# Loop-the-loop find height object falls out of the loop

1. Oct 22, 2011

### crazycool2

1. The problem statement, all variables and given/known data
m= 0.2Kg
r=0.6m
k= 500N/m
x=0.08m
the spring is pressed together 0.08m before it's released. Friction = 0. the objekt goes through a loop. find the height when the object loses contact with the loop.

2. Relevant equations

3. The attempt at a solution

1/2kx2= 1/2mv2 + mgh (where v is between 0 and √gr )
I got 0.5 with v=0 and 0.81 m with v=√gr
answer is supposed to be 0.744m.

2. Oct 22, 2011

### Staff: Mentor

You need to determine the conditions for maintaining contact with the loop. A minimum velocity (greater than zero) is required at each point. (There's more to this than just conservation of energy.)

3. Oct 22, 2011

### crazycool2

that's all that is the question, but I tried using Forces. the conditions to stay in the loop on top is mg + n = mv2/r where n=0 because it doesn't touch the loop. then v= √gr but that's just on the top, I don't know how to do it in another random area . the point it lets go it becomes free fall but just before I don't know. do you have any more clues or soultion. thnx

4. Oct 22, 2011

### LawrenceC

See if you can relate the velocity as a function of height above the horizontal diameter. You can easily determine the velocity at the horizontal diameter, right?

5. Oct 22, 2011

### Staff: Mentor

You need the same basic idea, but in an arbitrary radial direction. So, if the point in question were an angle θ above the horizontal, how would you write the force equation along the radial direction? (What's the component of gravity in that direction?)

6. Oct 22, 2011

### crazycool2

that's where I get stuck mixing radial stuff with Newtons laws,
but I'm guess you mean v=rω
ω= delta(θ)/delta t?

7. Oct 22, 2011

### Staff: Mentor

No, I mean ƩF = ma in the radial direction. What forces act? (It's almost the exact same equation you had.)

8. Oct 22, 2011

### crazycool2

mg sin(theta) = ma
the forces that act are mg downwards and centripetal in the radial direction. then our mg sin must be our normal force which also points in the radial direction,am I going in the right direction?

9. Oct 22, 2011

### Staff: Mentor

Right!
The forces are mg and the normal force.
Almost there--let's clean up that thinking a bit. The full expression would be:
ƩF = ma
N + mgsinθ = ma

We want the condition for just starting to fall off the loop, which is where N = 0. So:
mgsinθ = ma

Keep going. (Here's where you can start to apply conservation of energy to solve for the height where that condition is met.)

10. Oct 22, 2011

### crazycool2

if I add theta then I will have two unknowns, h and theta, and since now i can find the speed of v at horizontal diameter. mgsin 45=mvv/r but how will it help?

11. Oct 22, 2011

### Staff: Mentor

They are related, so they are really only one unknown. Express everything in terms of h. (You might find it easier to use height above the midpoint, then convert it to height above the bottom later.)
Not sure what you're doing here. Instead: Use conservation of energy to find the acceleration (v2/r) as a function of height. Then you can plug it into your other equation to solve for the height that makes N = 0.

12. Oct 22, 2011

### crazycool2

v = sqrt(g r sin (theta))
0.5mv*v = mg h
h= r sin (theta) /2
then i use my equation
1/2* kx^2 = 1/2m g sin (theta) + 1/2mg sin (theta). but i get my 1/2 kx^2 > mgr. should that be possible?

13. Oct 22, 2011

### Staff: Mentor

OK, but you don't really need to isolate v.
This isn't true.
Not exactly. Show how you arrived at that.
You need to combine this equation (once you express θ in terms of height):
mg*sinθ = mv2/r

With conservation of energy (also expressed in terms of height):

Initial energy (of spring) = final energy (KE + Gravitational PE)

14. Oct 24, 2011

### crazycool2

thanx I solved it! I expressed my v as √g r sinθ and my height as r(1+sin(θ) and then put it in the consevation of energy equation!
,but there was a part b) in the question. if you assume the object goes out off the loop at any given P and lands on the loop again at E which is the same height as the centre. show that the angle must be 60° and how much must the spring pressed together.

Attempt

initial speed when the object comes off the track. can be expressed as Vsin ω in y direction and Vcos ω in x direction. but this angle is the same as sin ω = sin (pi/2-θ) = cos θ then that gives less unknowns .
and I combined projectile motion equations
x = V0x*t
y= V0y*t+ 0.5gt2 y0 where x= r(1+cosθ) and y = r and y0= r(1+sinθ)

r = v*Cot[\[Theta]]*r (1 + Cos[\[Theta]]) + (
g*r^2 (1 + Cos[\[Theta]])^2)/(2*v^2*(Sin^2)[\[Theta]]) +
r (1 + Sin[\[Theta]])
but I'm not sure if I need to combine with the forces and how.