Maxim velocity at the top of the loop (rollercoaster pb)

  • Thread starter Thread starter shuby
  • Start date Start date
  • Tags Tags
    Loop Velocity
AI Thread Summary
The discussion revolves around calculating the maximum velocity a roller coaster should have at point A to ensure passengers remain in contact with their seats, with a specific radius of 7 meters. Participants clarify that friction is not a relevant factor in this scenario, focusing instead on the forces acting on the passenger, namely the normal and gravitational forces. The question is debated, with some suggesting it should be framed as a minimum velocity problem, as the maximum velocity could lead to structural concerns beyond the scope of the question. The conversation highlights confusion over the problem's wording and its implications for safety. Overall, the thread emphasizes understanding the forces involved in roller coaster dynamics.
shuby
Messages
11
Reaction score
0

Homework Statement


What's the MAXIMUM velocity the roller coaster should have at pt A in order for the passengers to stay in contact with their seats ?
calculate the velocity for r=7

It kinda looks like this
[img=http://img94.imageshack.us/img94/6994/image39.gif]


Homework Equations



The Attempt at a Solution


The friction is not mentioned anywhere, so let's forget about it, i thought that it has something to do with the normal force, so it should be like this:
Nmax + mg = mv^2/r

does it need some advanced stuff about the normal force to solve this?
does it have something to do with friction between the passenger and the seat ?

Thanks for clearing this up for me, and sorry for my poor English.
 
Physics news on Phys.org
Your English is fine. Friction between passenger and seat is nor relevant. At point A, what forces are acting on the passenger?
 
Normal and gravitational force(mg)

Usually we want to know what's the minimum velocity so that the can make it (N>0) but here I don't understand the question, what happens if it's going too fast ? will the rail break ? if yes what does that mean ?
 
shuby said:
Normal and gravitational force(mg)

Usually we want to know what's the minimum velocity so that the can make it (N>0) but here I don't understand the question, what happens if it's going too fast ? will the rail break ? if yes what does that mean ?

Unless you're taking a course on structural engineering, that's irrelevant. :P

I think your question was worded poorly. It should have said, MINIMUM velocity, since the only maximum there is to consider is the strength of the materials involved, which is completely out of the scope of the data of the question.
 
Yeah that's what I thought when I saw the question, but then I said maybe I'm wrong since the problem has 3 questions : a)what's the minimum V b)maximum V c)calculate the value for r=7

It's a instructor made problem... :p

Thanks for the help ! and I guess you will often see me in this board in the future, maybe not in this sub forum because I'll be in Electrical Engineering next year.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top