# Project 7: Part B: Breakneck -- The new roller-coaster at Michigan's Adventure

You are a team of engineers and scientists who are in the process of designing Michigan's Adventures exciting a new attraction “Breakneck”. This ride involves a roller-coaster like car that can hold 8 passengers, with a mass of 700 kg. This car starts by effectively free falling 60m. The track curves at the bottom so that the car can slide up an 8-m-high hill before hitting a horizontal straightaway 50m long. In the middle of the straightaway is a section of track that is used to slow the car down. You can have the car brake over any or all of the 10m length of that section.

Then at the end of the straightaway, a spring-like device hooks under the car. This device changes the car’s direction just in time to prevent it from apparently falling over the end of the track, sending it back over the braking section again, stopping neatly at the end of the braking area. The car should stop at the end of the braking section, on the way back towards the launch pad.

Your tasks are (a) to decide on an appropriate braking force and length of the braking region needed on the straightaway to stop the car at the right location, and (b) the effective spring constant of the turnaround device. The car has special accelerometers mounted which relay that information, and adjust the braking force to provide the acceleration you request. The most important piece of information for these devices is that a person can safely sustain accelerations of 3-4 “g’s” for a brief time, but not more than that.

# Project 7 Solution: Part B: Breakneck -- The new roller-coaster at Michigan's Adventure

Initially, we have a cart and passengers with total mass $M$, sitting a height $H$ above ground level. So, the initial energy \begin{eqnarray*}E_{\rm top}&=& U_{\rm top}+K_{\rm top}\\ &=& U_{\rm top}\\ &=& MgH.\end{eqnarray*} As the coaster falls to the ground, it climbs back up a hill of height $h$. Thus at the level of the track, we must have \begin{eqnarray*}E_{\rm track}&=&U_{\rm track}+K_{\rm track}\\ E_{\rm top}&=& Mgh+K_{\rm track},\end{eqnarray*} where we have used energy conservation ($E_{\rm total}=E_{\rm top}=E_{\rm track}={\rm constant}$). Accordingly, the total kinetic energy (which must be sapped from the coaster to make it stop) is given by $$K_{\rm track}=Mg(H-h).$$ That is, we ned the braking force $\mathbf{F}_{B}$ to do an amount of work over a distance $2B$, where $B$ is the length of the braking region, equal to that of the kinetic energy: \begin{eqnarray*}W&=&\,\big|\mathbf{F}_{B}\cdot\mathbf{B}_{1}\big|+\big|\mathbf{F}_{B}\cdot\mathbf{B}_{2}\big| \\ &=& 2F_{B}B \\ &=& K_{\rm track} \\ &=& Mg(H-h).\end{eqnarray*} Solving this for the braking force in terms of the brake distance, we have $$F_{B}=\frac{Mg(H-h)}{2B}.$$

Now, the most pressing condition here is that the acceleration felt by the passengers cannot exceed $3g$'s. We will take this as our limit: $F_{B}=3Mg$. So, $$3Mg=\frac{Mg(H-h)}{2B}\quad\longrightarrow\quad B=\frac{H-h}{6}\geq 8.66\,{\rm m}.$$ The students could just as well have used the maximum distance and solved for the minimum force.

Finally, in compressing the spring, we know that half of the kinetic energy of the coaster must be transferred to spring potential energy: \begin{eqnarray*}\frac{K}{2}&=&U_{\rm spring}\\ \frac{Mg(H-h)}{2}&=&\frac{1}{2}k x^{2}_{\rm max}.\end{eqnarray*} Also, we can say that the maximum force is experienced by the spring when compressed maximally: \begin{eqnarray*}F_{\rm max}&=&F_{\rm spring}\\ M3g&=&kx_{\rm max}.\end{eqnarray*} Solving these for the spring constant gives the maximum spring constant $$k=\frac{9Mg}{(H-h)}.$$ Any spring constant below this will do the job, but any spring constant above this will kill the riders. That is, for a spring constant above this maximum, the riders will go further than the allowed stretch, resulting in a larger than acceptable force (acceleration).

##### Tutor Questions:

**Question:**When thinking about this problem, what was your “system?”**Expected Answer:****The cart and the Earth**need to be considered for the drop. This way we they can consider the potential energy between the cart and the Earth. Also,**the spring**must be included when they use the spring potential energy.

**Question:**Why do we only need to worry about sapping the kinetic energy? What about the potential energy at the level of the track?**Expected Answer:**Since the entire process occurs on the track, we can basically ignore that potential energy.

**Question:**If energy is always conserved, where did the energy go during braking?**Expected Answer:**It got stolen from the kinetic energy of the cart. It moved into other things, like heating up the molecules in the brake pads.

**Question:**Can you talk me through the entire story of the movement of energy during this process?**Expected Answer:**Initially, all the energy is gravitational potential energy. This transforms into kinetic energy as it falls down the hill. At the bottom of the hill, the kinetic energy is a maximum. On the horizontal track, the energy is “sapped” from the braking region. Energy sloshes from the cart into the spring and back into the cart.**Question:**Why can't we just arbitrarily choose a spring constant and solve for the resulting stretch from $3Mg=kx$?**Expected Answer:**The actual stretch depends on both the spring constant**and**the initial energy coming into the spring. So, we need to make sure that for a given spring constant we won't go past the maximum stretch which would give the maximum force allowed.

**Question:**Is this spring constant that you solved for a maximum or minimum?**Expected Answer:**Maximum. Since we used the limit of $3g$'s, any spring weaker would give less $g$'s.

**Question:**How fast is the cart moving at the level of the track before the braking region? What does the speed depend on?**Expected Answer:**Since $K=\frac{1}{2}Mv^{2}$ and $K_{\rm track}=Mg(H-h)$, we see that $v=\sqrt{2g(H-h)}$. So, the speed depends only on the strength of gravitational acceleration and the difference in height from top to track.

**Question:**Could you analyze the “falling part” of the problem using work instead of energy conservation?**Expected Answer:**If instead of using energy conservation we use work, we need to choose our system to be just the cart. The net work done by the surroundings in that case, form the work-kinetic-energy theory, tells us that $W_{\rm surr}=\Delta K$. Now, we also know that $dW_{\rm surr}=\vec{F}_{\rm net}\cdot d\vec{r}$. Accordingly, we must have that \begin{align*}W_{\rm surr}&=\int\,dW_{\rm surr}\\&=\int\,\vec{F}_{\rm net}\cdot d\vec{r}\\&=\int(\vec{F}_{g}+\vec{F}_{\rm N})\cdot d\vec{r}\\&=\int_{\rm start}^{\rm finish}\vec{F}_{g}\cdot d\vec{r}\\&=\int_{\rm top}^{\rm bottom}\vec{F}_{g}\cdot d\vec{y}+\int_{\rm bottom}^{\rm track}\vec{F}_{g}\cdot d\vec{y}\\&=F_{g} \Delta y_{\rm bottom,top}+F_{g}\Delta y_{\rm track,bottom}\\&=F_{g}H-F_{g}h\\&=mg(H-h)\end{align*} where we have used the fact that $\vec{F}_{\rm N}\cdot d\vec{r}=0$ (the normal force is always perpendicular to the motion) and that the gravitational force only points in the “$y$”-direction. So, if the cart starts from rest, we must have $W_{\rm surr}=\Delta K=K_{f}$ and so $K_{\rm track}=Mg(H-h)$ as expected.

**Question:**Could you analyze the “spring part” of the problem using work instead of energy conservation?**Expected Answer:**If instead of using energy conservation we use work, we need to choose our system to be just the cart. The net work done b the surroundings in that case, from the work-kinetic-energy theory, tells us that $W_{\rm surr}=\Delta K$. Now, we also know that $dW_{\rm surr}=\vec{F}_{\rm net}\cdot d\vec{r}$ and that the spring force actually depends depends on the position: $F_{\rm spring}=-kx$. Accordingly, we must have that \begin{align*}W_{\rm surr}&=\int\,dW_{\rm surr}\\&=\int\vec{F}_{\rm net}\cdot d\vec{r}\\&=\int(\vec{F}_{\rm spring}+\vec{F}_{g}+\vec{F}_{\rm N})\cdot d\vec{r}\\&=\int_{\rm start}^{\rm finish} \vec{F}_{\rm spring}\cdot d\vec{r}\\&=\int_{0}^{x_{\rm max}}-k\vec{x}\cdot d\vec{x}+\int_{x_{\rm max}}^{0}-k\vec{x}\cdot d\vec{x}\\&=\int_{0}^{x_{\rm max}}-k\vec{x}\cdot d\vec{x}-\int_{0}^{x_{\rm max}}-k\vec{x}\cdot d\vec{x}\\&=0,\end{align*} where we have used the fact that $\vec{F}_{g}\cdot d\vec{r}=0$ and $\vec{F}_{\rm N}\cdot d\vec{r}=0$ ($\vec{F}_{g}$ and $\vec{F}_{\rm N}$ point perpendicularly to the displacement) and the cart begins and ends at the same location.

##### Common Difficulties:

- Students have a tendency to draw on previous concepts (kinematic equations) to solve this problem. Push them to make an “energy paradigm shift” and use energy conservation ideas.

##### Main Points:

- Recognize that energy conservation is a useful approach to solving this problem.
- Be able to decompose the total energy into kinetic and potential terms for various stages of the problem.