Toy car loop physics problem

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  • #1
xXmarkXx
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A toy car rolls down a hill at with a height of h and goes through a loop with a radius r. What is the height the toy car has to start off from rest at, if everything is frictionless and is able to make it throught the loop.

I believe the answer it 2r because of the energy involved. If everything is frictionless, then the toy car must start from the height of the top of the loop. My professor wants equations, but i don't know how to break this one down. Thanks for the help.
 

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  • #2
Doc Al
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I believe the answer it 2r because of the energy involved. If everything is frictionless, then the toy car must start from the height of the top of the loop.
That would give the car zero speed at the top of the loop--which is not enough for it to maintain contact. (It would have fallen away from the track before it made it to the top.) Hint: Consider the forces acting on the car at the top of the loop and the acceleration of the car.
 
  • #3
xXmarkXx
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That would give the car zero speed at the top of the loop--which is not enough for it to maintain contact. (It would have fallen away from the track before it made it to the top.) Hint: Consider the forces acting on the car at the top of the loop and the acceleration of the car.

Ok, that makes sense. so h>2r? So how would i achieve that answer by using equations. Gravity is acting on the car at the top of the loop, and a normal force.
 
  • #4
Doc Al
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Apply Newton's 2nd law.
 
  • #5
xXmarkXx
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Apply Newton's 2nd law.

Fnet=ma?
ok...i know the formula. How do i apply it though. I don't have mass, i don't have acceleration. What are you talking about. I'm dealing with potential energy.
 
  • #6
Doc Al
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Energy methods alone will not solve this problem. Another hint: What kind of motion is the car undergoing as it loops the loop?
 
  • #7
xXmarkXx
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Energy methods alone will not solve this problem. Another hint: What kind of motion is the car undergoing as it loops the loop?

+ and - motion?? i don't understand what your asking.
 
  • #8
Doc Al
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Yet another hint: It's going in a circle!
 
  • #9
xXmarkXx
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Yet another hint: It's going in a circle!


yeah...ok. I know it's going in a circle which is why I'm given a radius=R. I don't know where you are going with this.
 
  • #10
turdferguson
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The sum of the forces must keep the car in a circle. But what is the minimum force acting at the top of the loop?
 
  • #11
xXmarkXx
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The sum of the forces must keep the car in a circle. But what is the minimum force acting at the top of the loop?

The normal force
 
  • #12
mybsaccownt
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how do you get that normal force?
 
  • #13
xXmarkXx
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how do you get that normal force?

By the gravitational force or weight. But in this case. Wouldn't N be just barely over zero or zero because both of the forces are pointed downward?
 
  • #14
mybsaccownt
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well, if you had the car upright on a flat surface, yes the normal force would be the one that's caused by gravity

but in this case, the car is upside-down, if gravity wants to pull it off the track...what is resisting that?

certainly not the weight of the car, right?

you can't have N = 0 if you want the car to stay on the track, otherwise it's just in freefall, right? if gravity is the only thing acting on it...it'll do what it wants

so, something is causing a force here...read through the previous posts (doc al's hints) carefully

and keep Newton's second law in mind
 
Last edited:
  • #15
Luke1294
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I'm doing this same problem, and have approached it like this-

At the beginning, the car has potential energy = mgh. At the bottom, this potential has been converted into kinetic energy, 1/2mv^2.

1/2mv^2=mgh, or, 2mgh=mv^2.


During the loop, the car will undergo circular motion, where the net forces are equal to = (mv^2)/r. At the very top of the loop, the car will only feel two forces- normal force and the force of gravity.

So...

N + mg = mv^2/R

Subsititue 2mgh in for mv^2, which gives us

N + mg = 2mgh/R

N=2mgh/R - mg

N=(mg)(2h/R - 1)

Now, this is where I am not positive what I'm doing is legal- I rewrite normal as mg

mg = (mg)(2h/R-1)

1 = 2h/R - 1
2= 2h /R
2R=2h
r=h


Which would mean that the height only needs to be as tall as the radius of the loop? Am I to assume, then, that inertia is what carries the car through the rest of the loop? Or have I made a silly mistake?
 
  • #16
xXmarkXx
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I'm doing this same problem, and have approached it like this-

At the beginning, the car has potential energy = mgh. At the bottom, this potential has been converted into kinetic energy, 1/2mv^2.

1/2mv^2=mgh, or, 2mgh=mv^2.


During the loop, the car will undergo circular motion, where the net forces are equal to = (mv^2)/r. At the very top of the loop, the car will only feel two forces- normal force and the force of gravity.

So...

N + mg = mv^2/R

Subsititue 2mgh in for mv^2, which gives us

N + mg = 2mgh/R

N=2mgh/R - mg

N=(mg)(2h/R - 1)

Now, this is where I am not positive what I'm doing is legal- I rewrite normal as mg

mg = (mg)(2h/R-1)

1 = 2h/R - 1
2= 2h /R
2R=2h
r=h


Which would mean that the height only needs to be as tall as the radius of the loop? Am I to assume, then, that inertia is what carries the car through the rest of the loop? Or have I made a silly mistake?

I'm fairly positive you have made a silly mistake. I know the answer is 5/2R but I'm trying to figure out why. The answer is not just R. If the height was only half of the loop, do you actually think it would make it though the whole loop?? doubt it.
 
  • #17
mybsaccownt
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At the very top of the loop, the car will only feel two forces- normal force and the force of gravity.

So...

N + mg = mv^2/R


Or have I made a silly mistake?

what is the N in this equation?
 
  • #18
Luke1294
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what is the N in this equation?
N was meant to stand for the normal force.
 
  • #19
turdferguson
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The way you set up the problem, h is the height above the top of the loop, which makes the total height 3R. Youre wrong, though, and not with the algebra. The minimum centripetal force is not N + mg
 
  • #20
Luke1294
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The way you set up the problem, h is the height above the top of the loop, which makes the total height 3R. Youre wrong, though, and not with the algebra. The minimum centripetal force is not N + mg
I guess I don't see what other forces are acting on the car at the very top of the loop. It would seem at the top, the point I am looking at, there would only be gravity and the normal force. What other forces would be acting on it at this point?
 
  • #21
xXmarkXx
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I guess I don't see what other forces are acting on the car at the very top of the loop. It would seem at the top, the point I am looking at, there would only be gravity and the normal force. What other forces would be acting on it at this point?


no,your right luke, there are only two forces
 
  • #22
mybsaccownt
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luke, you are actually very close

but try this instead

consider the energy at the top of the loop

it will have some potential and some kinetic right?

so Ui + Ki = Uf + Kf

and, at the top of the loop, you want the car's weight to be equal to the force due to the normal acceleration

mg = mv^2/r

rearrange, to solve for mv^2, and plug this into your energy equation and that's that
 
  • #23
Luke1294
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Okay, I think I got it! Just check me real quick to make sure I'm not wayyy off base.

I solved mg = mv^2/r to be mgr = mv^2.

Ui + Ki = Uf + Kf

mgh(i) + 0 = mgh(f) + .5mv^2(f)

Or..

mgh(i) = mgh(f) + .5mgr(f)

I divided everything by mg, giving me

h(i) = h(f) + .5r

Well, the top of the loop is the same as the diameter from the ground to the top of the loop, or two radii, so...

h(i) = 2r + .5r

h(i) = 5/2r


Look right to everyone?
 
  • #24
turdferguson
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Yes, at the top of the loop there's only weight
 

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