Calculating Energy and Speed in a 2-Dimensional Roller Coaster

[HawKMX2004]

Ok...i am confused on some equations..please help me get a start, I am not asking for the answer, just something to go off of please

So your on a 2-Dimensional Roller coaster, and you start at a height of H with an initial velocity of 0 m/s and a total distance of 0 m The track moves up and down in varying places, ending at a distance of 300 m. Find Equations for 2 things. Initial Energy of each individual track segment (each track segment needs to be able to vary) and the Speed with Centripetal Acceleration added for each track segment. I don't know where to start on either of these. Please help? Thankz a bunch

 

[Sirus]

The question is a little unclear. Do you mean the initial energy of the coaster at the beginning of each track segment, and the speed of same with centripetal acceleration added?

I don't see centripetal acceleration as being particularly important here for determining energy or speed. Remember that:
$$E_{\mbox{Total}}=E_{\mbox{kinetic}}+E_{\mbox{potential}}$$
The cart will carry the same total mechanical energy throughout its path assuming an isolated system (no non-conservative forces such as friction present). You can substitute in for terms according to:
$$E_{\mbox{kinetic}}=\frac{1}{2}mv^2$$
$$E_{\mbox{potential}}=mgh$$
So basically the speed and initial energy of the coaster at the beginning of each track segment depends on its height relative to its initial starting position.

 

[HawKMX2004]

So initial energy is the same as total energy?and centripetal acceleration matters because the track moves up and down, creating g-forces.. I am still very confused To clarify more, The roller coaster is like going down and up hills. Except the hills can change so height and g-forces and speed and acceleration all have to change...i really appreciate all the help i can get..im still confused

 

[Sirus]

Think about it in terms of energy. The coaster sits at a height H above the ground initially, at rest. So its total energy is the gravitational potential energy it has, which is a function of H. We are assuming no frictional or applied forces are present here; we just let the cart go. Can you see that total energy must be conserved? Try to reread my earlier post and use those equations. I'll be here to guide you if you need help.

 

[HawKMX2004]

Ok, I am still confused, so let's get a hypethetical question...

1.) The coaster weight is 6400 kg, and is released from rest at a height of 95.8 meters. FInd the Initial energy after the coaster travels 20 meters distance and has a height of 7.3 meters, and find the Velocity at the same distance and height with Centripetal Acceleration Added. Coeffiecient of Friction is .1

So here is how I would work this.

Vf = Vi - (Fg + Ff) + (Vi^2/radius)

Vf = 0 - (ma + FnUk) + (0^2 / radius)
Vf = - (6400a + 6400a(.1) + 0


Thats where i get stuck, and I assume that IE = KE + GPE ?? or is IE = Energy Transfer at that point? I am still very confused...please help

 

[Sirus]

If you review my posts I am speaking of an isolated system. The system you described above is not isolated; a frictional force is present. The frictional force would continually be removing energy from the system during the 20-meter movement.

In order to answer that question, you would need to know exactly what the coaster's path looks like over the 20 meters: the radii of any circular portions of the track, etc. You cannot now how much energy the frictional force removes (I'm assuming the friction constant you gave is for kinetic friction) without radii since it is a function of the normal force and thus the angle of incline/decline of the track according to
$$\vec{F}_{\mbox{kinetic friction}}=\mu_{k}\vec{F}_{N}$$
This question becomes complicated when a frictional force and curved track is included.

 

[HawKMX2004]

well, i can get the angle of the track, and possily the length of the arc, but i do not know how to get the radii, and how does it remove energy? turn it into heat? i could show you the problem on excell if that helps