Question

You want to design a rollercoaster that will be able to go around a loop-the-loop on earth (where the value of g is 10 N/kg). If the mass of the rollercoaster is 6,000 kg and it is going 40 m/s at the bottom of the loop, calculate the maximum height of the top of the loop. Assume there is no energy input (motors, etc) or friction or any other energy waste in this system.

Answer 1

Answer:

$64\; {\rm m}$, assuming that the rollercoaster is not attached to the track.

Explanation:

Let $r$ denote the radius of the loop. The height of the top of the loop would be $2\, r$.

Let $a$ denote the acceleration of the rollercoaster. At any position in the loop, if the speed of the rollercoaster is $v$, the (centripetal) acceleration of the rollercoaster would be:

$\displaystyle a = \frac{v^{2}}{r}$.

Let $m$ denote the mass of the rollercoaster. The net force on the rollercoaster would then be:

$\displaystyle F_\text{net} = m\, a = \frac{m\, v^{2}}{r}$.

The rollercoaster would stay on the track (and goes around the loop without falling off) only if the normal force $F_\text{normal}$ between the track and the rollercoaster is non-negative. In other words: it is necessary that $F_\text{normal} > 0$ for the rollercoaster to stay on the track.

At the top of the loop, $F_\text{normal}$ and the weight of the rollercoaster $m\, g$ are in the same direction as the centripetal acceleration (downwards towards the center of the loop.) Hence:

$F_\text{net} = F_\text{normal} + m\, g$.

Let $v_\text{top}$ denote the speed of the rollercoaster at the top of the loop.

$\begin{aligned} F_\text{normal} &= F_\text{net} - m\, g \\ &= \frac{m\, v_\text{top}^{2}}{r} - m\, g\end{aligned}$.

If $F_\text{normal} > 0$ at the top of the loop, then:

$\displaystyle \frac{m\, v_\text{top}^{2}}{r} - m\, g > 0$.

$\displaystyle v_{\text{top}}^{2} > g\, r$.

At the same time, by the conservation of energy, the sum of the kinetic energy $\text{KE}$ and gravitational potential energy $\text{GPE}$ of the rollercoaster should stay the same during the entire ride. Assuming that $\text{GPE}\!$ is $0$ at the bottom of the loop:

$\begin{aligned}& \text{KE at the bottom}\\ &= \text{KE at the top} + \text{GPE at the top}\end{aligned}$.

Let $v_{0}$ denote the speed of the rollercoaster at the bottom of the loop.

$\begin{aligned}\text{KE at the bottom} = \frac{1}{2}\, m\, {v_{0}}^{2}\end{aligned}$.

The speed of the rollercoaster at the top of the loop $v_\text{top}$ is at least $(g/r)$. Therefore:

$\begin{aligned}& \text{KE at the top}= \frac{1}{2}\, m\, {v_{\text{top}}}^{2} > \frac{1}{2} \, m\, g\, r\end{aligned}$.

Since the height of the loop is $2\, r$, the $\text{GPE}$ of the rollercoaster at the top of the loop would be:

$\text{GPE at the top} = m\, g\, h = 2\, m\, g\, r$.

Thus:

$\begin{aligned} \frac{1}{2}\, m\, {v_{0}}^{2} &= \text{KE at the bottom}\\ &= \text{KE at the top} + \text{GPE at the top} \\ & > \frac{1}{2}\, m\, g\, r + \text{GPE at the top}\\ & = \frac{1}{2}\, m\, g\, r+ 2\, m\, g\, r \\ &= \frac{5}{2}\, m\, g\, r\end{aligned}$.

In other words:

$\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} > \frac{5}{2}\, m\, g\, r$.

Rearrange and simplify to obtain a bound on $r$:

$\begin{aligned} r < \frac{(1/2)\, m\, {v_{0}}^{2}}{(5/2)\, m\, g}\end{aligned}$.

$\begin{aligned} r < \frac{{v_{0}}^{2}}{5\, g}\end{aligned}$.

Given that $v_{0} = 40\; {\rm m\cdot s^{-1}$ and $g = 10\; {\rm N \cdot kg^{-1}} = 10\; {\rm m\cdot s^{-2}$:

$\begin{aligned} r & < \frac{{v_{0}}^{2}}{5\, g} \\ &= \frac{(40\; {\rm m\cdot s^{-1}})^{2}}{5 \times (10\; {\rm m\cdot s^{-2}})} \\ &= 32\; {\rm m}\end{aligned}$.

Hence, the radius of this loop is at most $32\; {\rm m}$, such that the height of the top of this loop is at most $2 \times 32\; {\rm m} = 64\; {\rm m}$.