Non conservative energies on a Loop

[pulcroman]

Hello, i appreciate your help!

Homework Statement

In this website the drawing of the situation in the first exercise can be seen, for better understanding.
http://wwwprof.uniandes.edu.co/~gtellez/fisica1/ejercicios-semana10.pdf

The problem asks me to determine the minimum height h to allow an amousement park car to complete a loop with radius r without falling. The inicial velocity is 0. In the beginning, i had to do it without friction, with no problem. The horribleness started when i tried to do it with friction, more specifically in the loop part.
I use v1 for the speed after it falls the height h, v2 for the time it continues by the flat, and v3 for the speed in the tallest part of the loop. (Question in the last sentencie, for the impatient readers :) )

Homework Equations

E2-E1=-W; W=Fk dx; Fk=µk N. E(1,2)= Ek+Ep

The Attempt at a Solution

For the beginning of the movement, the fall from the height h to the floor, assuming is a straight plane is given by:
m g h - 1/2 m v1^2 = -µk m g cosø (h/senø)<---that last is the distance from h.
v1=√(2 g h-g h µk cotø) <---- i'll go fast, in he first steps. I just simplified v1.

for the second part, when the car moved a small part on a flat, i had
1/2 m v2^2 - 1/2 mv1^2=-µk m g cos ø x <--- x=dx, distance.
v2=√(2gh - g h µk cotø -2 g x µk)

Now, in the loop.
I've tried everything, and can't come up with a reliable solution. I know i can use
1/2 m v3^2+ mg(2r) - 1/2 m v2^2=-Fk x2

well, x2 is πr because it'll just half the circle, but I am having a real bad time finding the normal force for the Fk=µk N. Because, for a circular movement i got:
N= (m v^2)/r + m g cos ø
is v changing? if i integrate for ø from 0 to π it'll give 0! if i want to find the tangential acceleration i'll complicate myself even more. So, in a summary, the big question i want to ask is

How do you estimate the work made by friction on a loop with given inicial velocity, so the car won't fall? Thank you very much!

 

[EkaterinaAvd]

N= (m v^2)/r + m g cos ø - why did you place gravitational force projection to the right side instead of left?

ma = N + mg_r
(mg_r - radial component of gravitational force)

yes, v during the loop is changing according with the energy-work theorem

are you given mu in this problem? You have to know the shape of the initial slope then to determine work done during the slope...

 

[pulcroman]

hi, thanks for your help!
well, according to the free body diagram, from 0 to π/2 gravitacional force is negative and normal force is positive.
N-mgcosø=mv^2/r <--- and that negative sign will change after π/2 because normal and gravitational will have the same sign!

i have to give the height h in all the parameters i want, and µ is effectively there.
about the tangential acceleration:
will -Fk - mgsenø =m dv/dt (ecuación for the other cordinate in the circle) help in something? i can get Fk from here, but that dv/dt is a real headache.

Fk (kinetic fricción force)

 

[pulcroman]

Well, i gave it a bit more thought and came up with this:
for the loop:
Ek(final)^2+Ep(final)-Ek(inicial)=-Fk dx
Fk=µk N; dx=pi R
N ?=mv^2/R+mgcosø (the + doesn't matter, after quarter the circle it'll change from - to +)

Ek(final)^2+Ep(final)-Ek(initial)=-µk (mv^2/R + ∫mgcosø 0->pi*) pi R
Ek(final)^2+Ep(final)-Ek(initial)=-µk mv^2 pi

is it ok what i did with the normal force in the right?


*because of the change in angle, if this is ok, the integral equals 0, so this factor disappears

 

[EkaterinaAvd]

I don't think that you want to have t in your system of equations

If you use

1) second Newton's law for position with arbitrary angle theta (which you already have)
2) relationship between Fk and N
3) energy-work theorem in differential form (how much kinetic and potential energy change once your car moves over angle dø)

if you plug-in (1) and (2) into (3), you will get linear differential equation for v^2 of the first order which is possible to solve. If you solv it, you will have relashionship between v(ø=0) (velocity at the bottom) and v(ø=pi) (velocity at the top)

I've never solved this problem before. Maybe, there is an easier way to do that exists, I don't know

 

[EkaterinaAvd]

Ek(final)^2+Ep(final)-Ek(inicial)=-Fk dx
Ek(final)^2+Ep(final)-Ek(initial)=-µk (mv^2/R + ∫mgcosø 0->pi*) pi R

I think that

Ek(final)+Ep(final)-Ek(inicial)=-∫Fk dx=-µk ∫(mv^2/R + mgcosø)dx=-µk ∫(mv^2/R + mgcosø)rdø

your v under integral depends on theta, it is not a constant over the loop

that is why I think you need a differential equation

yes, now I see that you choose ø=0 at the bottom and ø=pi, so sign near mg projection has sense to me