# Would an empty roller coaster and a full roller coaster take the same amount of time for a single trip? Explain.

May 27, 2016

#### Explanation:

A typical roller coaster works on the interchange of Gravitational potential energy with the kinetic energy of the roller coaster and vice-versa, during the course of ride. No motors are used to run the coaster during a single trip.

Some of the stored GPE is lost due to friction. The reason why it's impossible for a roller coaster to return to its original height once the ride is over.

Thereafter, roller coaster uses a motorized lift to return to its initial position at the top of the hill, for the next ride. See the figure below Gravitational Potential Energy, $P E = m g h$
Kinetic Energy $K E = \frac{1}{2} m {v}^{2}$
We use law of conservation of energy.
At the initial position, $P E$ is maximum and $K E$ is zero.
At the bottom points where $h = 0$, $P E = 0$; and $K E$ is maximum.
At this point, assuming there is no friction
$\frac{1}{2} m {v}^{2} = m g h$ ......(1)
$\implies v = \sqrt{2 g h}$

We see that velocity of the roller coaster is independent of its mass and is solely dependent on local $g$ and initial $h$.
Therefore, for an ideal roller coaster an empty roller coaster or a full roller coaster will take the same amount of time for a single trip.

Let's consider real life situation. If ${m}_{e}$ is the mass of empty roller coaster and ${m}_{f}$ is mass of full one, then equation (1) becomes
$\frac{1}{2} {m}_{e} {v}^{2} = {m}_{e} g h + {\text{Loss due to Friction}}_{e}$ .....(2)
$\frac{1}{2} {m}_{f} {v}^{2} = {m}_{f} g h + {\text{Loss due to Friction}}_{f}$ .......(3)
Now we know that $\text{Loss due to Friction}$ has two components:
1. Loss due to coefficient of dynamic friction, which is dependent on normal reaction. This term in both (2) and (3) will be dependent on the mass of roller coaster in each case. (Loss is a $- v e$ quantity).

Therefore, will be inconsequential for determining the velocity of the roller coaster as it appears in both equations; as we determined for the no friction case.
2. Loss due to drag. We know that this term is directly proportional to ${v}^{2}$ and to the area of cross section $A$, all other factors being same. Since full roller coaster will have larger area of cross-section exposed to drag, it follows that

$\text{Loss due to Drag"_fprop} {v}^{2} A$
We need to appreciate that for roller coaster this need to be divided by ${m}_{e} \mathmr{and} {m}_{f}$ in either case.
$\therefore \text{Loss due to Drag"_fprop} {v}^{2} \frac{A}{m}$.

To simplify@ we assume that effect on loss due to drag on account of increase in area of cross section is less pronounced than the combined effect due to change in velocity and increase in the mass of the roller coaster,
$\implies {\text{Loss due to Drag"_f<"Loss due to Drag}}_{e}$
$\implies v \text{ of full roller coaster is" > v " of empty roller coaster}$

Therefore, for a real life an empty roller coaster will take more amount of time as compared to a full roller coaster for a single trip.
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@This is an ad-hoc assumption, and it is really difficult to draw any fruitful inference unless we have more information about weight of roller coaster, number of riders in a full ride, change in velocity.

It may be remembered that the trip time is not at all important for a roller coaster ride.
What is important is:
Riders experience.
The main type of acceleration on a roller coaster is centripetal acceleration due to construction of the ride. This type of acceleration can produce strong $g$-forces, which can either push the rider into seat or make him feel like as if he was going to fly out of it.