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