# Centripetal Acceleration and induction roller coaster

• Ronnin
In summary, the net acceleration is the vector sum of the centripetal and linear accelerations. After P, is there any centripetal acceleration?Yes, there is a centripetal acceleration at point P.
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.

Ronnin said:
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.
Use the fact that the distance traveled 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

Andrew Mason said:
Use the fact that the distance traveled 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

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

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

Ronnin said:
How did you get from $$2ad/r to 2a/theta$$?
$\theta$ = d/r where d is the distance traveled (it prescribes a circular path immediately from the start)

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

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

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

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.

OlderDan said:
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.

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

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

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

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:
OlderDan said:
That does not look right. If the tangential acceleration is taken to be constant at g, and you ignore gravity, 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.

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

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

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

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.

OlderDan said:
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.

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 a lot of time. Thanks for everyones help

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular motion. It is always directed towards the center of the circle and its magnitude is equal to the square of the object's speed divided by the radius of the circle.

## 2. How is centripetal acceleration related to roller coasters?

In roller coasters, centripetal acceleration is responsible for keeping the riders on the track as the car moves along a curved path. Without this acceleration, the riders would continue moving in a straight line and fly off the track.

## 3. What role does gravity play in centripetal acceleration on a roller coaster?

Gravity is the force that creates the centripetal acceleration in a roller coaster. As the car moves along the track, the force of gravity pulls it towards the center of the circle, creating the necessary acceleration to keep the riders on the track.

## 4. How is induction used in roller coasters?

Induction is used in roller coasters to power the cars and keep them moving along the track. Electromagnetic induction is used to create a changing magnetic field, which then induces an electric current in the car. This current powers the motors that propel the car along the track.

## 5. What factors affect the centripetal acceleration experienced on a roller coaster?

The centripetal acceleration on a roller coaster is affected by the speed of the car, the radius of the track, and the mass of the car and its riders. A higher speed or smaller radius will result in a greater centripetal acceleration, while a heavier car will require more centripetal force to keep it on the track.

• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
28
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
6K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
4K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
8K
• Introductory Physics Homework Help
Replies
1
Views
2K