# Centripetal Acceleration and induction roller coaster

1. Nov 25, 2006

### Ronnin

We are to design the track for an induction roller coaster. To create an inital thrill, we want each passenger to leave the loading point with acceleration g along the horzontal track. To increase the thrill, we also want that first section of track to form a circular arc, so that the passenger also experiences a centripetal accleration. When the magnitutde a of the net acceleartion reaches 4g at some point P and angle theta P along the arc, we want the passenger then to move in a straight line, along a tangent to the arc.

a)What anle theta P should the arc subtend so that a is 4g at point P
b)What is the magnitude a of the passengers net accleration at the point P and after point P.

I'm really stuck on this one. I have tried to relate this to a pendulum but without an R I don't know how to begin.

2. Nov 25, 2006

### Andrew Mason

Use the fact that the distance travelled is: $d = .5 at^2$ and the speed at time t is v = at

$$a_c = 4g = v^2/r = (at)^2/r = 2ad/r = 2a\theta$$

The net acceleration is the vector sum of the centripetal and linear accelerations. After P, is there any centripetal acceleration?

AM

3. Nov 27, 2006

### Ronnin

How did you get from $$2ad/r to 2a/theta$$?

4. Nov 29, 2006

### Ronnin

Wouldn't this be 4g=(v^2)/r+alpha(r). Isn't there a tangental and radial acceleration at work here?

5. Nov 29, 2006

### Andrew Mason

$\theta$ = d/r where d is the distance travelled (it prescribes a circular path immediately from the start)

Quite right. One has to add the two accelerations as vectors. Since they are mutually perpendicular, the equation should be:

$$a_c^2 + g^2 = (4g)^2$$

$$a_c = \sqrt{15}g = 2a\theta = 2g\theta$$

so

$$\theta = \sqrt{15}/2$$ radians

AM

6. Nov 30, 2006

### Ronnin

Can this be solved algebraicly using alpha? I can't seem to solve to resolve the radii in the equation? Just curious.

7. Nov 30, 2006

### OlderDan

I am finding the problem statement to be ambiguous. Are you assuming a constant tangential acceleration of g or a constant tangential force? I assume the arc starts immediately, so that the car is basically at the bottom of the arc initially.

8. Nov 30, 2006

### Ronnin

Isn't the magnitude of tan acceleration constant? Yes the car starts at the beginning of the arc. I'm gonna post my algebra, I got Theta=3 rads

Last edited: Nov 30, 2006
9. Nov 30, 2006

### Ronnin

$$4a_{net}=a_{tan}=a_{rad}$$
$$4a_{net}=(\alpha)r+v^2/r$$
$$4a_{net}=(\omega)r/t+(\omega^2rt)/t$$
$$4a_{net}=v/t+v/t\theta$$
$$4a_{net}=a+a\theta$$
$$3=\theta$$

Last edited: Nov 30, 2006
10. Nov 30, 2006

### OlderDan

That does not look right. If the tangential acceleration is taken to be constant at g, then Andrew's revised calculation should be good. He is saying there is a constant angular acceleration by sayng there is a constant tangential acceleration. You can express that in terms of alpha if you choose. Either way should be fine.

I do find the result a bit disturbing, since it is more than pi/2 and the car will be upside down at point P.

I don't see where you get your first equation (maybe the second = is a +??), and the accelerations must be added as vectors.

Last edited: Nov 30, 2006
11. Nov 30, 2006

### Ronnin

It is supposed to be a plus, still getting the latex down. Guess I can't add them algebraicly since they are vectors.

12. Nov 30, 2006

### Ronnin

So is my thinking completely wrong by saying the net force is the same "a" after I take all the omega's into account

13. Nov 30, 2006

### OlderDan

I don't understand what you are asking. You can express the tangential acceleration as α = a_t/r = g/r and the centripetal acceleration as a_c = ω²r. It does not matter whether you use α & ω or a & v.

14. Nov 30, 2006

### Ronnin

I'll use Andrew's example, I just thought it could be worked out that way. I tried using my algebra and putting into the form (4a)^2=a^2+(atheta)^2, and got root(15) so I guess I just wasted alot of time. Thanks for everyones help