Gyros and Angular Motion

Calculations

Before setting up the equation, it seems like a good idea to see if it is even accurate to say that the drag coefficient is relatively unchanged over the range of velocities that the roller coaster will travel over, going from 1 m/s to the hypothetical maximum given by

v=\sqrt{gh}

. We'll use the roller coaster in the example on the first page. The reason the Reynolds number is being taken from 1 m/s instead of 0 m/s is that taking it from 0 m/s significantly expands the range, throwing off predictions for what the value will be for most of the roller coaster's motion, which will be clear in a moment.

v=\sqrt{2*9.81\frac{m}s*45m}=29.7\frac{m}s

Plugging this value into the Reynolds number equation gives

R_e=\frac{1.225 \frac{kg}{m^3}*29.7 \frac{m}s*\frac{4*(1.5m)^2}{3m}}{.00001827 Pa*s}=7.99E6

While a value of 1 m/s gives a prediction of

R_e=\frac{1.225 \frac{kg}{m^3}*1.0 \frac{m}s*\frac{4*(1.5m)^2}{3m}}{.00001827 Pa*s}=2.01E5

While these are close in magnitude, because the cross sectional area is rather large, the roller coaster moves between flow types, assuming that the drag behaves similarly to that of a sphere. This means that the drag is going to fluctuate significantly towards the end of the hill.

Now comes the actual calculation. As we all know, work is given by the equation

W=\integral{F*ds}

Normally we can simply multiply the force by the distance moved, but in this case the force actually changes, so an integral is required. Also, the distance is instead the length of the roller coaster track, which is not given in the example. Continuing on the list of challenges is that the force is itself dependent on the velocity, which is going to differ with the force applied.

Now now velocity can be taken with respect to distance covered if the slope of the track is known. This still becomes an exercise in futility, since the equation for force requires velocity to be known, and velocity can't be calculated without knowing the forces acting on the body!

F_D=\frac{1}2*ρ*v^2*C_D*A

dv^2=\sum\frac{F}mds

This creates a problem, and reaches the limits of my mathematical abilities. Even given an equation that graphs the path of the car, or the change in angle, it is not possible.

Thankfully, there are people who are smarter than myself and have done the derivations to figure out how to get velocity of an object accelerating through a fluid. To get this, we need three equations. The first calculates the terminal velocity of a body and looks rather familiar.

v_\infinity=\sqrt{\frac{2mg}{\rhoAC_D}}

The second gives the amount of time that it takes for an object to cover a given distance through a fluid, assuming a start from rest.

t=\frac{v_\infinity}{g}*\arccosh{e^{\frac{\Deltay}{v_\infinity^2}g}}

Finally, there is the equation for final velocity itself.

v=v_\infinity*\tanh\frac{gt}{v_\infinity}

As a note, those are hyperbolic functions, the cousins of sine, cosine and tangent. Also, we've been forced back into the simplification of simple vertical movement, rather than over a distance of track, but we can at least get an answer for the problem that we set out to solve.

v_\infinity=\sqrt{\frac{2*800kg*9.81\frac{m}{s^2}}{1.225\frac{kg}{m^3}1.5^2*\pi1.8}}=31.7\frac{m}{s}

t=\frac{31.7\frac{m}{s}}{9.81\frac{m}{s^2}}*\arccosh{e^{\frac{45m}{(31.7\frac{m}{s})^2}9.81\frac{m}{s^2}}}=3.25s

v=-31.7\frac{m}{s}*\tanh\frac{9.81\frac{m}{s^2}*3.25s}{31.7\frac{m}{s}}=23.8\frac{m}{s}

And so, we now have a more accurate answer to the roller coaster's final velocity. Rather than the normally assumed 29.7 meters per second, the ride will move closer to 23.8 meters per second at the bottom.

dv^2=\sum\frac{F}mds

The Physics of Roller Coasters

Calculations