This is the typeset summary of my handwritten notes when I took and finished the course 3AMX0 - Mechanics in 2018/2019. It contains the topics covered in the course (book, slides), along with some worked out example exam problems.
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This summary is based on the course 3AMX0, its lectures and the book Classical Mechanics by John
R. Taylor. This text is based on the handwritten notes made when I took the course during the year
2018-2019. Use this summary at your own risk; I am not responsible for your exam and
sections in this summary may contain errors, redundant information or absence of impor-
tant information. Information may be outdated and not part of the course anymore. My
recommendation is that you also study the book, lecture slides and exercises as well as
exam questions thoroughly.
The chapters and sections will be named after the sections in the book. This document will feature a
summary of the concepts along with some example exam questions and another version will be available
that’s trimmed down a bit on the content. I will also not fix typos, mistakes or whatever.I will state
here what chapters are treated in which weeks of the course.
1.1 Newton’s second law in Cartesian coordinates
Week 1, chapter 1.6 in Classical Mechanics (T) book
Newton’s second law is given by equation (1),
~ = m~¨r,
F (1)
where ~r is the position vector as a function of time, and the dots indicate the time derivatives. For
Cartesian coordinates we have for the net force equation (2),
~ = Fx x̂ + Fy ŷ + Fz ẑ
F (2)
and for the position vector equation (3),
~
r = xx̂ + yŷ + zẑ. (3)
Differentiating twice, we get equation (4),
Fx x̂ + Fy ŷ + Fz ẑ = mẍx̂ + mÿŷ + mz̈ẑ. (4)
So, by equation (5),
Fx = mẍ
~ = mr̈ ⇐⇒ Fy = mÿ
F (5)
Fz = mz̈
1.2 Two-dimensional polar coordinates
Week 2, Chapter 1.7 in T
The transformations from 2D Cartesian coordinates to polar coordinates is given by equation (6),
x = r cos φ
y = r sin φ
p (6)
r = x2 + y 2
y
φ = arctan
x
The unit vectors of the polar coordinates are given by equation (7),
r̂ = cos φx̂ + sin φŷ
(7)
φ̂ = − sin φx̂ + cos φŷ
and
dr̂
= φ̂
dφ
dφ̂
= −r̂
dφ
4
,The velocity in polar coordinates is given by equation (8),
~
v = ṙr̂ + rφ̇φ̂, (8)
where vr = dr
dt and vφ = r φ̇ = rω.
The acceleration in polar coordinates is given by equation (9),
a = (r̈ − rφ̇2 )r̂ + (rφ̈ + 2ṙφ̇φ̂.
~ (9)
a = −ω 2 rr̂.
Applying this to uniform circular motion, we have φ̇ = ω is constant and ṙ = r̈ = φ̈ = 0 so ~
5
,Chapter 2
Momentum and Angular Momentum
Chapter 3 in T
2.1 Conservation of momentum
Week 1, Chapter 3.1 in T
Principle of conservation of momentum
If theP ~ ext on an N -particle system is zero, the system’s total mechanical momentum
net external force F
~ =
P mα~v α is constant.
The sum of forces is given by equation (1),
d~
v dm~v ~
~ = dP .
X X
~ = m~
F a=m = =⇒ F (1)
dt dt dt
Different momenta are additive,
~ =p
P ~1 + p
~2 + . . .
then
dP~ d~p1
= + ...
dt dt
so, the total external force is given by equation (2),
dP~ X
= ~ i,ext ≡ F
F ~ ext . (2)
dt i
~ ext = 0, the momentum is conserved, so ~
dP
When F dt = 0.
2.2 Rockets
Week 1, Chapter 3.1 in T
See Figure 1. At time t, p = mv. At time t + ∆t, p = −vfuel + (v + ∆v)(m + ∆m) = ∆m(vex + ∆v) +
m(v + ∆v).
The equation of motion of the rocket, the rocket equation is given by equation (3),
dv dm(t)
m(t) = −vex . (3)
dt dt
The solution is given by equation (4),
m(0)
v(t) = v(0) + vex ln . (4)
m(t)
6
,Figure 1: A rocket of mass m travels to the right with speed v and ejects spent fuel exhaust speed vex
relative to the rocket.
2.3 The center of mass
Week 1, Chapter 3.3 in T
The position vector of the center of mass is given by equation (5),
P
~ m1 ~r 1 + m2~
r2 + . . . mi ~
ri
R= = Pi , (5)
m1 + m2 + . . . i mi
with the three Cartesian components given by equation (6)
1 X
X= mi xi
M
1 X
Y = mi yi (6)
M
1 X
Z= m i zi .
M
See Figure 2.The center of mass position for two particles is given by equation (7),
~ = m1 ~
R
r 1 + m2 ~
r2
. (7)
m1 + m2
The center of mass position for a continuous position is given by equation (8),
~ 1~
Figure 2: The CM of two particles lies at the position R(m r 1 + m2~
r 2 )/M . You can prove that this
lies on the line joining m1 to m2 , as shown, and that the distances of the CM from m1 and m2 are in
the ratio m2 /m1 .
Z Z
~ = 1
R ~
r dm =
1
ρ~
r dV. (8)
M M
7
,The velocity of the center of mass is given by equation (9),
~ = m1~
V
v 1 + m2~
v2 + . . . p
~ +p
= 1
~2 + . . . p
~ +p
= 1
~2 + . . .
(9)
m1 + m2 + . . . m1 + m2 M
Then,
~ =p
MV ~1 + · · · + p ~.
~n = P
The total external force is then given by equation (10),
~
d2 R dP~
~=M
MA = ~ ext .
=F (10)
dt 2 dt
2.4 Angular momentum of a single particle
Week 1, Chapter 3.4 in T
The angular momentum vector of a single particle is given by equation (11),
~l = ~
r×p
~. (11)
See Figure 3. Angular momentum describes the ’rotation’ around the origin.
Figure 3: Angular momentum is zero (left) and angular momentum is maximum (right).
The change of angular momentum is given by equation (12),
d~l d~
r d~
p d~
p ~.
= r×
+~ r×
=~ =F (12)
dt dt × p
~ dt dt
The first term is zero because d~
r
dt × p v ×~
~=~ v m = 0.
The torque is given by equation (13),
~
~
Γ=~ r×F ~ = dl . (13)
dt
See Figure 4. The magnitude of the torque may be expressed by any of the forms given in equation (14),
|~
Γ| = rF sin θ
|~
Γ| = rF⊥ (14)
|~
Γ| = r⊥ F
~ , then Γ = 0. When ~
rkF
When ~ ~ then Γ is maximum.
r⊥F
8
, Figure 4: Torque.
2.5 Angular momentum for several particles
Week 1, Chapter 3.5 in T
Principle of conservation of angular momentum
If ~
Pthe net external torque on an N -particle system is zero, the systems total angular momentum L =
rα × p
~α . is constant.
For a collection of point particles, the total torque is given by equation (15),
~
dL X ext
= ~ ~i = ~
ri × F Γ . (15)
dt i
The change of angular momentum of a system equals the net external torque on the system.
2.6 Partially a 3NBB0 recapitulation
dθ
ωz =
dt (16)
v = ωr
The moment of inertia is given by equation (17),
X
2
I= mi ri⊥ , (17)
where ri⊥ is the distance from the object perpendicular to the rotation axis.
The angular momentum about a fixed z-axis is given by equation (18),
Lz = Iω. (18)
The torque can be given by equation (19),
dω
Γext = I = Iα, (19)
dt
9
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