Roller Coaster Example

A roller coaster ride at an amusement park starts from rest at a height of \(\text{50}\) \(\text{m}\) above the ground and rapidly drops down along its track. At some point, the track does a full 360° loop which has a height of \(\text{20}\) \(\text{m}\), before finishing off at ground level. The roller coaster train itself with a full load of people on it has a mass of \(\text{850}\) \(\text{kg}\).

Roller Coaster

If the roller coaster and its track are frictionless, calculate:

1. the velocity of the roller coaster when it reaches the top of the loop

2. the velocity of the roller coaster at the bottom of the loop (i.e. ground level)

Step 1: Analyse the question to determine what information is provided

• The mass of the roller coaster is \(m = \text{850}\text{ kg}\)

• The initial height of the roller coaster at its starting position is \({h}_{1} = \text{50}\text{ m}\)

• The roller coaster starts from rest, so its initial velocity \({v}_{1} = \text{0}\text{ m·s$^{-1}$}\)

• The height of the loop is \({h}_{2} = \text{20}\text{ m}\)

• The height at the bottom of the loop is at ground level, \({h}_{3} = \text{0}\text{ m}\)

We do not need to convert units as they are in the correct form already.

Step 2: Analyse the question to determine what is being asked

• the velocity of the roller coaster at the top of the loop

• the velocity of the roller coaster at the bottom of the loop

Step 3: Calculate the velocity at the top of the loop

From the conservation of mechanical energy, We know that at any two points in the system, the total mechanical energy must be the same. Let's compare the situation at the start of the roller coaster to the situation at the top of the loop:

\begin{align*} {E}_{M1} & = {E}_{M2} \\ {E}_{K1} + {E}_{P1} & = {E}_{K2} + {E}_{P2} \\ 0 + mg{h}_{1} & = \frac{1}{2}m{\left({v}_{2}\right)}^{2} + mg{h}_{2} \end{align*}

We can eliminate the mass, \(m\), from the equation by dividing both sides by \(m\).

\begin{align*} g{h}_{1} & = \frac{1}{2}{\left({v}_{2}\right)}^{2} + g{h}_{2} \\ {\left({v}_{2}\right)}^{2} & = 2\left(g{h}_{1} - g{h}_{2}\right) \\ {\left({v}_{2}\right)}^{2} & = 2\left(\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{50}\text{ m}\right)-\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{20}\text{ m}\right)\right) \\ {v}_{2} & = \text{24.25}\text{ m·s$^{-1}$} \end{align*}

Step 4: Calculate the velocity at the bottom of the loop

Again we can use the conservation of energy and the total mechanical energy at the bottom of the loop should be the same as the total mechanical energy of the system at any other position. Let's compare the situations at the start of the roller coaster's trip and the bottom of the loop:

\begin{align*} {E}_{M1} & = {E}_{M3} \\ {E}_{K1} + {E}_{P1} & = {E}_{K3} + {E}_{P3} \\ \frac{1}{2}{m}_{1}{\left(0\right)}^{2} + mg{h}_{1} & = \frac{1}{2}m{\left({v}_{3}\right)}^{2} + mg\left(0\right) \\ mg{h}_{1} & = \frac{1}{2}m{\left({v}_{3}\right)}^{2} \\ {\left({v}_{3}\right)}^{2} & = 2g{h}_{1} \\ {\left({v}_{3}\right)}^{2} & = 2\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{50}\text{ m}\right) \\ {v}_{3} & = \text{31.30}\text{ m·s$^{-1}$} \end{align*}