What is the minimum turn radius for a roller coaster with given conditions?

[freshcoast]

Homework Statement

A roller coaster car (m = 750kg with passengers) starts with near zero velocity at the top of a 80m section of track that is inclined at -38 degrees from the horizontal. During the descent the average drag force acting on the car is 550 N. At the bottom of the descent is a horizontal turn to the right. In the turn the track is banked at a 33 degree angle with respect to the horizontal.

a) If the occupants are to experience a maximum centripetal acceleration of 1.5 times Earth gravity at the start of the turn, what is the minimum turn radius allowable? Assume the only forces acting on the car are its weight, drag and normal forces.

b) Assuming the same drag forces act on the car during the turn through what angle does the car travel before the centripetal acceleration decreases to 1g?

Homework Equations

PEo + KEo = PEf + KEf
ƩF = mV^2/r
Eo + Ef = W(non conservative)

PE = mgh
KE = 1/2 mv^2

The Attempt at a Solution

So what I did is, during the descent I used the equation Eo + Ef = W(non conservative) where I set initial kinetic energy is 0, initial potential energy would be mgh where h = Length(sinθ) subtracted by the non conservative work which is (drag force x length of track) and set that equal to the final kinetic energy. I then solved for Vf which I would be using for the ascent. Now during the ascent of part a, if the occupants are to experience a maximum centripetal acceleration of 1.5 times the gravity I assumed that it would mean that the normal force would equal to 1.5g, so with that said i drew a free body diagram of the ascent and applied the Forces( in the radial direction) where Fn - Fg = mv^2/r, which I then solved for R, using the V final I obtained from earlier.

ok now for part b..i'm a little stuck. I assume I just draw the same Free body diagram but this time add the drag force and use the same radius I found in part a and then just solve for the new θ?

thanks for all the suggestions and input!

 

[haruspex]

I then solved for Vf which I would be using for the ascent.

There is no ascent. The turn is in the horizontal plane. This alters your forces equation somewhat.

 

[freshcoast]

well the turn is banked right? so my free body diagram would remain the same wouldn't it?

 

[haruspex]

The diagram's ok, but ...

if the occupants are to experience a maximum centripetal acceleration of 1.5 times the gravity I assumed that it would mean that the normal force would equal to 1.5g

No, it says centripetal, not normal or total. What is the centripetal acceleration here?

i drew a free body diagram of the ascent and applied the Forces( in the radial direction) where Fn - Fg = mv^2/r

That would be wrong (where's the bank angle?), it's different from what you have in the image, but that's wrong too. And you don't even need to worry about the normal force to answer the question.

 

[freshcoast]

so the centripetal acceleration would just be 1.5g? but I don't understand how I would be able to find the radius without using the force equation? it says during the turn the bank is angled at 33 degrees to the horizontal, does that mean I just multiply the centripetal acceleration with the angle?

 

[haruspex]

so the centripetal acceleration would just be 1.5g?

That's how I read the question. In fact the question is somewhat strange. It says to ignore all forces except weight, drag and normal. That implies no lateral friction, so the bank angle is something you could calculate (and it would have to diminish as the curve progresses).

but I don't understand how I would be able to find the radius without using the force equation?

The centripetal acceleration, as you've already quoted, is v²/r. What's the difficulty?

 

[freshcoast]

Well the bank angle is given, and I need to solve for the radius which I am not able to do unless I use the force equation that I noted.

 

[haruspex]

You have v²/r = 1.5g, and you've already computed v.

 

[freshcoast]

So for part b, do I just apply the forces in the lateral direction and use the centripetal acceleration equation to find the angle??

 

[haruspex]

use the centripetal acceleration equation to find the angle??

How do you propose to do that?
What does the speed need to get down to for the centripetal acceleration to be 1g?

 

[freshcoast]

Oh i see, do I set mv^2/ r equal to 1g, solve for V, then draw the free body diagram and use the force of gravity to find the angle?

 

[haruspex]

Oh i see, do I set mv^2/ r equal to 1g, solve for V, then draw the free body diagram and use the force of gravity to find the angle?

Solve for v, yes. But I don't see how a free body diagram is going to help you calculate the turn angle. What is it that is making the speed drop?

 

[freshcoast]

the drag force and the x component of gravity? I was thinking since the x component of gravity has an angle to it I can do some algebra isolate it and to solve for the angle?

 

[haruspex]

the drag force and the x component of gravity? I was thinking since the x component of gravity has an angle to it I can do some algebra isolate it and to solve for the angle?

Through the turn, it's moving in a horizontal plane. How is gravity slowing it down?

 

[freshcoast]

Well when the turn is made the bank is at an angle so the x component of the gravity is with the same direction as the drag force which is in the negative x direction so I assumed it contributed to the drag to help slow the roller coaster down? Am I just to account for the drag force only for the force equation? But I don't see how else am I going to get an angle from that?

 

[haruspex]

Well when the turn is made the bank is at an angle so the x component of the gravity is with the same direction as the drag force which is in the negative x direction so I assumed it contributed to the drag to help slow the roller coaster down? Am I just to account for the drag force only for the force equation? But I don't see how else am I going to get an angle from that?

Not sure what you're taking as the x direction.
The gravitational force acts straight down; the drag force acts tangentially to the curve, so is horizontal; the centripetal acceleration is radial to the curve. These three vectors are mutually perpendicular. The speed can only be altered by a component of acceleration in the direction of movement. The gravitational force and the normal force are both perpendicular to the direction of movement, so the speed here is only affected by the drag.
The angle you are looking for is nothing to do with the bank angle. If you look down onto the path from above, it's following a circular arc. You want the angle that arc subtends at the centre of curvature. In other words, you're just looking for how far along the curve it goes to the point where it has slowed down so much that centripetal acceleration is only 1g. So calculate in this order:
- the speed it will have slowed to
- the KE it has therefore lost along the curve
- knowing the drag, the distance it has gone along the curve to lose that energy
- knowing the radius, the angle of arc it has gone along the curve to travel that distance

 

[freshcoast]

Still a little shakey on what you mean by the angle of arc the object has gone along the curve.. but this is what I've got so far, for the speed it slowed down to I apply the mv^2/r set that equal to 1g and solve for V to find the speed it is traveling when it's max centripetal acceleration is at 1g. For the kinetic energy lost, do I just plug in the new V in the formula 1/2mv^2 to find the kinetic energy at that point? for the distance, do I apply Newtons kinematic equation (Vf^2 = Vo^2 + 2ax) where Vo is the velocity computed in part A which is the objects speed once the object reached the bottom of the hill, and the Vf is the speed the object is traveling once making the turn, and the the a is the drag force over the mass?

sorry if I am still having a hard time grasping this but your reply's and input is very much appreciated thank you

 

[haruspex]

Still a little shakey on what you mean by the angle of arc the object has gone along the curve..

The OP says

during the turn, through what angle does the car travel

If you drive around an arc of a circle, you start off pointing in one direction and finish pointing in a different direction. The angle between those directions is the angle you have turned through. If you join the endpoints of your path to the centre of the circle, the angle the radii make at the centre of the circle is the same angle. If you express the angle in radians, the distance traveled along the arc equals the angle multiplied by the radius.

but this is what I've got so far, for the speed it slowed down to I apply the mv^2/r set that equal to 1g and solve for V to find the speed it is traveling when it's max centripetal acceleration is at 1g. For the kinetic energy lost, do I just plug in the new V in the formula 1/2mv^2 to find the kinetic energy at that point? for the distance, do I apply Newtons kinematic equation (Vf^2 = Vo^2 + 2ax) where Vo is the velocity computed in part A which is the objects speed once the object reached the bottom of the hill, and the Vf is the speed the object is traveling once making the turn, and the the a is the drag force over the mass?

That's all correct.

 

[freshcoast]

Oh I see, I think I'm starting to understand now, but I still have a couple questions so is the bank angle given not used at all for part b? And also the kinetic energy u told me to find what was the purpose of that? And for the angle I just divide the distance traveled by the radius and then convert to degrees if it is expressed in radians? I drew a picture just to see if I understood you correctly

 

[haruspex]

is the bank angle given not used at all for part b?

It appears to be irrelevant.

And also the kinetic energy u told me to find what was the purpose of that?

Decrease in KE tells you the work energy lost to drag. Work = force x distance, so knowing the drag force you can calculate the distance.

And for the angle I just divide the distance traveled by the radius and then convert to degrees if it is expressed in radians?

Yes, but the distance is the length around the arc, not the straight line you've drawn connecting the endpoints.

 

[freshcoast]

Oh, instead of using Newtons kinematic equation I can just set the change in kinetic energy equal to the force times the distance to solve for the distance? (KEf - KEo) = fd ? And then just divide that distance by the radius to find the angle?

 

[haruspex]

Yes.

 

[freshcoast]

Just to make sure I did can you look over my work? I'm wondering if the drag force is negative since it is pointing the opposite direction of the motion but if it is it gives me a negative number under the radical which makes it unsolvable.

 

[haruspex]

Vf2 < vf1, so you need the drag negative to give you a positive distance.

 

[freshcoast]

Is the drag for vf1 negative? Because if it is it will be a negative under the square root..

 

[haruspex]

In going from your OP to the working you posted most recently you've dropped the PE term. The KE at the bottom of the slope is what's left of the PE after taking out drag, so yes, the drag acts negatively there, but the value to be square rooted is positive because of the KE contribution.

 

[freshcoast]

Okay I see where the mistake was, I didn't include the initial PE in the system. Ok just to make sure I'm clear on finding the angle.. once I've found the distance I divide that by the radius which gives me an answer in radians? And then I covert the to an angle by multiplying that answer by 180/(pie)?

 

[haruspex]

Yes.