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\centerline{\bf ME 115(b): Homework $\#5$ Soltution}
\vskip 0.4 true in


{\bf Problem $\#1$}:

Recall that the definition of the evolute, $\beta(t)$, of a {\it planar}
curve, $\alpha(t)$, is: 
   $$\beta(t)=\alpha(t) + {1\over{\kappa(t)}} \vec{n}(t)$$
where $\kappa(t)$ is the curvature of $\alpha(t)$ and $\vec{n}(t)$ is the unit
normal vector of $\alpha(t)$ at $t$.  Let's assume that the curve is
arc-length parametrized, i.e., $t$ is the arc-length parameter.

The tangent to the evolute is simply derived by taking the derivative of
the evolute equation.
\begin{equation}
{ {d\beta}\over{dt}} = \vec{u} + {d\over{dt}}
	\left({1\over{\kappa(t)}}\vec{n}(t)\right) 
     = \vec{u} + { {\kappa(t)\vec{n}^{'}(t)-\vec{n}(t)\kappa^{'}(t)}
	\over{\kappa^2(t)}} 
\end{equation}
Recall that for a regular curve, $\vec{u}(t)=\alpha^{'}(t)$, $\vec{u}^{'}(t)
=\kappa(t)\vec{n}(t)$.  

What is $\vec{n}^{'}(t)$ for a planar curve?  Since $\vec{n}(t)$ is a unit
length curve, then $\vec{n}^{'}(t)$ must be a vector orthogonal to
$\vec{n}(t)$--i.e., a vector in the direction of $\vec{u}(t)$.  Thus, assume
that $\vec{n}^{'}=\gamma\vec{u}$, where $\gamma$ is some proportionality
constant, which is to be determined.

From the relationship $\vec{u}\cdot\vec{n}=0$, we can obtain (by taking
derivatives):
\begin{eqnarray}
      0 &=& \vec{u}^{'} \cdot\vec{n} + \vec{u}\cdot\vec{n}^{'} \\
         &=& \kappa + \vec{u}\cdot\vec{n}^{'} \\
	 &=& \kappa + \vec{u}\cdot(\gamma\vec{u}) \\
        &=& \kappa + \gamma
        \end{eqnarray}
This implies that:
  $$\vec{n}^{'} = -\kappa\vec{u} $$

Combining these results, gives:
\begin{eqnarray}
   {{d\beta(t)}\over {dt}} &=& \vec{u}  + 
    { {\kappa(t)\vec{n}^{'} - \vec{n}(t)\kappa^{'}}\over{\kappa^2}}\\
  &=& \vec{u} + {{-\kappa^2\vec{u} -\kappa^{'} \vec{n}}\over{\kappa^2}}\\
  & =& {{-\kappa^{'}}\over\kappa}\vec{n} r
  \end{eqnarray}
  
Thus,  $d\beta/dt$ is a vector in the direction of the normal to $\alpha(t)$.

{\bf Part (b):} 
The evolute of a circle is simply a point.

For a circle:
\begin{equation}
\alpha(t)=\begin{bmatrix} r\cos(\frac{t}{r})\\ r \sin(\frac{t}{r})\end{bmatrix}
\end{equation}

\begin{equation}
\beta(t)=\begin{bmatrix} r \cos(\frac{t}{r})\\r\sin(\frac{t}{r})\end{bmatrix}+r\begin{bmatrix}-\cos(\frac{t}{r})\\\-sin(\frac{t}{r})\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\end{equation}

{\bf Problem $\#$2}

{\bf Part (a):} Let's define the object frames on the two bodies as follows:
%%------------------------------------------------------------------------%%
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%\centerline{\psfig{figure=ellipse2.ps,width=0.65\hsize}}
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%%------------------------------------------------------------------------%%

Recall that the boundary of the ellipse can thus be parametrized as:
  $$c_1(\mu_1) =	\begin{bmatrix}a \cos(\mu_1) \\
		              b \sin(\mu_1) \end{bmatrix} $$
where $\mu_1$ is the ``curve parameter'' of the ellipse boundary.  It is
not necessarily the arc-length parameter.  Note, with this parametrization,
the normal vector (see below) is pointing {\it inward} into the object.
The circle can be paramterized as:
  $$c_2(\mu_2) = \begin{bmatrix} r \cos(\mu_2)\\ - r \sin(\mu_2)\end{bmatrix}$$

Thus, the tangent vectors (not necessarily unit length) to the two surfaces
are:
  \begin{equation}
  \vec{t}_1(\mu_1) =\begin{bmatrix}-a\sin(\mu_1) \\ b\cos(\mu_1) \end{bmatrix}
  \end{equation}
  
  \begin{equation}
     \vec{t}_2(\mu_2) = \begin{bmatrix}  - r \sin(\mu_2)\\ -r \cos(\mu_2)\end{bmatrix}
     \end{equation}
Similarly:
  \begin{eqnarray}
       M_1 &=& |\vec{t}_1| = (a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{1/2} \\
       M_2 &=& |\vec{t}_2| = r \end{eqnarray}

Recall that the curvature of the $i^{th}$ planar surface could be computed as:
  $$\kappa_i = -M_i^{-2} {({{\partial c_i}\over{\mu_i}})}^T
	{ {\partial \vec{n}_i}\over{\partial\mu_i}}$$
For planar curves, the normal can be easily computed as the unit tangent vector rotated by
$\pm\pi/2$
  $$\vec{n}_1 = - {1 \over{(a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{1/2}}}
	\begin{bmatrix} b\cos\mu_1 \\ a\sin\mu_1 \end{bmatrix}$$
Consequently:
  $$ { {\partial\vec{n}_i}\over{\partial\mu_1}} = 
	- {{ab}\over{(a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{3/2}}}
	\begin{bmatrix} -a \sin\mu_1 \\ b\cos\mu_1 \end{bmatrix} $$
Thus:

  $$\kappa_1 = {{ab}\over {(a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{3/2}}}$$


  $$\vec{n}_2 =
	\begin{bmatrix}  -r\cos(\mu_2)\\  -r\sin(\mu_2)  \end{bmatrix}$$
Consequently:
  $$ { {\partial\vec{n}_2}\over{\partial\mu_2}} = R $$
Thus:

  $$\kappa_2 = \frac{1}{r}$$


Assuming that $v_t$ and $\dot{\theta}$ are given, the contact equations are:

\begin{eqnarray}
\dot{\mu_1} &=& {(\kappa_1 +\kappa_2)}^{-1} {M_1}^{-1}(-\dot{\theta} + \kappa_2v_t)\\
&=&{ ( {{ab}\over {(a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{3/2}}}+\frac{1}{r})}^{-1} {(a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{1/2}}^{-1}(-\dot\theta + \frac{1}{r}v_t)\\
\dot\mu_2 &=& {(\kappa_1+\kappa_2)}^{-1}{M_2}^{-1}(\dot\theta +\kappa_1v_t)\\
&=&{ ( {{ab}\over {(a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{3/2}}}+\frac{1}{r})}^{-1}\frac{1}{r}(\dot\theta +{{ab}\over {(a^2\sin^2\mu_1 + b^2\cos^2\mu_1)^{3/2}}} v_t)
\end{eqnarray}
\bigskip
The relative curvature is ill defined when $\kappa_1+\kappa_2=0$.
In other words, when:
\begin{equation}\frac{ab}{{(a^2\sin^2\mu_1+b^2\cos^2\mu_1)}^\frac{3}{2}}=-\frac{1}{r}
\end{equation}

\end{document}




