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

( ) ( ) GGdtdvm

dtdvmF

====∑

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

Σ

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.

⇒=∑ dtGdF

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

tx GGGmvmvdtF ∆=−=−=∫ 12

2

1

12

( ) ( ) yyyyy

t

ty GGGmvmvdtF ∆=−=−=∫ 12

2

1

12

( ) ( ) zzzzz

t

tz 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.

dtFt

t∫∑2

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

( )

( )

smvv

v

vmg

mvmvFdt

vsmg

stsmvkgm

/ 1.29.36

620016008001620

62002)800(221)800()5(

? , / 62.1

, 5 , / 6 , 200

2

2

2

2

12

22

1

=−=−

−=−−

−=−

−

−=Σ

==

=== 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

µk=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

24.0

0

4.0

2.0

2.0

0

10

2

22

1

1

=⇒−=−+

−=−+

−=Σ

==⇒==

=−⇒=Σ

∫∫∫∫

==

=

=

µ

motion x

y

P

W=mg

N Ff=µkN

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 β made

by R with the horizontal.

( ) ( )

( ) ( )

( ) ( )

( ) ( )

°=⇒==

==

−=−

−=Σ

==

−−=

−=Σ

∫

∫

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

05.00

05.0

0

10

2

05.00

10

2

ββx

y

yy

y

y

t

yy

xx

x

x

t

xx

RR

NR

NRR

ttR

mvmvdtF

NRR

tR

mvmvdtF

SOLUTION

Rx

Ry R

W=mg

Rx

Ry R β

in x direction

in y direction

x

1v

2v

10°

20°

xv1

yv1

xv2

yv2 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

44.22440005.081.94505.0030sin540

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

( ) ( ) ( )kyvxvmjxvzvmizvyvmvvvzyxkji

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

amrmro vmrvmrvmr

dtdH ×+×=×=

=× 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

Ho

t

to HHHHddtM

o

o

∆=−== ∫∫∑ 12

2

1

2

1

( ) ( ) omomentumangularinchange

impulseangulartotal

t

to 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.

( ) ( )212

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

to

−=

−=

∫∑

∫∑

θ

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

linkspherepulleyv

rmrT

t

tzzz

08.15

4.06021504.0341.020

2

112

=

=

−=Σ∫

π

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 ω=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 ω′ which the

pendulum has immediately after impact and find the maximum deflection θ of

the pendulum.

SOLUTION Angular momentum is conserved during impact;

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

)(/77.2

2.02.34.02.3050.064.02.362.02.320cos4.0300050.0 2222

ccwsrad=′

′+′+=−−

ω

ωω

(1)

(2)

( ) ( )21

0

0

, 02112

vmrvmrM

HHHHdtM

O

OOOOt

O

×=×⇒=

==−=∫

1 2

2 1

θ

θ

θ

θ

ω

ω´

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

81.94.005.02.381.92.02.377.22.02.32177.24.02.305.0

21 22

=

+−+=

+−+⋅+⋅+

θ

θθ

1 2

2 1

θ

θ

θ

θ

ω

ω´

v1

v1´

v2´

v2

O