Centripetal Acceleration and induction roller coaster

[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.

 

[Andrew Mason]

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

 

[Ronnin]

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?

 

[Ronnin]

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

 

[Andrew Mason]

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)

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

 

[Ronnin]

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

 

[OlderDan]

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.

 

[Ronnin]

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

 

[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

 

[OlderDan]

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.

 

[Ronnin]

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.

 

[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

 

[OlderDan]

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.

 

[Ronnin]

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