about the origin O is
defined as the angular momentum of P about O and is given by the
cross-product relation for the moment of a vector
GrvmrHo
The angular momentum is a vector perpendicular to the plane A defined
by and . The sense of is clearly defined by the right-hand rule for
cross products.
rv vmG
vm
OH
r
v
OH
The scalar components of angular momentum may be obtained from the
expansion
kyvxvmjxvzvmizvyvm
vvv
zyx
kji
mH xyzxyz
zyx
o
In SI units, angular momentum has the units kg.m2/s =N.m.s.
so that
yzox zvyvmH zxoy xvzvmH xyoz yvxvmH
vmrHo
If represents the resultant of all forces acting on the particle P, the
moment about the origin O is the vector cross product
F
vmrFrM o
oM
We now differentiate with time, using the rule for the
differentiation of a cross product and obtain
vmrHo
oM
amrmr
o vmrvmrvmrdt
dH
0
The term is zero since the cross product of parallel vectors is
zero.
vmv
The scalar components of this equation is
oxox HM oyoy HM ozoz HM
Substitution into the expression for moment about O gives
oo HM
The Angular Impulse-Momentum Principle
To obtain the effect of the moment on the angular momentum of the
particle over a finite period of time, we integrate from time
t1 to t2.
oo HM
ooo
H
H
o
t
t
o HHHHddtMo
o
12
2
1
2
1
o
momentumangularinchange
impulseangulartotal
t
t
o HvmrvmrdtM
1122
2
1
or
The total angular impulse on m about the fixed point O
equals the corresponding change in angular momentum of
m about O.
21
2
1
o
t
too HdtMH
Alternatively, we may write
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.
1122
12
2
1
2
1
sin dmvdmvdtFr
HHdtM
t
t
oo
t
t
o
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
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.
210 OOo HHH
This equation expresses the principle of conservation of angular
momentum.
OH
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
st
t
HHdtM
sphere
linkspherepulley
v
rmrT
t
tzzz
08.15
4.060
21504.0341.020
2
112
z
v
v
v
v
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 =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 which the pendulum has immediately after impact and find
the maximum deflection of the pendulum.
SOLUTIONAngular momentum is conserved during impact;
)(/77.2
2.02.34.02.3050.064.02.362.02.320cos4.0300050.0 2222
ccwsrad
(1)
(2)
21
0
0
, 02112
vmrvmrM
HHHHdtM
O
OOOO
t
O
12
21
´
v1
v1´
v2´
v2
O
SOLUTION
Energy considerations after impact; 2211 gg VTVT (Datum at O)
o1.52
cos81.94.005.02.3cos81.92.02.30
81.94.005.02.381.92.02.377.22.02.32
177.24.02.305.0
2
1 22
12
21
´
v1
v1´
v2´
v2
O