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

In summary, the homework statement is that the spring is pressed together 0.08m before it's released. Friction = 0. the object goes through a loop. find the height when the object loses contact with the loop.
  • #1
crazycool2
16
0

Homework Statement


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.

1006757.jpe

Homework Equations





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.
 
Physics news on Phys.org
  • #2
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
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 let's go it becomes free fall but just before I don't know. do you have any more clues or soultion. thnx
 
  • #4
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
crazycool2 said:
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 let's go it becomes free fall but just before I don't know. do you have any more clues or soultion. thnx
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
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
crazycool2 said:
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?
No, I mean ƩF = ma in the radial direction. What forces act? (It's almost the exact same equation you had.)
 
  • #8
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
crazycool2 said:
mg sin(theta) = ma
Right!
the forces that act are mg downwards and centripetal in the radial direction.
The forces are mg and the normal force.
then our mg sin must be our normal force which also points in the radial direction,am I going in the right direction?
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
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
crazycool2 said:
if I add theta then I will have two unknowns, h and theta,
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.)
and since now i can find the speed of v at horizontal diameter. mgsin 45=mvv/r but how will it help?
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
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
crazycool2 said:
v = sqrt(g r sin (theta))
OK, but you don't really need to isolate v.
0.5mv*v = mg h
This isn't true.
h= r sin (theta) /2
Not exactly. Show how you arrived at that.
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?
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
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.
 

1. What is a loop-the-loop?

A loop-the-loop is a type of roller coaster or amusement park ride where the track forms a complete loop, with the train or object traveling through the loop and coming back to the starting point.

2. How does a loop-the-loop work?

A loop-the-loop works by using the principles of centripetal force and inertia. As the train or object travels through the loop, it experiences a centripetal force that keeps it moving in a circular path. This force is provided by the track, which is designed to have a larger radius at the top of the loop and a smaller radius at the bottom, allowing for a smooth transition and constant speed throughout the loop.

3. What is the height at which an object falls out of a loop-the-loop?

The height at which an object falls out of a loop-the-loop depends on a few factors, including the speed of the object, the radius of the loop, and the force of gravity. Generally, the object will fall out of the loop when the centripetal force is no longer strong enough to keep it moving in a circular path, which can be calculated using the formula Fc = mv^2/r, where m is the mass of the object, v is the speed, and r is the radius of the loop.

4. What happens if an object falls out of a loop-the-loop?

If an object falls out of a loop-the-loop, it will continue to move in a straight line due to inertia, potentially hitting the ground or other objects. This can result in injury or damage, which is why loop-the-loops are carefully designed and tested to ensure the safety of riders or objects.

5. Can you change the height at which an object falls out of a loop-the-loop?

Yes, the height at which an object falls out of a loop-the-loop can be changed by adjusting the speed, radius, or other factors such as the angle of the loop or the force of gravity. By changing these variables, it is possible to design a loop-the-loop that can accommodate different types of objects or riders of varying weights and sizes.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
10K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
2K
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
Back
Top