Chapter 1
Newtonian Mechanics - Single
Particle
1.1
Newton’s Laws (Review)
Newton’s Laws:
I: A body remains at rest or in uniform motion unless acted upon by a force.
Treat it as a definition of inertial frame.
II: A body acted upon by a force moves in such a manner that the time rate of change
of the momentum equal to the force.
Treat it as a definition of force.
#»
F = #»
p˙ = m #»
a
We use the dot notation for time-derivatives that ẋ = dx/dt and ẍ = d2 x/dt2
III: If two bodies exert forces on each other, these forces are equal in magnitude and
opposite in direction.
#»
#»
F 12 = −F 21
For isolated system, one can write
#»
#»
F 12 + F 21 = 0.
#»
˙1 +p#»
˙2 = 0
Since F = #»
p˙ , the sum of the momenta of the bodies are constant in time, as p#»
#»
#»
⇒ p 1 + p 2 = constant, which is the statement of conservation of linear momentum.
This is also referred to as the weak form of 3rd Law. The strong form of 3rd Law
requires the directions of the action and reaction to lie on the same straight line. The
consequence is that it guarantees the conservation of angular momentum as well.
1.1.1
More on Newton’s 3rd Law
We mentioned that the First Law and the Second Law can be regarded as the definitions
of inertial frame and force. Newton’s Third Law is indeed a law which concern the physical
world and has implications beyond definition. Moreover, it can be verified and falsified by
observations. In fact, the Third Law is not a general law of nature. It holds only in limited
situations. The followings are two examples where the Third Law does not apply to.
1
CHAPTER 1. NEWTONIAN MECHANICS - SINGLE PARTICLE
2
ˆ Forces do not act instantaneously: It takes 8 minutes for the gravitational force to
transmit between the Sun and Earth. Suppose a mass is created on Earth recently,
this mass will feel the gravitational force by the Sun immediately but the Sun will
only feel the force by the mass after 8 minutes. In these 8 minutes, the Third Law
does not hold.
ˆ Magnetic force between moving charges: Consider two positive charges on the xy
plane: at a certain moment, charge A is at (0, 0) and charge B at (1, 0), while Charge
A is moving along the y-axis vA = (0, 1) and charge B is moving along the x-axis
#»
#»
vB = (1, 0). According to Lorentz Force Law, F = q #»
v × B + qE, charge A will
experience no magnetic force from B since the B-field at (0,0) due to charge B is
zero. But there is a non-zero magnetic force acting on B along the y-direction due to
the magnetic field generated by charge A. Therefore, the same of the forces between
A and B is not zero. (see discussion by J. M. Keller, ”Newton’s Third Law and
Electrodynamics,” American Journal of Physics 10, 302 - 307. )
1.2
Frames of Reference
A particle P is observed by two observers: A and B. The position vectors of P relative
to A and B, respectively, are #»
r P/A and #»
r P/B . The position of B relative to A is #»
r B/A .
According to coordinate transformation, the position vectors are related by
#»
r P/B = #»
r P/A − #»
r B/A
(1.1)
#»
v P/B = #»
v P/A − #»
v B/A
#»
a
= #»
a
− #»
a
(1.2)
so as
P/B
P/A
B/A
(1.3)
Suppose there is no net force acting on P and in the point of view from A the particle is has
a constant velocity, #»
a P/A = 0 (Newton’s 1st Law holds.). But B has non-zero acceleration
with respect to A, #»
a B/A ̸= 0. Then according to B, P would have non-zero acceleration
while there is no net force acting on P.
#»
#»
a P/B = − #»
a B/A ̸= 0 but F net = 0
This violates the 1st Law!
Inertial frame: if Newton’s laws are valid in that frame, namely, if a body subject to no net
force moves in a straight line with constant speed described in a certain coordinate system,
then this coordinate system (S) is an inertial frame.
Any other reference frames that is in uniform motion with respect to the frame S are also
inertial frames.
Otherwise, they are non-inertial frames and Newton’s law, as it is in the form stated above,
will not hold in non-inertial frames.
1.3
Equation of Motion
Newton’s 2nd Law
d #»
v
d
#»
v) = m
F = (m #»
= m #¨r»
dt
dt
(1.4)
CHAPTER 1. NEWTONIAN MECHANICS - SINGLE PARTICLE
3
Assume: mass is constant in time.
This is a second order ordinary differential equation (2nd order ODE) to find #»
r (t).
1.3.1
Typical Forces
Fundamental Forces:
ˆ Gravity
ˆ E&M
ˆ Weak Nuclear Force
ˆ Strong Nuclear Force
Empirical Forces: - description is based on observing the effect of the force
ˆ Contact frictions
ˆ Fluid resistance
ˆ Spring force (Hooke’s Law)
ˆ etc.
Constraint Forces: - force that exists only to restrict the way how to particles can move
ˆ Normal force (particle cannot go through the surface)
ˆ Tension on idea string (particle must be at a fixed distance away from a certain point)
1.4
Problem Sovling
Problem Solving Techniques:
1. Free-body Diagram: make a sketch of the problem indicating force, coordinates and
acceleration.
2. Write down the given quantities and what to be determined (target quantities).
3. Base on principles of physics (Newton’s Law, conservation laws, kinematics constraints), write down useful (maybe later not-so-useful) equations that connect the
quantities together.
4. Derive, derive and derive
5. Put actual numerical values into the symbolic expression of the target quantities.
6. Evaluvate the result: check dimension, take limit, vary the parameter/variables to see
if the answer make sense...
CHAPTER 1. NEWTONIAN MECHANICS - SINGLE PARTICLE
1.4.1
4
Problem with Constant Force
For a given initial condition, x(t0 ) = x0 and v(t0 ) = v0 in 1D, we have
mẍ = F = constant
(1.5)
F
m
F
dv = dt
m
Z v
Z
F t
dv =
dt
m t0
v0
F
v = v0 + (t − t0 )
m
v̇ = ẍ =
(1.6)
(1.7)
(1.8)
(1.9)
Integrating it once, one obtains x
F
ẋ = v = v0 + (t − t0 )
m
Z x
Z t
F
dx =
v0 + (t − t0 )dt
m
x0
t0
F
x = x0 + v0 (t − t0 ) +
(t − t0 )2
2m
(1.10)
(1.11)
(1.12)
One can also work from Eq. (1.6) to get v(x) instead of v(t) and x(t). First multiply 2ẋ on
both sides,
F
ẋ
m
d
d
F dx
(ẋ)2 = v 2 = 2
dt
dt
m dt
2F
2
2
v − v0 =
(x − x0 )
m
2ẋẍ = 2
(1.13)
(1.14)
(1.15)
F
.
The last equation is the well-known result in constant accelerating motion as a = m
1.4.2
Problem with Time-dependent Force
For F = F (t), all one needs to do is integration in time.
F (t)
m
Z v
Z
1 t
⇒
v̇dt =
F (t′ )dt′
m
v0
t0
Z
1 t
⇒ v(t) = v0 +
F (t′ )dt′
m t0
ẍ =
(1.16)
(1.17)
(1.18)
CHAPTER 1. NEWTONIAN MECHANICS - SINGLE PARTICLE
5
For position x(t)
Z
1 t
ẋ = v(t) = v0 +
F (t′ )dt′
m t0
#
Z t"
Z ′
Z x
1 t
v0 +
F (t′′ )dt′′ dt′
ẋdt =
⇒
m
t0
t0
x0
Z Z ′
1 t t
⇒ x(t) = x0 + v0 (t − t0 ) +
F (t′′ )dt′′ dt′
m t0 t0
1.4.3
(1.19)
(1.20)
(1.21)
Problem with Velocity-dependent Force
If the force is velocity dependent, for example, the fluid resistance can be represented by
#»
f = −km #»
v , one can use the method of separation of variable to separate v and t.
1
F (v)
m
dv
1
= dt
F (v)
m
Z v
⇒ t = t0 + m
ẍ = v̇ =
1
dv ′
′)
F
(v
v0
(1.22)
(1.23)
(1.24)
Example: Suppose we drop a mass m in some liquid which acts a resistance to the mass by
F (v) = −kmv. Including gravity, there are two forces acting on the mass
Fy = −mg − kmvy .
Using the result obtained above in Eq. (1.24) and let t0 = 0,
Z vy
1
t=0−
dvy′
′
kv
+
g
v0
y
kvy + g
1
= − ln
k
kv0y + g
(1.25)
(1.26)
(1.27)
So that
g
(1.28)
1 − e−kt
k
As t → ∞, vy → − kg . To find y(t), we can simply integrate vy (t) with respect to t.
Z t h
i
g
′
′
y = y0 +
v0y e−kt −
1 − e−kt
dt′
(1.29)
k
t0 =0
gt v0y k + g −kt
= y0 −
+
1
−
e
.
(1.30)
k
k2
Application to 2D Projectile:
We can apply the solution to projectile motion under air resistance. Consider a projectile
is launched at the origin at an angle θ0 above the x-axis with initial speed v0 . The equation
of motion is given by
vy = v0y e−kt −
m #¨r» = −mg ĵ − km #»
v
(
ẍ = −kvx
⇒
ÿ = −kvy − g
(1.31)
(1.32)
CHAPTER 1. NEWTONIAN MECHANICS - SINGLE PARTICLE
6
Given the initial velocities (define them as U and V ) and (x0 , y0 ) = (0, 0):
v0x = v0 cos θ0 ≡ U
(1.33)
v0y = v0 sin θ0 ≡ V
(1.34)
we have the solutions of both x and y because one can simply consider that the horizontal
motion obeys the same equation of motion with gx = 0. Therefore,
U
1 − e−kt
k
gt V k + g −kt
y(t) = − +
1
−
e
.
k
k2
x(t) =
(1.35)
(1.36)
We can see that the vertical velocity approaches to −g/k while the horizontal velocity
approaches to zero. Therefore, the trajectory approaches to the vertical line after a long
time. As a result, the trajectory is not a perfect parabola. In order to find the range of
the projectile, we can set y(T ) = 0 to solve for the time of flight, T . The range, R, is then
given by R = x(T ). However, the equation for T is
T =
kV + g 1 − e−kT
gk
(1.37)
which cannot be solve analytically. One may need to use numerical method or perturbation method (see textbook Example 2.7 for details).
1.4.4
Problem with Position-dependent Force
When the force is a function of position, it is more convenient to find v(x) instead of v(t).
1
F (x)
m
2
⇒ 2ẋẍ = F (x)ẋ
m
d 2
2
dx
⇒ ẋ = F (x)
dt
m
dt
ẍ =
Integrating both sides,
v
2
− v02 =
2
m
Z x
F (x′ )dx′
(1.38)
(1.39)
(1.40)
(1.41)
x0
This is actually the Work-Energy Theorem: W = ∆K. After obtaining v(x), we use
separation of variable again, if possible, to solve for x(t) in
dx
= v(x)
dt
such that
Z x
t = t0 +
1
dx′ .
′)
v(x
x0
(1.42)