Skule Final Exam Cheat Sheet.pdf

Formula Booklet MIE100 Final Exam | Revision 2.1 Apr 20, 2019 | 1

Velocity and acceleration Special cases Rectangular coordinates SUVAT equation (a const) Projectile motion

v =ds dt a =dv

dt= 𝑣𝑑𝑣 𝑑𝑠

(a = 0) 𝑠 = 𝑠0+ 𝑣₀𝑡

𝑣²= 𝑣₀²+ 2 ∫ 𝑎(𝑠)

𝑠 𝑠0

𝑑𝑠

𝑟⃑ = 𝑥𝑖⃑ + 𝑦𝑗⃑ + 𝑧𝑘⃑⃑

𝑣⃑ = 𝑥̇𝑖⃑ + 𝑦̇𝑗⃑ + 𝑧̇𝑘⃑⃑

𝑎⃑ = 𝑥̈𝑖⃑ + 𝑦̈𝑗⃑ + 𝑧̈𝑘 ⃑⃑

|𝑣⃑| = √𝑥̇

²

+ 𝑦̇

²

+ 𝑧̇

²

v = v₀+ a₀t s = s0+ 𝑣0𝑡 +1

2𝑎0𝑡² 𝑣²= v₀²+ 2𝑎₀(𝑠 − 𝑠0)

𝑠 =(𝑣 + 𝑣0)𝑡 2

𝑣_0𝑥= 𝑣₀cos 𝜃 𝑣0𝑦= v0sin 𝜃

𝑎𝑥= 0 𝑥 = 𝑥0+ 𝑣0𝑡

𝑎_𝑦= 𝑎_0𝑦 y = y₀+ 𝑣_0𝑦𝑡 +¹

2𝑎_0𝑦𝑡² (const a) v = v0+ a0t

s = s₀+ 𝑣₀𝑡 +1 2𝑎₀𝑡² 𝑣²= v02+ 2𝑎0(𝑠 − 𝑠0)

𝑣 = 𝑣₀+ ∫ 𝑎(𝑡)₀^𝑡 𝑑𝑡 s = s0+ 𝑣0𝑡 +

∫ ∫ 𝑎(𝑡)₀^𝑡 ₀^𝑡 𝑑𝑡𝑑𝑡

Normal-tangential system Cylindrical 𝐫 − 𝛉 system Circular motion Dependent motion

𝑣_𝑛−𝑡

⃑⃑⃑⃑⃑⃑⃑⃑ = 𝑣𝑢⃑⃑⃑ _𝑡 𝑢𝑏

⃑⃑⃑⃑ = 𝑢⃑⃑⃑ × 𝑢𝑡 ⃑⃑⃑⃑ 𝑛

v =^ds

dt

𝑎 = 𝑣̇𝑢⃑⃑⃑ + 𝑣𝜃̇𝑢𝑡 ⃑⃑⃑⃑ 𝑛

= 𝑣̇𝑢⃑⃑⃑ +_𝑡 ^𝑣2

𝜌𝑢⃑⃑⃑⃑ _𝑛 ρ =⁽¹⁺⁽

dy dx)²)

1.5

|^𝑑2𝑦_𝑑𝑥2|

𝑟⃑ = 𝑟𝑢⃑⃑⃑⃑⃑ 𝑟

𝑣⃑ = 𝑟̇𝑢⃑⃑⃑⃑⃑ + 𝑟𝜃̇𝑢𝑟 ⃑⃑⃑⃑⃑𝜃

𝑎⃑ = (𝑟̈ − 𝑟𝜃̇²)𝑢⃑⃑⃑⃑⃑ +𝑟

(2𝑟̇𝜃̇ + 𝑟𝜃̈)𝑢⃑⃑⃑⃑⃑_𝜃 𝑢_𝑟

⃑⃑⃑⃑⃑̇ = 𝜃̇𝑢⃑⃑⃑⃑⃑ _𝜃 𝑢_𝜃

⃑⃑⃑⃑⃑̇ = −𝜃̇𝑢⃑⃑⃑⃑⃑_𝑟

- Rope has constant length - Define good datum lines (fixed position)

- Find fixed length if possible - Divide the rope into sections if needed

LT= 𝑠𝐴+ 𝑠𝐵

Then 𝑣A+ 𝑣𝐵= 0 𝑎_A+ 𝑎_𝐵= 0 Relative motion Gravitational force Frictional force (oppose motion) Spring force Equilibrium (x-y-z) Equilibrium (n-t)

𝑟_𝐵

⃑⃑⃑ = 𝑟⃑⃑⃑ + 𝑟_𝐴 ⃑⃑⃑⃑⃑⃑⃑ _𝐵/𝐴 𝑣𝐵

⃑⃑⃑⃑ = 𝑣⃑⃑⃑⃑ + 𝑣𝐴 ⃑⃑⃑⃑⃑⃑⃑⃑ 𝐵/𝐴

𝑎𝐵

⃑⃑⃑⃑ = 𝑎⃑⃑⃑⃑ + 𝑎𝐴 ⃑⃑⃑⃑⃑⃑⃑⃑ 𝐵/𝐴

g⃑ = −9.81

𝑗

⃑ms⁻² F = mg⃑

Static 𝐹𝑓𝑠𝑚𝑎𝑥=𝜇𝑠𝐹𝑁

Kinetic 𝐹_𝑓𝑘=𝜇_𝑘𝐹_𝑁

F_s= −𝑘𝑥 k is spring constant x is deviation from rest

∑ 𝐹_𝑥= 𝑚𝑎_𝑥= 𝑚

𝑥

^̈

∑ 𝐹_𝑦= 𝑚𝑎_𝑦= 𝑚

𝑦

^̈

∑ 𝐹𝑧= 𝑚𝑎𝑧= 𝑚

𝑧

^̈

∑ 𝐹_𝑛= 𝑚𝑎_𝑛= 𝑚𝑣𝜃̇

=mv² 𝜌

∑ 𝐹_𝑡= 𝑚𝑎_𝑡= 𝑚𝑣̇

Equilibrium (𝐫 − 𝛉) Work and motion Work by gravitational F Work by kinetic friction Work by spring Kinetic energy

∑ 𝐹_𝑟= 𝑚𝑎_𝑟= 𝑚 (𝑟̈ − 𝑟𝜃̇²)

∑ 𝐹_𝜃= 𝑚𝑎_𝜃= 𝑚 (2𝑟̇𝜃̇ + 𝑟𝜃̈)

dU = 𝐹 ∙ d𝑟 = 𝐹 cos 𝜃 𝑑𝑟 UP>P′= ∫ 𝑑𝑈_𝑃^𝑃′ =

∫ 𝐹 _𝑃^𝑃′ d𝑟 = ∫ 𝐹 cos 𝜃_𝑃^𝑃′ ds

Ug= −𝑊∆𝑦

= −𝑚𝑔(𝑦₂− 𝑦₁)

*Always negative

*Against motion> negative

U_f= −𝐹_𝑓∆𝑥 Us= ∫ −𝑘𝑥

𝑥₂ 𝑥₁

𝑑𝑥

= −1

2𝑘(𝑥22− 𝑥12)

T =¹

2𝑚𝑣² 𝑇1+ 𝑈1>2= 𝑇2 1

2𝑚𝑖𝑣𝑖12+ ∫^𝑠𝑖2𝐹⃑⃑⃑⃑ 𝑖𝑡 𝑠_𝑖1 ds +

∫^𝑠𝑖2𝑓⃑⃑⃑⃑ _𝑖𝑡

𝑠_𝑖1 ds =¹

2𝑚_𝑖𝑣_𝑖2² Internal force is zero Work done by force Potential energy Conservation of energy Linear momentum Elastic collision If particles connected

by inextensible cable

∫^𝑠𝑖2𝑓⃑⃑⃑⃑ _𝑖𝑡

𝑠_𝑖1 ds = 0

Ug= −𝑚𝑔∆𝑦 𝑈𝑠= −¹

2𝑘(𝑠22− 𝑠12) U_f= −𝐹_𝑓𝑘∆𝑠

Vg= 𝑚𝑔ℎ 𝑉𝑠=1

2𝑘𝑥²

T1+ 𝑉1+ 𝑈1>2= 𝑇2+ 𝑉2

If (𝐔_𝟏>𝟐= 𝟎) T₁+ 𝑉₁= 𝑇₂+ 𝑉₂

L⃑ = mv⃑ m₁v_i1+ 𝑚₂𝑣_𝑖2

= 𝑚1𝑣𝑓1+ 𝑚2𝑣𝑓2

Inelastic collision Conservation of

momentum: Constant force

Conservation of momentum:

Avg force

Conservation of momentum:

∑ 𝑭 = 𝟎 || ∆𝐭 = 𝟎 Multiple particles Moment m₁v_i1+ 𝑚₂𝑣_𝑖2

= (𝑚1+ 𝑚2)𝑣𝑓 ∫ 𝐹

𝑡₂ 𝑡₁

dt = 𝐹 ∆𝑡 ∫ 𝐹

𝑡₂ 𝑡₁

dt = 𝐹⃑⃑⃑⃑⃑⃑⃑⃑ ∆𝑡𝑎𝑣𝑔 mv⃑⃑⃑ = mv1 ⃑⃑⃑ 2

L1

⃑⃑⃑ = L⃑⃑⃑⃑ ₂ ∑ 𝑚_𝑖( v⃑⃑⃑⃑⃑ ) + ∑ ∫ 𝐹_𝑖1 ^𝑡2⃑⃑ 𝑖 𝑡₁ dt =

∑ 𝑚_𝑖( v⃑⃑⃑⃑⃑ ) _𝑖2

𝑀0

⃑⃑⃑⃑⃑ = 𝑟⃑⃑⃑ × 𝐹 ₀

Angular momentum Principle of angular momentum and impulse

Conservation of linear momentum

Rigid body motions Instantaneous centre of zero velocity

(Point where perpendicular vectors of velocities meet)

𝐻

₀

⃑⃑⃑⃑ = 𝑟 ⃑⃑⃑ × 𝑚𝑣

0

|𝐻 ⃑⃑⃑⃑ | = 𝑟

0 0

𝑚𝑣 sin 𝜃

= 𝑟

₀

𝑚𝑣

_𝜃

H01

⃑⃑⃑⃑⃑⃑ + ∫ ∑ 𝜇_𝑡^𝑡2 ⃑⃑⃑⃑ 𝑑𝑡0 1 = H⃑⃑⃑⃑⃑⃑ 02

𝜇⃑⃑⃑⃑ =₀ ^{𝑑H⃑⃑⃑⃑⃑ 0}

𝑑𝑡

∑ 𝐻_01𝑖= ∑ 𝐻_02𝑖 - Translation - Fixed rotation - General motion

Fixed rotation General motion Translation

Angular displacement θ⃑ Angular velocity 𝜔⃑⃑

Angular acceleration 𝛼 v_𝑃

⃑⃑⃑⃑ = 𝜔⃑⃑ × 𝑟⃑ 𝑎_𝑃

⃑⃑⃑⃑ = 𝜔⃑⃑ × 𝑣⃑⃑ + 𝛼 × 𝑟 If 𝛼 constant,

ω = ω0+ 𝛼𝑐𝑡 𝜃 = 𝜃0+ ω0𝑡 +1

2𝛼𝑐𝑡² ω²= ω02+ 2𝛼𝑐𝜃

Decompose the motion Translation > rotation

ω =

^{𝑣𝐵−𝑣𝐴}

𝑟_𝐵/𝐴

vB

⃑⃑⃑⃑ = 𝑣⃑⃑⃑⃑ + 𝜔𝐴 ⃑⃑ × 𝑟⃑⃑⃑⃑⃑⃑⃑⃑ 𝐵/𝐴

𝑎

_B

⃑⃑⃑⃑

= 𝑎 ⃑⃑⃑⃑ + 𝜔

_𝐴

⃑⃑ × 𝑣 ⃑⃑⃑⃑⃑⃑⃑⃑⃑

_𝐵/𝐴

+ 𝛼 × 𝑟 ⃑⃑⃑⃑⃑⃑⃑⃑

𝐵/𝐴

= 𝑎 ⃑⃑⃑⃑ − 𝜔

_𝐴 ²

𝑟 ⃑⃑⃑⃑⃑⃑⃑

_𝐵/𝐴

+ 𝛼𝑟 ⃑⃑⃑⃑⃑⃑⃑⃑

𝐵/𝐴

|𝑎𝐵/𝐴𝑡| = 𝑟𝛼

|𝑎_{𝐵/𝐴𝑛}| = 𝜔²𝑟

𝑟𝐵

⃑⃑⃑ = 𝑟⃑⃑⃑ + 𝑟𝐴 ⃑⃑⃑⃑⃑⃑⃑ 𝐵/𝐴

𝑣_𝐵

⃑⃑⃑⃑ = 𝑣⃑⃑⃑⃑ _𝐴 𝑎⃑⃑⃑⃑ = 𝑎𝐵 ⃑⃑⃑⃑ 𝐴

Magnitude don’t change Direction don’t change

Force, Moment, Angular Momentum Parallel Axis Theorem Square moment of inertia Circle moment of inertia Donut moment of inertia 𝐹 = m𝑎⃑⃑⃑⃑ = ∑ 𝑚_{𝐺 𝑖 𝑖}𝑎⃑⃑⃑ _𝑖

HA

⃑⃑⃑⃑⃑ = r⃑⃑⃑⃑⃑⃑⃑⃑ × mv_{p A}⁄ ⃑

= 𝑟⃑⃑⃑⃑⃑⃑⃑ × 𝐿_{𝑝 𝐴}⁄ ⃑ 𝐻_𝐺

⃑⃑⃑⃑⃑ ̇ = ∑ 𝑀⃑⃑⃑⃑⃑ _𝐺 (𝐻⃑⃑⃑⃑⃑ ̇ = ∑ 𝑀𝐺 ⃑⃑⃑⃑⃑ _𝐺 X)

∑ 𝐻 = 𝑤𝐼 H₀= 𝑤𝐼_𝑜 H_𝐺= 𝑤𝐼_𝐺

∑ 𝑀 = 𝐼𝛼 𝑀𝐺= 𝐼𝐺α 𝑀0= 𝐼0α (pinned at O) 𝑀_𝐴= 𝐼_𝐴α (rolling, no slip)

I_p= I_𝐺+ 𝑚𝑑² d – distance between P and G Moment of inertia can be added/subtracted

I𝐺= 1 12𝑚𝑙² I𝐴=1

3𝑚𝑙² G – centre of gravity A – end part of square

IG=¹

2𝑚𝑅² I_G= 𝑚𝑘_𝐺² I =¹

2𝜌𝑡𝜋𝑅⁴ 𝑘𝐺= √^𝐼𝐺

𝑚=^𝑅

√2 𝑘𝐺− radius of gyration

IG=1 𝑅₀ – outer radius 2 𝑅𝑖 – inner radius

I_G=¹

2𝜌𝑡𝜋(𝑅₀⁴− 𝑅_𝑖⁴) m = 𝜌𝑡𝜋(𝑅02− 𝑅_𝑖²)

Kinetic Energy (𝐓_𝟏+ 𝑼_𝟏→𝟐= 𝑻_𝟐) Work by forces Conservation of energy

T_Body=¹

2𝐼_𝐼𝐶𝑤²

=¹

2𝑤²(𝐼_𝐺+ 𝑚𝑑²)

T = Trotate+ 𝑇𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑒

=1 2𝐼_𝐺𝑤²+1

2𝑚𝑣_𝐺²

Ug= −𝑚𝑔∆ℎ 𝑈𝑒= −1

2𝑘(𝑠22− 𝑠12)

𝑀⃑⃑ = 𝑟 × 𝐹 𝑈𝑀= 𝑀(𝜃2− 𝜃1)

T1+ 𝑉1+ U1→2

= 𝑇2+ 𝑉2 U_1→2 – work of nonconservative F

T1+ 𝑉1

= 𝑇2+ 𝑉2 If no

nonconservative F

V = V𝑒+ 𝑉𝑔

V𝑒=¹

2𝑘𝑠² 𝑉_𝑔= 𝑚𝑔ℎ_𝐺

Momentum, impulse and angular momentum Second order differential equations

𝐿⃑ = 𝑚𝑣 L₁

⃑⃑⃑ + ∑ ∫ 𝐹^𝑡2⃑⃑ 𝑖 𝑡₁ dt = L⃑⃑⃑⃑ ₂

H⃑⃑ = r × mv⃑ = Iw 𝑀⃑⃑ = 𝑟 × 𝐹 = 𝐼α 𝐻_𝐺1

⃑⃑⃑⃑⃑⃑⃑ + ∑ ∫ 𝑀^𝑡2⃑⃑⃑⃑⃑ 𝐺 𝑡1 dt = 𝐻⃑⃑⃑⃑⃑⃑⃑ _{𝐺 2} 𝐻_𝑂1

⃑⃑⃑⃑⃑⃑⃑ + ∑ ∫ 𝑀^𝑡2⃑⃑⃑⃑⃑ 𝑂 𝑡₁ dt = 𝐻⃑⃑⃑⃑⃑⃑⃑ _𝑂2

∫ 𝐹 𝑑𝑡 = 0 → L⃑⃑⃑ = L1 ⃑⃑⃑⃑ ₂

∫ Σ𝑀⃑⃑⃑⃑⃑ 𝑑𝑡 = 0 → 𝐻𝐺 ⃑⃑⃑⃑⃑⃑⃑ = 𝐻_𝐺1 ⃑⃑⃑⃑⃑⃑⃑ _{𝐺 2}

∫ Σ𝑀⃑⃑⃑⃑⃑ 𝑑𝑡 = 0 → 𝐻𝑂 ⃑⃑⃑⃑⃑⃑⃑ = 𝐻_𝑂1 ⃑⃑⃑⃑⃑⃑⃑ _𝑂2

m𝑥̈ + c𝑥̇ + 𝑘𝑥 = 0 𝑥 = 𝑒^𝜆𝑡 𝑒^𝜆𝑡(𝑚𝜆²+ 𝑐𝜆 + 𝑘)

= 0

𝜆 =^{−𝑐±√𝑐2−4𝑘𝑚}

2𝑚

𝑐²− 4𝑘𝑚 > 0

|c| > √4𝑘𝑚 𝑥

= 𝐴𝑒^𝜆1𝑡+ 𝐵𝑒^𝜆2𝑡 2 real λs

𝑐²− 4𝑘𝑚 = 0

|c| = √4𝑘𝑚 𝑥

= (𝐴 + 𝐵𝑡)𝑒^𝜆𝑡 one real λ

𝑐²− 4𝑘𝑚 < 0 2 complex λs

λ₁= 𝑎 + 𝑏𝑖 λ₂= 𝑎 − 𝑏𝑖 𝑥 = 𝐴𝑒^𝑎𝑡𝑒^𝑏𝑡𝑖+ 𝐵𝑒^𝑎𝑡𝑒^{−𝑏𝑡𝑖}

= 𝑒^𝑎𝑡(𝛼𝑐𝑜𝑠𝑏𝑡 + 𝛽𝑠𝑖𝑛𝑏𝑡)

Formula Booklet MIE100 Final Exam | Revision 2.1 Apr 20, 2019 | 2 Undamped free vibration – spring motion (horizontal) Vertical spring Parallel / Series spring Undamped free vibration – pendulum motion

𝑥̈ +𝑘 𝑚x = 0 𝑥̈ + 𝑤𝑛2x = 0 w_n= √^𝑘

𝑚

x

= A sin 𝑤_𝑛𝑡 + 𝐵 cos 𝑤_𝑛𝑡

= 𝐶 sin(𝑤_𝑛𝑡 + 𝜃) C = √𝐴²+ 𝐵² 𝜏 =^2𝜋

𝑤_𝑛=¹

𝑓 𝜃 = tan^{−1 𝐵}

𝐴 wn= 2𝜋𝑓

∑ 𝐹_𝑦= 0

𝑚𝑔 − 𝑘(𝑙 − 𝑙₀) = 0 𝛿_𝑒𝑞= 𝑙 − 𝑙₀

=^𝑚𝑔

𝑘

Parallel Series −mg sin 𝜃 = 𝑚𝑎_𝑡 𝑠 = 𝑙θ m𝑙𝜃̈ = −mg sin 𝜃

= −𝑚𝑔𝜃 𝜃̈ +𝑔

𝑙𝜃 = 0

w_n= √𝑔 𝑙 𝜃̈ + 𝑤_𝑛²𝜃 = 0 k_eq= ∑ 𝑘_{𝑖 𝑖}

1 k_eq=

∑ ¹

𝑘_𝑖 𝑖 𝑥̈ +𝑘_𝑒𝑞

𝑚x = 0

Bar pendulum Square pendulum Undamped forced vibration Damping coefficient

w_n= √^3𝑔

2𝑙 wn= √_2√2𝑎^3𝑔 w0 forcing frequency

𝛿0=^𝐹0

𝑘 static deflection

𝑥̈ +𝑘 𝑚x =F0

msin 𝑤₀𝑡 x = x𝑐+ x𝑝

x𝑐= 𝐴 sin 𝑤𝑛𝑡 + 𝐵 cos 𝑤𝑛𝑡 (Transient) x𝑝= C sin 𝑤0𝑡 (steady) x𝑝𝑚𝑎𝑥= C = ^𝐹0

(𝑤_𝑛²−𝑤₀²)𝑚=

𝐹₀/𝑘 (1−(^𝑤0

𝑤𝑛)²)= ^𝛿0

(1−(^𝑤0 𝑤𝑛)²)

M =𝑥_{𝑝𝑚𝑎𝑥} 𝐹0/𝑘 = 1

1 − (𝑤₀ 𝑤_𝑘)² Magnification factor

F_d= −𝑐𝑥̇

(c is damping coefficient)

𝑐c= √^𝑘

𝑚2𝑚 = 2𝑚𝑤𝑛 (critical damping coeff) Damping equation Overdamped (𝐜 > 𝐜𝐜) Critically damped (𝐜 = 𝐜_𝐜) Underdamped (𝐜 < 𝐜𝐜) Electrical circuit analogy

m𝑥̈ + c𝑥̇ + 𝑘𝑥 = 0 𝑥 = 𝑒^𝜆𝑡 𝑒^𝜆𝑡(𝑚𝜆²+ 𝑐𝜆 + 𝑘) = 0

𝜆 =^{−𝑐±√𝑐2−4𝑘𝑚}

2𝑚

𝑥 = 𝐴𝑒^𝜆1𝑡+ 𝐵𝑒^𝜆2𝑡

2 real, negative λs No vibration

(c 2m)

2

−𝑘 𝑚> 0

one real λ No vibration

𝑐_c is smallest c which system won’t vibrate

(c 2m)

2

−𝑘 𝑚= 0 𝑥 = (𝐴 + 𝐵𝑡)𝑒^𝜆𝑡

2 complex λs (^c

2m)²−^𝑘

𝑚< 0 x = D [e^−2mc𝑡sin(𝑤_𝑑𝑡 + 𝜃)]

𝑤𝑑= 𝑤𝑛√1 − (^𝑐

𝑐_𝑐)² ln (^𝑥1

𝑥2) = ^2𝜋(

𝑐 𝑐𝑐)

√1−(^𝑐 𝑐𝑐)²

𝜏_𝑑=^2𝜋

𝑤_𝑑> 𝜏 ^{x = x0e}

−c 2m𝑡

Mass <-> inductance Damp coeff. <-> resistance Spring con. <-> 1/capacitance Force <-> voltage Displacement <-> charge Velocity <-> current