---
title: 'Practice Exam 1'
subtitle: ' '
summary: ''
date: "2024-02-27"
featured: true
draft: false

#   Associate this post with one or more of your projects.
#   Simply enter your project's folder or file name without extension.
#   E.g. `projects = ["internal-project"]` references `content/project/deep-learning/index.md`.
#   Otherwise, set `projects = []`.
projects: []
---

<style type="text/css"> @media print { body { display:none } } </style>

\newcommand{\bA}{\boldsymbol{A}}
\newcommand{\bB}{\boldsymbol{B}}
\newcommand{\bC}{\boldsymbol{C}}
\newcommand{\bX}{\boldsymbol{X}}
\newcommand{\bD}{\boldsymbol{D}}
\newcommand{\bG}{\boldsymbol{G}}
\newcommand{\bH}{\boldsymbol{H}}
\newcommand{\bM}{\boldsymbol{M}}
\newcommand{\br}{\boldsymbol{r}}
\newcommand{\be}{\boldsymbol{\hat{e}}}
\newcommand{\bu}{\boldsymbol{\hat{u}}}
\newcommand{\bv}{\boldsymbol{\hat{v}}}
\newcommand{\bw}{\boldsymbol{\hat{w}}}
\newcommand{\bx}{\boldsymbol{\hat{x}}}
\newcommand{\by}{\boldsymbol{\hat{y}}}
\newcommand{\bz}{\boldsymbol{\hat{z}}}
\newcommand{\bV}{\boldsymbol{V}}
\newcommand{\bU}{\boldsymbol{U}}
\newcommand{\bvv}{\boldsymbol{v}}
\newcommand{\buu}{\boldsymbol{u}}

\newcommand{\bha}{\boldsymbol{\hat{a}}}
\newcommand{\bhb}{\boldsymbol{\hat{b}}}
\newcommand{\bhc}{\boldsymbol{\hat{c}}}
\newcommand{\bze}{\boldsymbol{0}}

\newcommand{\te}{\theta}
\newcommand{\al}{\alpha}
\newcommand{\om}{\omega}
\newcommand{\bom}{\boldsymbol{\omega}}
\newcommand{\bal}{\boldsymbol{\alpha}}
\newcommand{\Si}{\Sigma}
\newcommand{\la}{\lambda}

\newcommand{\vel}{\boldsymbol{v}}
\newcommand{\cE}{{\cal E}}
\newcommand{\cF}{{\cal F}}
\newcommand{\cA}{{\cal A}}
\newcommand{\cB}{{\cal B}}
\newcommand{\cC}{{\cal C}}
\newcommand{\cI}{{\cal I}}
\newcommand{\cV}{{\cal V}}
\newcommand{\cR}{{\cal R}}

\newcommand{\ba}{\boldsymbol{a}}
\newcommand{\dte}{\dot{\theta}}
\newcommand{\ddte}{\ddot{\theta}}
\newcommand{\dr}{\dot{r}}
\newcommand{\ddr}{\ddot{r}}
\newcommand{\dpsi}{\dot{\psi}}
\newcommand{\ddpsi}{\ddot{\psi}}
\newcommand{\dphi}{\dot{\phi}}
\newcommand{\ddphi}{\ddot{\phi}}
\newcommand{\dx}{\dot{x}}
 
[[MEEG620]](/courses/620)

<hr>


**Problem **

Consider in Figure 1 two identical cylinders of radius $r$ 
denoted as **1** and **2** in planar motion
in a referential **0**.
Cylinder **1** rolls without slipping along axis $(O, \bx_0)$ of **0**. Denote by $I$ the corresponding contact point. Cylinder **2** rolls and slips
along axis $(O, \bz_0)$ of **0**. Denote by $J$ the corresponding contact point. The two cylinders are in contact at point $K$: assume no-slip at $K$.
Denote by $\theta$ the (oriented) angle which defines the orientation 
of the line connecting the centers $O_1$ and $O_2$ of the cylinders.

![Figure 1](/courses/620/exam1a/2cyls.png){width=50%}
**1.** Using a graphical method, find the IC $I_2$ of body **2**: conclude that slip must occur at $J$.

**2.** Find $\br_{OO_1}$ as a function of $\theta$. Deduce velocity
$\vel_{O_1 /0}$ by taking the time-derivative of $\br_{OO_1}$. Then impose no-slip at $I$ to relate $\bom_{1/0}$ to $\dte$. Also find $\vel_{K\in 1/0}$.

**3.** Find $\br_{OO_2}$ as a function of $\theta$. Deduce velocity $\vel_{O_2/0}$ by taking the time-derivative of $\br_{OO_2}$.
Find $\vel_{K\in 2/0}$ in terms of $\bom_{2/0}$.
Now impose no-slip at $K$ to relate $\bom_{2/0}$ to $\dte$.

**4.** Conclude this kinematic study by giving the expressions of the
kinematic screws of the two bodies \verb=1= and \verb=2= (relative to \verb=0=) in terms of $r$, $\theta$, and $\dot{\theta}$.

Note: all derivations can be done on fixed basis $(\bx_0,\by_0,\bz_0)$.



<hr>

## Formulas:

**1. Time-Derivative Formula.**
Given two rigid bodies (or referentials) $\cE$ and $\cF$, and an arbitrary vector
function $\bU(t)$,
\[
\Big(\frac{d\bU}{dt} \Big)_\cE =\Big(\frac{d\bU}{dt} \Big)_\cF + \bom_{\cF/ \cE}\times \bU
\]

**2. Triple product formula:** 
\[
\bA\times (\bB\times \bC)= (\bA\cdot \bC) \bB- (\bA\cdot \bB) \bC$
\]


**3. Rigid Body Kinematics:** Given two rigid bodies (or referentials) $\cE$ and $\cF$,  for any two points $P$ and $Q$, and at any given time,
\[
\vel_{Q \in \cF/\cE}=  \vel_{P \in \cF /\cE}+ \bom_{\cF/ \cE}\times \br_{PQ}
\]
\[
\ba_{Q \in \cF/\cE}=  \ba_{P \in \cF /\cE}+ \bal_{\cF/ \cE}\times \br_{PQ}
+ \bom_{\cF/ \cE}\times (\bom_{\cF/ \cE}\times \br_{PQ})
\]

**4. Change of Referential Formulas.** Given two rigid bodies (or referentials) $\cE$ and $\cF$, for any particle $P$, and at any given time,
\[
\vel_{P/\cE} = \vel_{P/\cF}  + \vel_{P\in \cF / \cE}
\]
\[
\ba_{P/\cE} = \ba_{P/\cF}  + \ba_{P\in \cF / \cE} + 2 \bom_{\cF/\cE} \times \vel_{P/\cF}
\]

**5. Loop formulas:** Given 3 rigid bodies $\cA$, $\cB$, $\cC$, and an arbitrary point $P$,
\[
\bom_{\cA/\cB}+ \bom_{\cB/\cC}+ \bom_{\cC/\cA} = \bze
\]
\[
\vel_{P\in \cA / \cB}+ \vel_{P\in \cB / \cC}+ \vel_{P\in \cC / \cA}=\bze
\]

<hr>

**Solution**

**1.** Using $I$ as the IC of body **1**, you can find the direction of velocity $\vel_{K\in 1}$ as the normal to line
$IK$. Since $\vel_{K\in 2}=\vel_{K\in 1}$, we find that $I_2$ (the IC of body **2**) is located on line $IK$. It is also
located on line $JO_2$: hence $I_2$ is at the intersection
of lines $IK$ and $JO_2$. $J$ is not the IC of body **2**.

**2.** We find the velocity of the center $O_1$ by taking the time-derivative of the position vector
\[
\br_{OO_1}= (r+2r\cos\te)\bx_0+ r\bz_0
\]
This leads to
\[
\boxed{\vel_{O_1/0} = -2r\dte\sin\te\bx_0}
\]
We must impose no-slip condition at $I$: denoting $\bom_{1/0}= \om_1 \by_0$, we find
\[
\vel_{I\in 1/0}= -2r\dte\sin\te\bx_0 +\om_{1}\by_0 \times (-r\bz_0) =\bze
\]
This gives
\[
\boxed{\om_{1}= -2\dte\sin\te}
\]
We can now find $\vel_{K\in 1/0}$ from $\vel_{O_1 /0}$:
\[
\vel_{K\in 1/0} = -2r\dte\sin\te\bx_0 -2\dte\sin\te \by_0 \times r(-\cos\te\bx_0+
\sin\te\bz_0)
= - 2r\dte\sin\te ( (1+\sin\te)\bx_0 +\cos\te\bz_0)
\]



![Figure 1](/courses/620/exam1a/2cylsS.png){width=40%}

**3.** We repeat this work for body **2**: from  the position vector
of $O_2$
\[
\br_{OO_2}=r\bx_0 + (r+2r\sin\te)\bz_0
\]
we find
\[
\boxed{\vel_{O_2 /0}=2r\dte\cos\te\bz_0}
\]
Note that the directions of the vectors $\vel_{O_1/0}$ and $\vel_{O_2/0}$
are compatible with the expected motion of the centers.
We can find $\vel_{K\in 2/0}$ from $\vel_{O_2/0}$ and $\bom_2 =\om_2 \by_0$
($\om_2$ is yet to found):
\[
  \vel_{K\in 2/0} = 2r\dte\cos\te\bz_0 + \om_{2}\by_0 \times r(\cos\te\bx_0-
\sin\te\bz_0)
=-r  \om_{2}\sin\te \bx_0 + r\cos\te (2\dte - \om_{2})\bz_0
\]
Now we impose the equality $\vel_{K\in 2/0} =\vel_{K\in 1/0}$ to find 2 equations
\[
2\dte(1+\sin\te)=   \om_{2} 
,\quad
2\dte\sin\te = (-2\dte + \om_{2})
\]
which give the same solution
\[
\boxed{\om_{2} =2\dte(1+\sin\te)}
\]

**4.** At this point, we have all we need to find the kinematic screws of the 2 bodies
\[
\boxed{\{\cV_{1/0}\} = \begin{Bmatrix}
  \bom_{1/0}
  \\
  \vel_{I\in 1/0}
                \end{Bmatrix}
                =\begin{Bmatrix}
  -2\dte\sin\te \by_0
  \\
  \bze
  \end{Bmatrix}_I}
\]
\[
\boxed{\{\cV_{2/0}\} = \begin{Bmatrix}
  \bom_{2/0}
  \\
  \vel_{O_2/0}
              \end{Bmatrix}
              =
\begin{Bmatrix}
  2\dte(1+\sin\te) \by_0
  \\
  2r\dte\cos\te\bz_0
\end{Bmatrix}_{O_2}    
}
\]

*Note 1: Be careful, don't confuse $\vel_{I\in 1/0}$ with $\vel_{I/0}$. Similarly, $\vel_{K/0}$ is not equal to $\vel_{K\in 1/0}=\vel_{K\in 2/0}$.*

*Note 2: Also do not set $\bom_{1/0}$ to $\dte \by_0$.*


<hr>

&copy; 2024 \ R. Val&eacute;ry Roy. 
