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\begin{document}

\newpage
\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\CIRCULAR ROLLER COASTER}


\markboth{Circular Roller Coaster}{Circular Roller Coaster}

\author{Kevin S. Huang\\
Centennial High School\\
Ellicott City MD 21042 USA
}

\makeStitlePDFLaTex

\begin{abstract}
Students will study the dynamics of a circular roller coaster and work out the equations of motion in the ideal case as well as considering the interesting complication of including kinetic friction. This problem is an excellent introduction for students to apply physics in real-life. A basic understanding of Newton's Laws, work, and energy is expected. Solving the problem will involve proficiency with the fundamental theorem of calculus, first-order linear differential equations,   integrating factors, and integration by parts.
\end{abstract}



\section*{SCENARIO DESCRIPTION}

\begin{flushleft}

Consider a cart of mass $m$ moving through a vertical circle of radius $R$. We will assume the size of the cart is much smaller than the circle and neglect the rotational inertia of the cart's wheels. The acceleration due to gravity is $g$ (Figure~\ref{fig:1-120-CoasterImage}).

\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.40]{1-120-rollercoaster.png} \\
\caption{\label{fig:1-120-CoasterImage}  Circular roller coaster.}
\end{center}
\end{figure}

A typical textbook problem is to calculate the minimum initial velocity of the cart for it to stay on the track throughout the loop  \cite{Serway2008}. 

\begin{enumerate}

\item Find this minimum initial velocity.

\item Derive the equation of motion for the cart and verify your answer to (1) is correct.

\item Now include the presence of kinetic friction and derive the equation of motion for the cart. 
\item Suppose the coefficient of friction between the wheels of the cart and circular track is $\mu$. Verify this equation reduces to the frictionless case (2) when $\mu=0$ and find the minimum initial velocity of the cart for it to stay on the track throughout the loop in the case of friction.

\end{enumerate}

\end{flushleft}

\begin{thebibliography}{98}

\bibitem{Serway2008} Serway, Raymond A., \& Jewett, Jr., John W. 2008. {\it Physics for Scientists and Engineers with Modern Physics, Ninth Edition\/}. Boston MA:  Brooks/Cole.

\end{thebibliography}

\teacher{
\section*{NOTES FOR TEACHERS}

\textbf{Answers}

\begin{enumerate}
\item $$v_0=\sqrt{5gR}$$.\vspace*{-.4 true cm}
\item $$\dot{\theta}=\sqrt{{\omega_0}^2-\frac{
2g(1-\cos{\theta})}{R}}$$.\vspace*{-.4 true cm}
\item $$\dot{\theta}=\sqrt{\left[{\omega_0}^2-\frac{2g(1-2\mu^2)}{R(1+4\mu^2)}\right]e^{-2\mu\theta}+\frac{2g[(1-2\mu^2)\cos{\theta}-3\mu\sin{\theta}]}{R(1+4\mu^2)}}$$.\vspace*{-.4 true cm}
\item $${v_0}=\sqrt{\frac{[3e^{2\mu\pi}+2(1-2\mu^2)]gR}{1+4\mu^2}}$$.\vspace*{-.4 true cm}
\end{enumerate}

For convenience, step-by-step solutions are given.


\textbf{Solutions}

\begin{enumerate}

\item The critical point to analyze is when the cart is at the top of the track. In that case, the normal force can only point downwards.

\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.40]{1-120-rollercoaster2.png} \\
\caption{\label{fig:1-120-Conserve} Conservation of energy.}
\end{center}
\end{figure}





Suppose the cart has initial velocity $v_0$. By conservation of energy, some of the cart's kinetic energy is converted to potential energy at the top of the loop (Figure~\ref{fig:1-120-Conserve}) and this leads to (\ref{eq:first}):

\begin{equation}\label{eq:first}
\frac{1}{2}m{v_0}^2=mg(2R)+\frac{1}{2}mv^2
\end{equation}

For circular motion at the top of track, the centripetal acceleration is provided by gravity and the normal force:

\begin{equation}\label{eq:NF}
mg+N=\frac{mv^2}{R}
\end{equation}

The key realization is that $N \geq 0$ and traveling at the minimum velocity will cause the cart to be just in contact with the track, that is $N=0$. In that case, we can substitute $N = 0$ in (\ref{eq:NF}) to obtain (\ref{eq:mg}):

\begin{equation}\label{eq:mg}
mg=\frac{mv^2}{R}
\end{equation}
and solve for the velocity at the top of the circle (\ref{eq:top}):
\begin{equation}\label{eq:top}
v=\sqrt{gR}\, .
\end{equation}

Using the value of $v$ from (\ref{eq:top}) in (\ref{eq:first}) yields (\ref{eq:next}):

\begin{equation}\label{eq:next}
\frac{1}{2}m{v_0}^2=2mgR+\frac{1}{2}mgR=\frac{5}{2}mgR
\end{equation}
from whence we derive the minimum initial velocity (\ref{eq:minvel}):

\begin{equation}\label{eq:minvel}
v_0=\sqrt{5gR}\, .
\end{equation}

\item To analyze the entire motion, we can conveniently describe the location of the cart with coordinate $\theta$, the angle between the cart and base of the circle (Figure~\ref{fig:1-120-Describe}):


\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.40]{1-120-rollercoaster3.png} \\
\caption{\label{fig:1-120-Describe} Describing the position of the cart with angle $\theta$.}
\end{center}
\end{figure}


By the conservation of energy principle, applied when the cart is at angle $\theta$ in Figure~\ref{fig:1-120-Describe} we have:

\begin{equation}\label{eq:CE}
\frac{1}{2}m{v_0}^2=mgR(1-\cos{\theta})+\frac{1}{2}m{v}^2\, .
\end{equation}

Simplifying (\ref{eq:CE})

\begin{equation}\label{eq:simp}
v^2={v_0}^2-2gR(1-\cos{\theta})\, .
\end{equation}

Note that $v=R\omega=R\dot{\theta}$ from the definition of angular velocity. We similarly define:

\begin{equation*}
\omega_0 \equiv \frac{v_0}{R}
\end{equation*}

Substituting in  (\ref{eq:simp}), we obtain the equation of motion in terms coordinate $\theta$:

\begin{equation}\label{eg:EOM}
\dot{\theta}=\sqrt{{\omega_0}^2-\frac{
2g(1-\cos{\theta})}{R}}
\end{equation}

We can determine whether the cart will stay on the track or when it will lose contact by analyzing the normal force which is subject to the constraint $N \geq 0$. From Figure~\ref{fig:1-120-Describe}, the normal force satisfies:

\begin{equation}\label{eq:NF1}
N-mg\cos{\theta}=\frac{mv^2}{R}
\end{equation}

Substituting from (\ref{eg:EOM}) in terms of $\theta$ we obtain an expression for $N$ in (\ref{eq:N}) :

\begin{equation}\label{eq:N}
N=m R{\omega_0}^2+(3\cos\theta-2)m g
\end{equation}

Now, applying our constraint that $N \geq 0$ in terms of $\omega_0$, we have (\ref{eq:om}):

\begin{equation}\label{eq:om}
\omega_0 \geq \sqrt{\frac{(2-3\cos\theta)g}{R}}
\end{equation}

Choosing $\cos\theta=-1$ maximizes the right-hand side and hence the minimum initial velocity needed is $v_0=R\omega_0=\sqrt{5gR}$ which verifies our intuition.

As a side note, we may also be interested in considering the acceleration of the cart over time (Figure~\ref{fig:1-120-Describe}):

\textbf{Centripetal acceleration}

\begin{equation*}
a_c={\omega}^2R={\omega_0}^2R-2g(1-\cos\theta)
\end{equation*}

\textbf{Tangential acceleration}

\begin{equation*}
a_t=\frac{-mg\sin{\theta}}{m}=-g\sin{\theta}
\end{equation*}

\item Now consider a more interesting track with friction present (Figure~\ref{fig:1-120-ConserveWithFriction}). Let's analyze this system as we did before.



\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.40]{1-120-rollercoaster4.png} \\
\caption{\label{fig:1-120-ConserveWithFriction} Conservation of energy including friction.}
\end{center}
\end{figure}

By conservation of energy, some of the kinetic energy is converted to potential energy while some is lost to friction. Thus (\ref{eq:first}) is now modified to become (\ref{eq:LOF}):

\begin{equation}\label{eq:LOF}
\frac{1}{2}m{v_0}^2=mgR(1-\cos{\theta})+\frac{1}{2}m{v}^2+W_f
\end{equation}

where $W_f$ is work done by friction which is dissipated energy in our system and  is given by (\ref{eq:Work}). Here $N = N(\theta)$ is the normal force to the track at angular position $\theta$.

\begin{equation}\label{eq:Work}
W_f=\int_{0}^{\theta}\mu NR\,d\theta'
\end{equation}

Substituting $W_f$ into  (\ref{eq:LOF}) and differentiating once with respect to $\theta$ leads to (\ref{eq:lead}):

\begin{equation}\label{eq:lead}
0=m g R \sin{\theta}+m v\frac{dv}{d\theta}+\mu N R
\end{equation}

Because friction only affects the tangential acceleration, we can substitute $N$ from (\ref{eq:NF1}) into (\ref{eq:lead}) to obtain (\ref{eq:leadnext}):

\begin{equation}\label{eq:leadnext}
0=m g R\sin{\theta}+mv\frac{dv}{d\theta}+\mu \left(mg\cos{\theta}+\frac{mv^2}{R}\right) R
\end{equation}

If we simplify and group terms together we find (\ref{eq:simp2}):

\begin{equation}\label{eq:simp2}
0=g R(\sin{\theta}+\mu\cos{\theta})+v\frac{dv}{d\theta}+\mu v^2
\end{equation}

We now utilize a helpful substitution, $u=v^2$:

\begin{equation*}
u'=2vv'
\end{equation*}
to obtain (\ref{eq:newform}):
\begin{equation}\label{eq:newform}
0=gR(\sin{\theta}+\mu\cos{\theta})+\frac{u'}{2}+\mu u
\end{equation}

This first-order linear differential equation is not separable, but we can make use of an integrating factor. As a reminder, for the first-order differential equation

\begin{equation*}
y'+p(t)y=g(t)
\end{equation*}

the appropriate integrating factor is:

\begin{equation*}
\mu(t)=e^{\int p(t)\,dt}
\end{equation*}

To avoid confusion, let's use $\delta(\theta)$ as our integrating factor. Our corresponding equation, through rearranging (\ref{eq:newform}) is:

\begin{equation*}
u'+2\mu u=-2gR(\sin{\theta}+\mu\cos{\theta}),
\end{equation*}
and thus:
\begin{equation*}
\delta(\theta)=e^{\int 2\mu d\theta}=e^{2\mu\theta}\, . 
\end{equation*}

Multiply both sides of our equaiton by our integrating factor to obtain:

\begin{equation*}
\frac{d}{d\theta}\left(e^{2\mu\theta}u\right)=-2gRe^{2\mu\theta}(\sin{\theta}+\mu\cos{\theta})\, ,
\end{equation*}
and simplifying yields:
\begin{equation}\label{eq:simp3}
e^{2\mu\theta}u=-2gR\int e^{2\mu\theta}(\sin{\theta}+\mu\cos{\theta})\,d\theta
\end{equation}

Integrating by parts:

\begin{equation}\label{eq:IP}
\int e^{2\mu\theta}\sin{\theta}\,d\theta=\frac{e^{2\mu\theta}}{2\mu}\sin{\theta}-\frac{1}{2\mu}\int e^{2\mu\theta}\cos{\theta}\,d\theta
\end{equation}

\begin{equation}\label{eq:subs}
\int e^{2\mu\theta}\cos{\theta}\,d\theta=\frac{e^{2\mu\theta}}{2\mu}\cos{\theta}+\frac{1}{2\mu}\int e^{2\mu\theta}\sin{\theta}\,d\theta\, .
\end{equation}

Substituting  (\ref{eq:IP}) in  (\ref{eq:subs}) and vice versa leads to:

\begin{equation*}
\left(1+\frac{1}{4\mu^2}\right)\int e^{2\mu\theta}\sin{\theta}\,d\theta=\frac{e^{2\mu\theta}}{2\mu}\left(\sin{\theta}-\frac{\cos{\theta}}{2\mu}\right)
\end{equation*}

\begin{equation*}
\left(1+\frac{1}{4\mu^2}\right)\int e^{2\mu\theta}\cos{\theta}\,d\theta=\frac{e^{2\mu\theta}}{2\mu}\left(\cos{\theta}+\frac{\sin{\theta}}{2\mu}\right)\, 
\end{equation*}

Simplifying results in:

\begin{equation*}
\int e^{2\mu\theta}\sin{\theta}\,d\theta=\frac{e^{2\mu\theta}(2\mu\sin{\theta}-\cos{\theta})}{1+4\mu^2}
\end{equation*}
and
\begin{equation*}
\int e^{2\mu\theta}\cos{\theta}\,d\theta=\frac{e^{2\mu\theta}(2\mu\cos{\theta}+\sin{\theta})}{1+4\mu^2}\, .
\end{equation*}

When we substitute these results into (\ref{eq:simp3})  and simplify we can determing $u$, our substituted variable, $u = v^2$:

\begin{equation*}
e^{2\mu\theta}u=-2gR\left(\int e^{2\mu\theta}\sin{\theta}\,d\theta+\mu\int e^{2\mu\theta}\cos{\theta}\,d\theta\right)\, ,
\end{equation*}

\begin{equation*}
e^{2\mu\theta}u=-2gR\left[\frac{e^{2\mu\theta}(2\mu\sin{\theta}-\cos{\theta})}{1+4\mu^2}+\frac{\mu e^{2\mu\theta}(2\mu\cos{\theta}+\sin{\theta})}{1+4\mu^2}\right]+C\, ,
\end{equation*}

\begin{equation*}
e^{2\mu\theta}u=-2gR\left[\frac{e^{2\mu\theta}(3\mu\sin{\theta}-\cos{\theta}+2\mu^2\cos{\theta})}{1+4\mu^2}\right]+C\, ,
\end{equation*}

\begin{equation*}
u=-2gR\left[\frac{(3\mu\sin{\theta}-\cos{\theta}+2\mu^2\cos{\theta})}{1+4\mu^2}\right]+Ce^{-2\mu\theta}\, ,
\end{equation*}
and
\begin{equation*}
u=Ce^{-2\mu\theta}+\frac{2gR}{1+4\mu^2}[(1-2\mu^2)\cos{\theta}-3\mu\sin{\theta}]\, .
\end{equation*}

Recall we used $u=v^2$:

\begin{equation*}
v^2=Ce^{-2\mu\theta}+\frac{2gR}{1+4\mu^2}[(1-2\mu^2)\cos{\theta}-3\mu\sin{\theta}]
\end{equation*}

We now apply our initial condition: $v=v_0$ at $\theta(0)=0$ to determine the constant of integration $C$:

\begin{equation*}
{v_0}^2=C+\frac{2gR}{1+4\mu^2}(1-2\mu^2)\, 
\end{equation*}

\begin{equation*}
C={v_0}^2-\frac{2gR(1-2\mu^2)}{1+4\mu^2}\,  .
\end{equation*}
Then we can determine $v^2$:
\begin{equation*}
v^2=\left[{v_0}^2-\frac{2gR(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\theta}+\frac{2gR}{1+4\mu^2}[(1-2\mu^2)\cos{\theta}-3\mu\sin{\theta}]\, 
\end{equation*}
and be on our way to determining $\dot{\theta}$ for our equaiton of motion (\ref{eq:rom}) of the cart:
\begin{equation*}
R^2\dot{\theta}^2=\left[R^2{\omega_0}^2-\frac{2gR(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\theta}+\frac{2gR}{1+4\mu^2}\, .[(1-2\mu^2)\cos{\theta}-3\mu\sin{\theta}]
\end{equation*}

Therefore, our equation of motion is:

\begin{equation}\label{eq:rom}
\dot{\theta}=\sqrt{\left[{\omega_0}^2-\frac{2g(1-2\mu^2)}{R(1+4\mu^2)}\right]e^{-2\mu\theta}+\frac{2g[(1-2\mu^2)\cos{\theta}-3\mu\sin{\theta}]}{R(1+4\mu^2)}}
\end{equation}

As a side note, you may also be interested in considering acceleration of the cart over time (Figure~\ref{fig:1-120-ConserveWithFriction}) where $\omega=\dot{\theta}$:

\textbf{Centripetal acceleration}

\begin{equation*}
a_c=\omega^2R
\end{equation*}

\textbf{Tangential acceleration}

\begin{equation*}
a_t=-\frac{mg\sin{\theta}+\mu N}{m}=-g(\sin{\theta}+\mu\cos{\theta})-\frac{\mu v^2}{R}
\end{equation*}

We verify our derivation is correct by substituting $\mu = 0$ in our new equation of motion (\ref{eq:rom}) and see it does reduce to the frictionless case:

$$\dot{\theta}=\sqrt{{\omega_0}^2-\frac{
2g(1-\cos{\theta})}{R}}$$


\item In order to determine the minimal initial velocity to keep the cart on the track in the presence of friction we will use the same approach as before, analyzing the constraint $N \geq 0$. For notational convenience, we will utilize natural units in which $R=1, g=1, m=1$ and define, 

\begin{equation*}
r_0 \equiv \frac{v_0}{\sqrt{gR}}\, .
\end{equation*}

\begin{flushleft}
Adapting from (\ref{eq:NF1})  while using (\ref{eq:rom})  we determine $N$ in (\ref{eq:Nfric})
\end{flushleft}

\begin{equation}\label{eq:Nfric}
N=\cos\theta+v^2=\cos\theta+\left[{r_0}^2-\frac{2(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\theta}+\frac{2}{1+4\mu^2}[(1-2\mu^2)\cos\theta-3\mu\sin\theta]\, .
\end{equation}
As in the case of no friction, we are looking for the critical point which both minimizes the normal force $\frac{dN}{d\theta}=0$ and this is when the cart is barely in contact with the track, i.e.~$N=0$. Then (\ref{eq:Nfric})  yields in this situation (\ref{eq:com1}) and (\ref{eq:com2}) :


\begin{equation}\label{eq:com1}
\cos\theta+\left[{r_0}^2-\frac{2(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\theta}+\frac{2}{1+4\mu^2}[(1-2\mu^2)\cos\theta-3\mu\sin\theta]=0
\end{equation}
and 
\begin{equation}\label{eq:com2}
-\sin\theta-2\mu\left[{r_0}^2-\frac{2(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\theta}-\frac{2}{1+4\mu^2}[(1-2\mu^2)\sin\theta+3\mu\cos\theta]=0\, .
\end{equation}
Substituting $\left[{r_0}^2-\frac{2(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\theta}$ from  (\ref{eq:com1}) in (\ref{eq:com2}) we obtain 


\begin{equation*}
2\mu\cos\theta-\sin\theta+\frac{4\mu}{1+4\mu^2}\left[(1-2\mu^2)\cos\theta-3\mu\sin\theta\right]-\frac{2}{1+4\mu^2}\left[(1-2\mu^2)\sin\theta+3\mu\cos\theta\right]=0\, ,
\end{equation*}
and with   a nice cancellation of terms we obtain:


\begin{equation*}
3(1+4\mu^2)\sin\theta=0\, ,
\end{equation*}
which  yields $\theta=\pi$. Notably, the critical point is not affected by friction! This confirms our intuition that gravity begins to speed up the cart right after passing the top of the track.

Substituting $\theta=\pi$ in (\ref{eq:com1}) gives


\begin{equation*}
-1+\left[{r_0}^2-\frac{2(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\pi}-\frac{2(1-2\mu^2)}{1+4\mu^2}=0\, 
\end{equation*}
and simplfying produces an expression for $r_0$,


\begin{equation*}
\left[{r_0}^2-\frac{2(1-2\mu^2)}{1+4\mu^2}\right]e^{-2\mu\pi}=\frac{3}{1+4\mu^2}\, 
\end{equation*}

\begin{equation*}
{r_0}^2=\frac{3e^{2\mu\pi}+2(1-2\mu^2)}{1+4\mu^2}\, 
\end{equation*}

\begin{equation*}
{r_0}=\sqrt{\frac{3e^{2\mu\pi}+2(1-2\mu^2)}{1+4\mu^2}}\, .
\end{equation*}
Transforming back to standard units we have our initial velocity $v_0$ necessary for the cart to stay on the track under friction,


\begin{equation}\label{eq:fin}
{v_0}=\sqrt{\frac{[3e^{2\mu\pi}+2(1-2\mu^2)]gR}{1+4\mu^2}}\, .
\end{equation}
We can see  our derivation is correct by substituting $\mu=0$ in (\ref{eq:fin})  to see the reduction to the frictionless case:


\begin{equation*}
{v_0}=\sqrt{5gR}
\end{equation*}
\end{enumerate}



\section*{APPENDIX -- REFEREE SUGGESTIONS}

In light of the fact that the derivations in this analysis are quite complex we have received  suggestions from referees and we offer several here.

This is a hard problem.  Getting lost in the notation, and not getting anywhere on the problem, is a concern.
I would start by giving the students an appropriate coordinate system.  I would pick planar polar coordinates because all students are familiar with it, but a tangential-normal coordinate system is equivalent.  So the first task is to relate the more familiar Cartesian expressions for position, velocity, and acceleration to their polar equivalents.   

I would make this part of project, but I would get them started by offering Figure~\ref{fig:1-120-ConserveWithFriction} along with   definitions of variables.

\begin{figure}[!h]
\begin{center}
\includegraphics[scale=0.40]{1-120-Ref1VectorFig.png} \\
\caption{\label{fig:1-120-ConserveWithFriction} Conservation of energy including friction.}
\end{center}
\end{figure}
 
\vspace*{-1 true cm}
 
\begin{eqnarray*}
x(t)&=&r(t)  \cos⁡(\theta(t))\\
y(t)&=&r(t)  \sin⁡(\theta(t))\\
\hat{\bf e}_r&=& \cos⁡(\theta(t)) \hat{i} + \sin⁡(\theta(t))\hat{j} \\
\hat{\bf e}_{\theta}&=&-\sin⁡(\theta(t))\hat{i}+ \cos⁡(\theta(t)) \hat{j}\\
\vec{\bf s}(t)&=& x(t) \hat{i}+y(t) \hat{j}\\
&=& r(t) \hat{\bf e}_r
\end{eqnarray*}




I am using $\vec{\bf s}(t)$ for position, and from here the students should be able to find the polar expressions (using the polar basis vectors) for the velocity and acceleration vectors.  I would also use $\dot{\theta}$ and $\ddot{\theta}$ rather than introducing $\omega$ and α$\alpha$, even though these are common expressions, just to limit the number of symbols.

The next order of business is to find the equation of motion – not the first order differential equation that is derived in the solution included in the Teacher Version.  An equation of motion comes directly from application of a balance law, balance of momentum in this case.  Most students will probably recognize it as Newton's Second Law.  By decomposing the weight vector into its polar components and applying Newton's Second law in the 
$\theta$  direction, we get the equation of motion:
\begin{equation}\label{eq:mom}
m R\ddot{\theta}=-m g \sin⁡(\theta)\, .
\end{equation}
We have ignored friction so that we can get to energy conservation.   We can derive conservation of energy from the equation of motion; it does not need to be assumed.  The components of energy, $E$, present are the kinetic energy and the potential due to height.

\begin{equation}\label{eq:energy}
E=\frac{1}{2}m v^2+m g h\, .
\end{equation}

After rewriting the energy in polar coordinates, the equation of motion can be used to show that
\begin{equation}\label{eq:equal}
\frac{dE}{dt}=0\, .
\end{equation}

When friction is introduced, this same process can be repeated to show that 
\begin{equation}\label{eq:less}
\frac{dE}{dt}<0\, .
\end{equation}
Once that energy conservation is established, the method used for determining the minimum velocity can proceed.   Combining the equation of motion in the radial direction:
\begin{equation}\label{eq:radial}
-m R(\dot{\theta})^2=-N+m g \cos⁡(\theta)\, ,
\end{equation}
and the expression for the conserved energy, with the insight written up in the solution that the normal force will just vanish at the top of the loop is enough information.  The formula for the centripetal acceleration does not need to be introduced (or memorized).


At this point I feel that the project can proceed in two possible directions

\begin{enumerate}
\item Numerical analysis of the equation of motion (the one in the $\theta$  direction) – Euler's method might not keep stable long enough to get to the top of the loop.  This would be a good introduction to recognizing when a numerical method has gone off the rails, and a jumping off point for more sophisticated methods, like Runge-Kutta.
\item Geometric analysis of the equation of motion (again, the one in the $\theta$ direction) – convert to a first order system to find information from the phase portrait.
\end{enumerate}

I think each path would make a full project.  Maybe following a project on the linearized system (from a linear spring, or small amplitude pendulum) would be a great transition to the world of nonlinear dynamics.


Here are two more comments:
I would spend a paragraph (3 or 4 sentences) on the work done by friction.  First, how to calculate it, then stating that all of this work is dissipated energy.  All of the students will have done calculations when energy is conserved, but I think that the energy bookkeeping at this level of detail will be a first for most of the students, so a short exposition is appropriate.

I feel that the analytical solution that is derived in detail in the Teacher Version is too much to expect for a typical sophomore/junior to  produce independently. [Editor's note:  This is a reasonable observation, but we leave it to the individual teacher as to how much guidance from the analytic solution in the COMMENTS of the Teacher Version is to be used.]  I would recommend putting the details in an appendix, its derivation should not be part of the project, but for the projects I suggested it is useful for:
\begin{enumerate}
\item Comparing with a numerical solution
\item Demonstrating how hard it is to extract the same information from the formula that can come from a geometric analysis of the differential equation.
\end{enumerate}

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