Chapter II
Newtonian Mechanics
Single Particle
Recommended problems: 2-2, 2-5, 2-6, 2-8, 2-9, 2-11, 2-12, 2-14, 2-16, 2-21,
2-22, 2-24, 2-25, 2-26, 2-27, 2-29, 2-30, 2-32, 2-34, 2-37, 2-38, 2-39, 2-41,
.2-42, 2-43, 2-44, 2-47, 2-51, 2-52, 2-53, 2-54.
2.2 Newton’s Laws
The First Law: A body remains at rest or in uniform motion unless acted upon
by a force.
The Second Law: A body acted upon by a force moves in such a manner that
the force is equal to the time rate of change of momentum, i.e.,

 dP
F
dt


with P  mv
(2.1)
(2.2)
The Third Law: If two bodies exert force on each other, these force are equal in
magnitude and opposite in direction.
In another word, if two bodies constitute an ideal, isolated system, then the
accelerations of these bodies are always in opposite directions, and the ratio of
the magnitudes of the accelerations is constant and equal to the inverse ratio of
the masses of the bodies.


F1   F2
Using Eq.(2.1) we get
(2.3)


dP1
dP
 2
dt
dt


dv1
dv2
m1
 m2
dt
dt


m1a1  m2 a2

(2.4) 
( with cons tan t mass)

a1
m2
 
a2
m1
(2.5)
Eq.(2.4) can be rearranged as
 
d P1  P2 
0
dt

 
P1  P2  constant
That is , for isolated system , the total momentum is conserved.

2.3 Frames of Reference
For Newton's laws of motion to have meaning, the motion must be measured
relative to some reference frame.
A reference frame is called an inertial frame if Newton’s laws are valid in that
frame.
If Newton’s laws are valid in one reference frame , then they are also valid in
any reference frame in uniform motion (not accelerated) with respect to the first
frame.
If Newton’s laws are valid in one reference frame , then they are also valid in
any reference frame in uniform motion (not accelerated) with respect to the first
frame.


This is because the Equation F  mv involves a time derivative of velocity: a
change of coordinates involving constant velocity doesn’t influence the
equation.
2.4 The Equation of Motion of a Particle
Newton’s second law can be expressed as



F  mv  mr
(2.6)
This is a second order differential equation, which can be solved to find r if the
force function F(v,r,t) and the initial values of r and v are known.
Example 2.1 If a particle slides without friction down a fixed, inclined plane with
=30o. What is the block’s acceleration.
Solution There are two forces acting on
the block: The gravitational force and the
normal force. If the block is constrained to
move on the plane, and taking the +ve xaxis down the plane the only direction the
block can move is the x-direction. Eq.(2.6)
now reads



Fg  N  mr
N
y
Fgsin
Fgcos
Fg
x

Applying the last equation in the two directions, we have
 Fg cos  N  my  0
y-direction:
Fg sin   mx
x-direction:
mg sin   mx

x  g sin   4.9 m / s 2

To find the velocity after time t, we have
dx
 g sin 
dt

v
t
vo
0
 dx  g sin   dt
v  vo  g sin  t

To find the velocity after it moves a distance x down the plane we have
dx
 2 x  2 g sin  x
dt
dx
 g sin 
dt
dx 2
dx
 2 g sin 
dt
dt

v

x
 dx  2 g sin   dx 
vo
2
xo
v 2  vo2  2 g sin   x  xo 
If at t=0, xo and vo are zero, then
v  2gx sin 
Example 2.2 If the coefficient of static friction between the block and the plane
is s=0.4, at what angle  will the block start sliding if it is initially at rest?
acceleration.
fs
N
Solution We have now an additional force,
the static frictional force which is parallel to
the plane. Applying Newton’s we have
y-direction:
x-direction:
Knowing that
 Fg cos  N  my  0
Fgsin
Fgcos
Fg
Fg sin   f s  mx
f s  f max   s N
At the verge of slipping fs reaches its maximum value, then we write

Fg sin   f max  mx

Fg sin    s N  mx
Substituting for Fg=mg and for N from the first equation we get
mg sin   mg s cos  mx
x  g sin    s cos 
Just before the block starts to slide, the acceleration is zero
sin    s cos  0

  tan 1  s   tan 1 0.4  22o

Example 2.3 After the block in the previous example begin to slide, the
coefficient of kinetic friction becomes k=0.3. Find the acceleration for the angle
=30o.
Solution Applying Newton’s again we have
y-direction:
x-direction:
Knowing that
 mg cos  N  my  0
mg sin   f k  mx
f k   k N  mg cos
 x  g sin    k cos   0.24 g
In general s> k. This can be observed by lowering the angle  below 16.7o,
we find that x 0 and the block stops. If we increase the angle  above 16.7o,
the block doesn’t sliding again until  exceeds 22o. This is because now the
force that retarding the motion is no longer the kinetic frictional force but rather
it is the static frictional force which is greater than the kinetic frictional force .
Effects of Retarding Forces
If a body is acted upon by a resisting force F(v) in addition, for instance, to the
gravitational force, the total force is then.
 



F  Fg  Fr  Fg  F v 


v
or F  mg  mkv n
v
(2.7)
(2.8)
Where k is a positive constant that specifies the strength of the retarding force
and v/v is a unit vector in the direction of v.
Example 2.4 Find the displacement and the velocity of horizontal motion of a
particle in a medium in which the retarding force is proportional to the velocity.
Solution Knowing that the only horizontal force acting on the particle is the
retarding force, applying Newton’s again we have
x-direction:
 kmv  mx
dv
  k  dt
v


dv
 kv
dt
ln v  kt  C1

If the initial velocity (at t=0) v=vo then
v  vo e kt
To find the displacement we have
v
dx
 vo e kt
dt


v
x  vo  e kt dt   o e kt  C2
k
vo
kt
x

1

e
If x(t=0)=0 then C2=vo/t 
k

We can find the velocity as a function of displacement by writting
dv dv dt dv 1


dx dt dx dt v
 dv  k  dx
dv dv
 v   kv
dx dt

v  kx  C3
If the initial vel x(t=0)=0 and v(t=0)=vo then C3=vo
v  vo  kx


dv
 k
dx

Example 2.5 Find the displacement and the velocity of a particle undergoing
vertical motion in a medium with a retarding force is proportional to the velocity.
Solution Considering the particle is falling downward with a initial velocity vo
from a height h in a constant gravitational field. The equation of motion is
z-direction:
dv
F  m  mg  kmv
dt

The minus sign in the retarding force (which is upward force) is due to the fact
that the velocity is downward. The last equation can be written as
dv
 dt

kv  g
Knowing that (at t=0) v=vo then integrating the last equation we get
1
lnkv  g   t  C
k
kv  g  kvo  g e kt
 1 ln kv  g   t

k kvo  g 
dz
g kvo  g  kt
v




e

2
dt
k
k
It is clear that as t, the velocity approaching the terminal value (-g/k). At this
value the net force vanish.
If vo exceeds the
terminal velocity in
magnitude, then the
body begins to slow
down
and
v
approaches
the
terminal speed from the
opposite direction.
To find the displacement we integrate again, with (at t=0) z=h to get
z h

gt kvo  g 

1  e kt
k
k

Example 2.6 Let us
study the projectile
motion in 2-dimensions
without considering air
resistance.
Solution The
equations of motion
are
x-direction:
0  mx
y-direction:
 mg  my
y
vy=0
Hmax
vo

x
The range R
Assuming x(t=0)= y(t=0) we get
x  vo cos 
y  vo sin   gt
x  vo cos t
y  vo sin  t  12 gt 2
Eliminating t from the above 2-equations we get

 2
g
x
y  vo tan  x   2
2
 2v cos  
 o

Which is the equation of a parabola
The speed and the total displacement are found to be
v  x 2  y 2  vo2  g 2t 2  2vo gt sin 
r  x 2  y 2  vo2t 2  14 g 2t 2  vo gt 3 sin 
The range can by found by determining the value of x when the projectile falls
back to ground, i.e., (y=0)

y  t vo sin 
 12 gt
 0
 t 0
2vo sin 
& t T 
g
Now the range R is found by
2vo2 sin  cos  vo2 sin 2
R  xt  T  

g
g
It is easy to show that the maximum range occurs at =45o.
Example 2.7 Let us study the effect of the air resistance to the projectile
motion in the previous example, assuming that the retarding force is directly
proportional to the projectile’s velocity (Fr=-kmv).
Solution The equations of motion are in this case
x-direction:
 kmx  mx
y-direction:
 mg  kmy  my
Assuming again x(t=0)= y(t=0) we get

v cos 
x o
1  e kt
k

gt kvo sin   g 
 kt
y 
1

e
k
k2


To find the range we need the time T when y=0 
T
kvo sin   g 
gk
1  ekT 
This is a transcendental equation so we can’t obtain an analytical expression
for T. It can be solved by approximation (perturbation) or by numerical
technique.
To apply the perturbation method we assume that k is relatively small. Now
rewrite the transcendental equation as
T
kvo sin   g 
gk
kT  12 k 2T 2  16 k 3T 3  

T  kvo sin  g  1T  12 kvo sin  g  1kT 2  16 kvo sin  g  1k 2T 3
0  kvo sin  g   12 kvo sin  g  1T  16 kvo sin  g  1kT 2
2vo sin  g
T
 13 kT 2
1  kvo sin  g
Using the identity 1  x 1  1  x  x 2  x 3  
1
 1  kvo sin  g  kvo sin  g 2 
1  kvo sin  g
To first order of k we then have
 
2vo sin   T 2 2vo2 sin 2  
2
T


k

O
k
2
 3

g
g






With no air resistance (k=0) we recover the same result as in the previous
example, i.e.,
T  To 
2vo sin 
g
If k is small (but nonvanishing), the flight time will be approximately equal to To.
Using this approximated value we get
2vo sin   4vo2 sin 2  2vo2 sin 2  
T


k
2
2


g
g
 3g

2vo sin   kvo sin  
 T
1 

g
3g 

Now to find the range we have


v cos
x o
kt  12 k 2t 2  16 k 3t 3  
k
To first order of k the range is obtained from
Substituting for T in the last equation

R  vo cos T  12 kT 2

vo2 sin 2  4kvo sin  
 4kvo sin  
R
1 
  Ro 1 

g
3g
3g




With Ro is the range without air resistance.
Example 2.9 Atwood’s machine consists of a smooth pulley with 2-masses
suspended from a light string at each end. Find the acceleration of the masses
and the tension of the string (a) when the pulley center is at rest and (b) when
the pulley is descending in an elevator with constant acceleration .
Solution (a) The equations of motion, for
each mass, are
m1g  T  m1x1
m2 g  T  m2 x2
If the string is inextensible, i.e.,
x1  x2  constant
x1   x2


m1g  m2 g  m2  m1 x1
x1 
m1  m2 g
  x2
m2  m1 

Solving the first two equations for T we get
2m1m2 g
T
m2  m1 
(b) The coordinate system with the origin at
the pulley center is no longer an inertial. So
we select the origin of the coordinates to be at
the top of the elevator shaft. The equations of
motions in such a system are
m1g  T  m1x1
m2 g  T  m2 x2
But, as it is clear from the figure,
m1g  T  m1 x1  x1 
x1  x1  x1, x2  x2  x2
m2 g  T  m2 x2  x2 
Knowing that x1   x2 & x1  x2   
m1g  T  m1   x1 
m2 g  T  m2   x1 

Solving the last 2-equations for the acceleration and the tension we get
m1  m2  
g  
x1   x2 
m2  m1 
2m1m2  g   
T
m2  m1 
Note that the result are just as if the acceleration of gravity were reduced by an
amount of the elevator acceleration. If the elevator is ascending rather than
descending we expect
x1   x2 
T
m1  m2  
g 
m2  m1 
2m1m2  g   
m2  m1 
Example 2.10 Consider a charged particle entering a region of uniform
magnetic field. Determine its subsequent motion.
Solution Let the magnetic field be parallel to the y-axis. The magnetic force is

 
F  qv  B
The equation of motion reads

 
qB x kˆ  z iˆ  m xiˆ  y ˆj  zkˆ


mx   qBz
(1)
my  0
(2)
mz qBx
(3)
Integrating Eq.(2) we get
Integrating Eq.(1&3) we get
y  constant  y o
(4) 
y  y ot  yo
(5)
x  
qB

z  C1  z  C1
z  x  C2
(6)
(7 )
Substituting for z Eq.(7) into Eq.(1) we get
x    2 x  C3

x   2 x   2 A
x  a  R cost   o 
x   2 x   2 A


(8)
(9)
Differentiating Eq. (9) w.r.t time and substituting into Eq. (6) 
  R sin t   o    z  C1 
z  b  R sin t   o 
(10)
Squaring Eqs(9+10) and then adding we get
 x  a 2   z  b 2  R 2
(11)
Then the path of the motion is a circle of radius R and centered at (a,b).
Now for z  constant the motion is helix with its axis in the direction of B.
Now from Eqs.(10&11) we have
x   R sin t   o 
y   R  cost   o 
Squaring the above two equations and
then adding we get
 R
v
mv
 qB

x 2  y 2  R 2 w2  v 2
2.5 Conservation Theorems
Recalling Eq.(1) and assuming that the net force is zero we get

 dP
F
0
dt


P  constant
I.
(2.9)
The total linear momentum of a particle is conserved when the total
force on it is zero.
 
Let s be some constant vector such that F  s  0 . Then
   
F  s  P  s  0
 
P  s  constant

(2.10)
The component of linear momentum in a direction in which the force vanishes is
constant in time.
Defining the angular momentum of a particle with respect to origin as
  
L rP
(2.11)
The torque with respect to the same origin is defined as
  
N rF
(2.12)
Where r is the position vector from the origin to the point where the force acts.
Now substituting for F from Eq.(2.1) into Eq.(2.12) we get
  
N  r  P
Now from Eq.(2.12) we have
(2.13)
 d r  P 
  
L 
 r  P  r  P
dt

But r  P  mr  r   0
   
L  r  P  N

(2.14)
If no torque acting on a particle then the angular momentum is constant, i.e.,
II.
The angular momentum of a particle subject to no torque is conserved.
Defining the work done on a particle by a force in transforming the particle from
point 1 to point 2 as
 
W   F  dr
2
(2.15)
1
 

 
dv dr
dv 
Now F  dr  m  dt  m  v dt
dt dt
dt
  1 d v  v 
F  dr  2 m
dt  d 12 mv 2
dt



W  12 m v22  v12  T2  T1



(2.16)
2
With T  12 mv is the kinetic energy of the particle.
If the work of a force is independent on the path, such a force is called
conservative. For every conservative force we associate a potential energy
according to
W  U
 
  F  dr  U1  U 2
2
(2.17)
1
From the last equation we conclude that the force can be written as


F  U
(2.18)
To prove Eq.(2.18) we have from Eq.(2.17)
2
2
 

 F  dr    U  dr    dU  U1  U 2
2
1
1
1
Potential energy has no absolute meaning; only differences of potential energy
are physically meaningful.
Now defining the total energy as the sum of kinetic and potential energies, i.e.,.
E  T U

dE dT dU


dt dt dt
(2.19)
(2.20)
But dT  d

1 mv 2
2

 
 F  dr
dT  
 F  r
dt

(2.21)

 U
dU
U xi U
U
U
And



xi 
 U   r 
dt
t
t
t
i xi t
i xi
(2.22)
Substituting Eqs.(2.21 & 2.22) into Eq.(2.20) we get
 
 U
dE
 F  U   r 
dt
t


 

Since F  U
 F  U  r  0

dE U

dt
t
If U is not an explicit function of time then the total energy is constant, i.e.,
III.
The total energy E of a particle is a conservative field is conserved.
Example 2.11 A mouse of mass m jumps on the outside edge of a freely
turning ceiling fan of rotational inertia I and radius R. By what ratio does the
angular velocity change?
Solution Here the angular momentum is conserved before and after the
mouse's jumping. Recalling that the angular momentum can be written as
L  I

Li  L f  Io  I  mvR
Knowing that
v  R


Io  I  mR 2   I  mR 2

I

o I  mR 2




2.6 Energy
In todays physics, energy is more popular than Newton’s laws. Most of physical
problems are solved by means of energy.
Consider a particle under the influence of a conservative, 1-dimensional, force.
The total energy is written as
E  T  U  12 mv 2  U  x 
dx
2
E  U  x 
v 
dt
m
x
dx
 dx
 t  to   
dt x 2
o
E  U  x 
m
(2.23)
(2.24)
(2.25)
If we know U we can solve Eq.(2.25) to get x as a function of time.
We can know a lot about the motion of a particle by examining the plot of U(x).
Let consider the plot of the following figure.
It is clear, from Eq.(2.19), and since T is always positive 
E  T  U  U x
(2.26)
The motion is bounded for E1 &
E2, i.e., it can't move off to .
For E1 the motion is periodic
between xa & xb, i.e., xa  x  xb.
For E2 the motion is periodic in
two possible regions: between
xc  x  xd and xe  x  xf .
For Eo the particle is at rest
since here E=U.
For E3 the particle comes from , stops and turns at x = xg and returns to .
Here the motion is unbounded.
For E4 the motion is unbounded and the particle may be at any position.
The point x = xo is called an equilibrium point. In general, the equilibrium state is
characterized by
dU  x 
0
dx
(2.27)
The equilibrium is said to be stable if
 d 2U  x  


0
2
 dx 

 x  xo
(2.28)
The equilibrium is said to be unstable if
 d 2U  x  


0
 dx 2 

 x  xo
(2.29)
An equilibrium is considered stable if the system always returns to equilibrium
after small disturbances. If the system moves away from the equilibrium after
small disturbances, then the equilibrium is unstable.
Example 2.12 Consider the system
of light pulleys, masses, and string
shown. A light string of length b is
attached at point A, passes over a
pulley at point B located a distance
2d away, and finally attaches to
mass m1. Another pulley with mass
m2 attached passes over the string,
pulling it down between A & B.
Calculate the distance x1 when the
system is in equilibrium, and
determine whether the equilibrium
is stable or unstable.
Solution Let U=0 along the line AB. 
But x2 
U  m1gx1  m2 g  x2  c 
b  x1 2 4  d 2  U  m1gx1  m2 g b  x1 2 4  d 2  m2 gc
To determine the equilibrium position we set
dU
0
dx1

 m1g 
m2 g b  x1 
4 b  x1 2 4  d 2
0 
b  x1 2 4  d 2  m2 b  x1  
4m1
4m12 b  x1 2  16m12 d 2  m22 b  x1 2
b  x1  
2
16m12 d 2

4m12
 m22



 b  x1 2 4m12  m22  16m12 d 2
4m1d
x0  x1  b 
4m12  m22



Notice that a real solution exists only when 4m12  m22
Now to determine wither the equilibrium is stable or no we have
d 2U
dx12

m2 g
4 b  x1 2 4  d 2


m2 g b  x1 2
 
16 b  x1 2 4  d 2
3

2
Now substituting for x1=xo we get
2
d U

2
dx1
g

4m12
 m22
4m22 d

3
2
 0 for 4m12  m22
The equilibrium is stable for real solutiion.


Example 2.12 Consider the potential
U x 

 Wd 2 x 2  d 2

x 4  8d 4
Sketch the potential and discuss the motion at various values of x. Is the motion
bounded or unbounded? Where are the equilibrium values? Are they stable or
unstable? Find the turning points for E=-W/8. W is a +ve constant.


U x  y 2  1
x
Z  y 
 4
with y 
W
d
y 8
Solution Rewrite the potential as
Let us first find the equilibrium points using Eq.(2.27)


dZ
 2 y 4 y3 y 2  1
 4

0
2
dy y  8
y4  8



 2 y y 4  8  4 y 3 y 2  1 y y 4  2 y 2  8

0

2
2
y 4  8
y 4  8
yo2  0,2
y y 4  2 y 2  8  y y 2  4y 2  2  0


x
Using y 
d

xo1  0
xo 2  2d
xo3   2d
We have 3-equilibrium points. Now sketch Z(y) versus y we get
As it is clear from the figure, the
equilibrium is stable at xo2 & xo3
but unstable fro xo1. The motion
is bounded for all energies E<0.
AT the turning points the speed
is zero. So T=0, i.e.
E U
E  U  y 



4
y 8
y4  8  8y2  8

y2 y2  8  0
  W
W y2 1
8

 y  0,  2 2


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