Dr. M I. Adam

Dynamics of Single Particle in 1 D, 2 D, and 3 Dimensions

Contents:

Newtons Laws of Motion

Equation of Motion: Constant Force, F

Time dependent , F(t), Velocity dependent, F(v),

and Position dependent, F(x) Forces

Objectives: After the lesson, students should be able to:

Write the equation of motion in certain deferential forms.

Solve differential and integral equations to obtain kinematic variables such as x(t), v(t).

Analyze motion of a particle via kinematic variables or qualitatively describe the motion of a single particle via

potential functions.

Newtons Laws:

I. A body remains at rest or in uniform motion unless acted

upon by a force.

II. A body acted upon by a force moves in such a manner

that the time rate of change of momentum equals the

force.

III. If two bodies exert forces on each other, these forces are

equal in magnitude and opposite in direction.

Dr. M I. Adam

The Equation of Motion for a Particle:

In 1-D case, the motion of a single particle is written as

= = = =

(1)

where m is mass, v is velocity, t is time, and, a is the acceleration of a

particle. P is momentum.

For a given dynamic system,

= = (, , ) (2)

Hence, the following cases will be studied

i. (, , ) = ii. (, , ) = () iii. (, , ) = () iv. (, , ) = ()

Useful Techniques for Problems Solving:

1- Represent all variables given in a simple sketch.

2- Write down the equations. Trust your mathematical

background and use it to manipulate the equations.

3- Use any given numerical values to get the value of the

unknown variables.

Dr. M I. Adam

i. Constant Force, F:

Determine the velocity and position of a particle of mass m at

any time when its acted upon by a constant force F. take

and as initial values of velocity and position, respectively. Hence, Eqn. of motion is:

= =

=

= 0, ,,(), ()

The Velocity v(t) and the Position x(t) of the particle are

obtained by integration as follows

" # = " # () = + (1)

() = " = " + , specifically, " #''( = " (

+ )# () = + + )* * (2)

ii. Velocity dependent Force = (): We already have

## =## = (),

We shall start with the expression shown above to solve Eqns.

given in the form of = ()or = ().

Dr. M I. Adam

First case: = () ()# = #,

By integrating both sides,

" #( = "()

( # = "

()

( #

If () is explicitly known, then = () can be determined. Hence, = (). We integrate () to determine () as follows:

() () = " ()( # (ii. 1) Second case: = (), then # =

()#

By integrating both sides, weve

" #''( = "()

( # = "

()

( # (ii. 2)

Note:

Friction force is experimentally found to follow power law

behavior of velocity,0 = 12, where k is geometry and medium dependent constant, m is

mass, and n = 1, 2, 3, .

n = 1 for small object travelling in air with ~ *5 . n = 2 for objects travelling in other medium with

~ 665 .

Dr. M I. Adam

iii. Time dependent Force, (): Determine the velocity, ()andposition, () from the Eqn. of motion = ()# =

() #.

Integrating for velocity,

" #( = "()

# () = + ) " ()

# (iii. 1)

Solution for position is obtained by integrating for velocity.

iv. Position dependent Force, (): Here we introduce the concept of Kinetic Energy;. But first, the Eqn. of motion is:

' = (); # = ()# " #( = " ()

''( #

Rearranging the Eqn. above,

' =

)**> = () (iv. 1)

Notice that the Kinetic Energy, ; = )** hence, Eqn. (iv. 1) can be written as?' = (). Next, we integrate as:

@ #;?

?(= @ ()

'

'(#whichDs; ; = @ ()

'

'(#

)** )** = " ()

''( # (iv. 2)

Dr. M I. Adam

Theorems of Conservation

Objective:

1- To drive the consequences of Newtons Laws of Dynamics

that applied to the Motion of a Single Particle.

Theory 1: The total linear momentum E of a Particle is conserved when the total force acting on it is zero. How?

From Newtons Second Law,

= F = () = EH + H

For = 0,E = 0; it means E is a vector constant in time. Now, let I be a constant vector such that I = 0 and since, I = EH I = 0 then,

KEH IL = "EH I = constant (1)

The component of linear momentum in the direction at

which the force vanishes is constant in time.

Theory 2: The angular momentum of a Particle subject to no

torque is conserved. How?

Definitions:

i. Angular Momentum, M E

Dr. M I. Adam

ii. Torque or Moment of Force, P = , is a position vector.

From Newtons Second Law, = H , hence Def. ii. Becomes P = H = EH

From Def. i., MH = ( E) = (H E) + ( EH ) , and since, H E = H = (H H) = 0

MH = 0 + EH = EH (2) This means that if P = 0, thenMH = 0andM is a vector

constant in time.

To solve Problems related to ii.:

1- Choose the origin of the coordinate system because

Torque P = 0 in coordinate systems centered along the resultant line of Force where M = constant.

Kinetic Energy, T:

Assuming a work W done on a Particle by a Force F resulted in

transforming the Particle from Point 1 to Point 2.

R)* = " *) #; is the resultant force. From Newtons Second Law, = ,

# = ## ## # =

## #

Dr. M I. Adam

= * ( )# =

*

(*)# = #(

)**)

Total force F acting on a Particle = Change in Kinetic Energy.

R)* = =)**>S)* = )*(v** v)*) = T* T) (1)

where T is the Kinetic Energy of the Particle, ; = )*v*. At ;)>;* R)* < 0; decrease in Kinetic Energy.

Potential Energy, W: Force, F has the property that depends only on the original and

final position. Hence, the capacity to carry a Particle from point

1 to point 2 without change in Kinetic Energy is:

" *) # X) X* = Difference in U at the two points, (Fig.). If F is the gradient of the Scalar function U:

= Y#X = X,then,

1

2

a

b

c

0

F is independent of

the path

Dr. M I. Adam

" *) # = " (U)*) # = " #\*) = X) X* (2) The Potential Energy is a function of position, time:X(, ).

Total Energy is the sum of the Kinetic and Potential Energies,

] = ; + X. Total time derivative of E is:

^ =

? +

_ (i)

We already have # = # =)**> = #; so; we divide this expression by # as follows

0 = H =? (ii)

From the Potential Energy derivative, we have:

_ =

a_a'bc

a'ba +

a_a =

a_a'bc Hc +

a_a = (X) H +

a_a (iii)

Substituting Eqns. (ii) and (iii) into (i), we get

#]# = H + (X) H +

dXd = ( + X) H +

dXd

+ X vanishes if total force = X. E = a_a (3)

Dr. M I. Adam

Hence, if U is not explicit function of time then, the force field

denoted F is conservative force and, under these conditions:

Theory 3: The total energy E of a Particle in a conservative force

field is constant in time.

No conservation laws can be applied in situations of:

1- Interaction of moving electric charges.

2- Quantum mechanical systems.

Dr. M I. Adam

Prob. 2.2: Find the velocity H as a function of the displacement for a particle of mass, which starts from the rest at = 0, subject to the following force functions:

(a) ' = + e (b) ' = fgh' (c) ' = cos e

Where and eare positive constants. Solution:

(a) dx dx dx dx

x xdt dx dt dx

= = = & & &

&& & ( )1dxx F cxdx m

= +o

&&

( )1xdx F cx dxm

= + o

& & 2

21 1

2 2

cxx F x

m

= +

o

& ( )1

2

2x

x F cxm

= + o

&

(b) 1 cxdxx x F e

dx m

= = o

&&& &

1 cxxdx F e dxm

= o

& &

( ) ( )21 1 12

cx cxF Fx e ecm cm

= = o o& ( )1

221

cxFx ecm

= o&

(c) ( )1 cosdxx x F cxdx m

= = o

&&& & cos

Fxdx cxdx

m= o& &

21sin

2

Fx cx

cm= o&

1

22sin

Fx cx

cm

=

o&

Dr. M I. Adam

Prob. 2.8: A projectile is fired with a velocity such that it passes through two points both a distance above the horizontal. Show that if the gun is adjusted for a maximum

range, the separation of the points is = (j k* 4Y . Solution:

The equations for the coordinates are

0cosx v t = (1)

20

1sin

2y v t gt= (2)

In order to calculate the time when a projectile reaches the

ground, we let y = 0 in (2):

20

1sin 0

2v t gt = (3)

02

sinv

tg

= (4)

Substituting (4) into (1) we find the relation between the range

and the angle as

2

0sin 2

vx

g= (5)

P Q

x

y

v0

h

Dr. M I. Adam

The range is maximum when22

= , i.e.,

4

= . For this value

of m the coordinates become

0

20

2

1

22

vx t

vy t gt

= =

(6)

Eliminating t between these equations yields

2 2

2 0 00

v vx x y

g g + = (7)

We can find the x-coordinate of the projectile when it is at the

height h by putting y = h in (7):

2 2

2 0 00

v v hx x

g g + = (8)

This equation has two solutions:

2 2

20 0

1 0

2 2

20 0

2 0

42 2

42 2

v vx v gh

g g

v vx v gh

g g

=

= +

(9)

where 1x corresponds to the point P and

2x to Q in the diagram.

Therefore,

20

2 1 04

vd x x v gh

g= = (10)

Dr. M I. Adam

Prob.2-21: Show directly that the time rate of change of the

angular momentum about the origin for a projectile fired from

the origin (constant g) is equal to the moment of force (or

torque) about the origin.

Solution:

Assume a coordinate system in which the projectile moves in

the 2 3x x plane. Then,

2 0

2

3 0

cos

1sin

2

x v t

x v t gt

= =

(1)

or,

( ) 22 2 3 3 0 2 0 31

cos sin2

x x v t v t gt = + = +

r e e e e (2)

The linear momentum of the projectile is

( ) ( )0 2 0 3cos sinm m v v gt = = + p r e e& (3)

and the angular momentum is

( ) ( ) ( ) ( )20 2 0 3 0 2 0 3cos sin cos sinv t v t gt m v v gt = = + + L r p e e e e (4)

x3

v0

x2

x1

Dr. M I. Adam

Using the property of the unit vectors that 3i j ijk =e e e , we

find

( )20 11 cos2mg v t =L e (5)

This gives

( )0 1cosmg v t = L e& (6)

Now, the force acting on the projectile is

3

mg= F e (7)

so that the torque is

( ) ( )

( )

2

0 2 0 3 3

0 1

1cos sin

2

cos

v t v t gt mg

mg v t

= = +

=

N r F e e e

e

which is the same result as in (6).

Dr. M I. Adam

Standard Examples: Motion of a Particle in 2-D:

1- Consider a block sliding on an inclined plane. If the angle

of the plane is n and the mass of the block is , sliding through a frictionless path. What is the block acceleration?

Two forces act on the block; j and P, the normal to the plane.

[(THE REST OF THIS LECTURE WILL BE DISCUSSED ON WHITE

BOARD IN CLASS]!

N

h

x

y

Fg sin

Fg cos

N

Fg

mg sin

mg cos mg

N

mg sin

mg

mg cos

N