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% Document Info
\author{Jaco (brandenz1229)}
\pdfinfo{
  /Title (midterm-3.pdf)
  /Creator (Cheatography)
  /Author (Jaco (brandenz1229))
  /Subject (MidTerm 3 Cheat Sheet)
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\noindent
\begin{multicols}{3}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{MidTerm 3 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Jaco (brandenz1229)} via \textcolor{DarkBackground}{\uline{cheatography.com/138824/cs/29996/}}}
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\noindent
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Jaco (brandenz1229) \\
  \uline{cheatography.com/brandenz1229} \\
  \end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 1st December, 2021.\\
   Updated 1st December, 2021.\\
   Page {\thepage} of \pageref{LastPage}.
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 9}}  \tn
% Row 0
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\mymulticolumn{2}{x{5.377cm}}{{\bf{Angular Velocity and Acceleration}}} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
Θ = angle (radians) & s = length \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
r = radius & 90\textasciicircum{}o\textasciicircum{} = π/2 rad \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
Θ = (s/r) & s = r ⋅ Θ \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
1 rad = (360\textasciicircum{}o\textasciicircum{} / 2π) = 57.3\textasciicircum{}o\textasciicircum{} & 180\textasciicircum{}o\textasciicircum{} = π rad \tn 
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\SetRowColor{white}
{\bf{Angular Velocity}} & (1st Derivative) \tn 
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\SetRowColor{LightBackground}
ω = (Θ`f`- Θ`i`) / (t`f`-t`i`) & ω = "velocity" \tn 
% Row Count 9 (+ 2)
% Row 7
\SetRowColor{white}
1 rev/s = 2π rad/s & 1 rev/min = 1 rpm = 2π/60 rad/s \tn 
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% Row 8
\SetRowColor{LightBackground}
{\bf{Angular Acceleration}} & (2nd Derivative) \tn 
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% Row 9
\SetRowColor{white}
α = (ω`f`- ω`i`) / (t`f`-t`i`) & α = "acceleration" \tn 
% Row Count 15 (+ 2)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\bf{Rotation w/ Constant Angular Acceleration}}} \tn 
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% Row 11
\SetRowColor{white}
α`f` = (ω`f`- ω`i`) / (t - 0) & α`f` = constant \tn 
% Row Count 18 (+ 2)
% Row 12
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\mymulticolumn{2}{x{5.377cm}}{ω`f` = ω`i` + α`f` ⋅ t} \tn 
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% Row 13
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{Θ`f`- Θ`i` = 1/2(ω`i`+ω`f`) ⋅ t} \tn 
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% Row 14
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Θ`f` = Θ`i`+ (ω`i`⋅ t ) + 1/2 (α`f`⋅ t\textasciicircum{}2\textasciicircum{})} \tn 
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% Row 15
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{ω`f`\textasciicircum{}2\textasciicircum{} = ω`i`\textasciicircum{}2\textasciicircum{} + 2⋅α`f`(Θ`f`- Θ`i`)} \tn 
% Row Count 22 (+ 1)
% Row 16
\SetRowColor{LightBackground}
{\bf{Relating Linear and Angular Kinematics}} & K = 1/2(m ⋅ v\textasciicircum{}2\textasciicircum{}) \tn 
% Row Count 25 (+ 3)
% Row 17
\SetRowColor{white}
{\bf{Linear Speed in Rigid-Body Rotation}} & s = r ⋅ Θ \tn 
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% Row 18
\SetRowColor{LightBackground}
Linear Speed & v = r ⋅ ω \tn 
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% Row 19
\SetRowColor{white}
{\bf{Linear Acceleration in Rigid-Body Rotation}} & a`tan` = r ⋅ α \tn 
% Row Count 31 (+ 3)
\end{tabularx}
\par\addvspace{1.3em}

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\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 9 (cont)}}  \tn
% Row 20
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{\bf{Centripetal Component of Acceleration}} & a`rad` = (v\textasciicircum{}2\textasciicircum{}/r) = ω\textasciicircum{}2\textasciicircum{}⋅r \tn 
% Row Count 3 (+ 3)
% Row 21
\SetRowColor{white}
{\bf{Energy in Rotational Motion}} & K`E`: 1/2⋅m⋅v\textasciicircum{}2\textasciicircum{} = 1/2⋅m⋅r\textasciicircum{}2\textasciicircum{}⋅ω\textasciicircum{}2\textasciicircum{} \tn 
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K = 1/2⋅m⋅r\textasciicircum{}2\textasciicircum{}⋅ω\textasciicircum{}2\textasciicircum{} & I = m⋅r\textasciicircum{}2\textasciicircum{} \tn 
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% Row 23
\SetRowColor{white}
{\bf{Gravitational Potential Energy for an Extended Body}} & U = M⋅g⋅y`cm` \tn 
% Row Count 11 (+ 3)
% Row 24
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{\bf{Moment of Inertia}} & I`p`=I`cm`+Md\textasciicircum{}2\textasciicircum{} \tn 
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\begin{tabularx}{5.377cm}{x{2.73735 cm} x{2.23965 cm} }
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 9 Cont:}}  \tn
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{\bf{Rotational Kinetic Energy}} & K = Joules \tn 
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K = 1/2⋅I⋅ω\textasciicircum{}2\textasciicircum{} & R = Radius \tn 
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\mymulticolumn{2}{x{5.377cm}}{M = mass pivoted about an axis} \tn 
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Perpendicular to the Rod & I = (M⋅L\textasciicircum{}2\textasciicircum{}) / 3 \tn 
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Slender Rod (Axis Center) & I = 1/12M⋅L\textasciicircum{}2\textasciicircum{} \tn 
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Slender Rod (Axis End) & I = 1/3M⋅L\textasciicircum{}2\textasciicircum{} \tn 
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Rectangular Plate (Axis Center) & I = 1/12M⋅(a\textasciicircum{}2\textasciicircum{}+b\textasciicircum{}2\textasciicircum{}) \tn 
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% Row 7
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Rectangular Plate (Axis End) & I = 1/3M⋅(a\textasciicircum{}2\textasciicircum{}) \tn 
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Hallow Cylinder & I = 1/2M(R`i`\textasciicircum{}2\textasciicircum{}+R`f`\textasciicircum{}2\textasciicircum{}) \tn 
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Solid Cylinder & I = 1/2MR\textasciicircum{}2\textasciicircum{} \tn 
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Hollow Cylinder (Thin) & I = MR\textasciicircum{}2\textasciicircum{} \tn 
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Solid Sphere & I = 2/5MR\textasciicircum{}2\textasciicircum{} \tn 
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Hollow Sphere (Thin) & I = 2/3MR\textasciicircum{}2\textasciicircum{} \tn 
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 11: Equilibrium and Elasticity}}  \tn
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{\bf{1st Condition of Equilibrium (at rest)}} & ΣF = 0 \tn 
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{\bf{2nd Condition of Equilibrium (nonrotating)}} & Στ = 0 \tn 
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{\bf{Center of Gravity}} & r`cm` = (m`1`⋅ r`1`) / m`1` \tn 
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{\bf{Solving Rigid-Body Equilibrium Problems}} & ΣF`x` = 0 \tn 
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1st Condition & ΣF`x` = 0 \{\{nl\}\} ΣF`y` = 0 \tn 
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2nd Condition (Forces xy-plane) & Στ`z` = 0 \tn 
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{\bf{Stress, Strain, and Elastic Moduli}} & Stress = Force Applied to deform a body \{\{nl\}\} Strain = how much deformation \tn 
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{\bf{Hooke's Law}} & (Stress / Strain) = Elastic Modulus \tn 
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A = Area & F = Magnitude of Force \tn 
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{\bf{Tensile Stress}} & F / A \tn 
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% Row 10
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1 Pascal = Pa = 1 N/m\textasciicircum{}2\textasciicircum{} & 1 psi = 6895 Pa \tn 
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l = length & 1 Pa = 1.450 ⋅ 10\textasciicircum{}-4\textasciicircum{} \tn 
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{\bf{Tensile Strain}} & (l`f`- l`i`) / (l`i`) \tn 
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\end{tabularx}
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\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 11: Equilibrium and Elasticity (cont)}}  \tn
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{\bf{Young Modulus}} & (Tensile Stress) / (Tensile Strain) \tn 
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{\bf{Pressure}} & p = F (Force Fluid is Applied) / A (Area which force is exerted) \tn 
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{\bf{Bulk Stress}} & (p`f` - p`i`) \tn 
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\SetRowColor{white}
{\bf{Bulk Strain}} & (V`f`- V`i`) / (V`i`) \tn 
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\SetRowColor{LightBackground}
{\bf{Bulk Modulus}} & Bulk Stress / Bulk Strain \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.68758 cm} x{2.28942 cm} }
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 10: Dynamics of Rotational Motion}}  \tn
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\mymulticolumn{2}{x{5.377cm}}{{\bf{Torque}}} \tn 
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\SetRowColor{white}
F = Magnitude of F & || || = Magnitude Symbol \tn 
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τ = F⋅l = r⋅F⋅sinΘ = F`tan`r & L = lever arm of F \tn 
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\mymulticolumn{2}{x{5.377cm}}{τ = ||r|| x ||F||} \tn 
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\mymulticolumn{2}{x{5.377cm}}{{\bf{Torque and Angular Acceleration for a Rigid Body}}} \tn 
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Newtons 2nd Law of Tangential Component & F`tan` = m`1`⋅a`1` \tn 
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\mymulticolumn{2}{x{5.377cm}}{{\bf{Rotational analog of Newton's second law for a rigid body}}} \tn 
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Στ`z` = l⋅α`z` & `z` = rigid body about z-axis \tn 
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\mymulticolumn{2}{x{5.377cm}}{{\bf{Combined Translation and Rotation: Energy Relationships}}} \tn 
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% Row 9
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\mymulticolumn{2}{x{5.377cm}}{K = 1/2M⋅v\textasciicircum{}2\textasciicircum{} + 1/2⋅I⋅ω\textasciicircum{}2\textasciicircum{}} \tn 
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{\bf{Rolling without Slipping}} & v = R⋅ω \tn 
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\mymulticolumn{2}{x{5.377cm}}{{\bf{Combined Translation and Rotation: Dynamics}}} \tn 
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Rotational Motion about the center of mass & Στ`z` = l⋅α`z` \tn 
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% Row 13
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{\bf{Work and Power in Rotational Motion}} & F = M ⋅ a \tn 
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% Row 14
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When it rotates from Θ`i` to Θ`f` & W = ∫ (Θ`f` to Θ`i`) τ`f` dΘ \tn 
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% Row 15
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When the torque remains constant while angle changes & W = τ`f`(Θ`f` to Θ`i`) \tn 
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% Row 16
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Total WorkDone on rotating rigid body & W = 1/2(ω`f`\textasciicircum{}2\textasciicircum{}) - 1/2(ω`i`\textasciicircum{}2\textasciicircum{}) \tn 
% Row Count 31 (+ 2)
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\begin{tabularx}{5.377cm}{x{2.68758 cm} x{2.28942 cm} }
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 10: Dynamics of Rotational Motion (cont)}}  \tn
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Power due to torque on rigid body & P = τ`z`⋅ ω`z` \tn 
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Angular Momentum & L = r x p (r x m ⋅ v) \tn 
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Angular Momentum of a Rigid Body & L = m`i`⋅r`i`\textasciicircum{}2\textasciicircum{}⋅ω \tn 
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Chapter 11: Equilibrium and Elasticity (cont.)}}  \tn
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 & F = Force acting tangent to the surface divided by the Area \tn 
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% Row 1
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{\bf{Shear Stress}} & F / A \tn 
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h = transverse dimension {[}bigger{]} & x = relative displacement (empty) {[}smaller{]} \tn 
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{\bf{Shear Strain}} & x / h \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
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% That's all folks
\end{multicols*}

\end{document}