# Loop-de-Loop with function of angle

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1. Nov 11, 2015

### Indira_Clueless

1. The problem statement, all variables and given/known data

On a friction free loop de loop, a object (m=600kg with passengers) moves with a velocity of vu = 120km/h at the lowest point of the loop. The loop has a radius of r=15m. The position of the object is given with the angle α.

Derive the formula for the velocity v(α). Calculate whether or not the object passes through the highest point of the loop.

2. Relevant equations
* Fg = m * g
* FZP = (m * v2) / r
* Ekin = 1/2 * m * v2
* Epot = m * g * h
* cos(α) = adjacent / hypotenuse

3. The attempt at a solution

So I know that I can calculate the speed with the help of the rule of conservation of energy.
Generally, the formula for the speed at any given height (with hmax = 2r) would be:

1/2 * m * vh 2 = 1/2 * m * vu2 - m * g * h
I can divide that through the mass and take the whole thing times 2. That results in:
vh2 = vu2 - 2 * g * h
and by taking the squareroot, the final formula v of x would result in:
vh = √(vu2 - 2 * g * h)
Now I thought: Hey, why not write h so that it is dependent on the angle α. But that turned out to be way more difficult than I thought ... and I am sure I didn't do it right but here it goes:

On the picture "Idea" it's kinda explained:

The heigth of the object is always a part of 2r. So if I drew a straight line from the point the object is on the loop/circle and let it intersect, the actual height of the object would be that part of the 2r. I used the idea that the angle α could create a segmentline s between the startpoint and the current location of the object. Wikipedia said the length of a segment is calculated with s = 2 * r * sin(α/2). But the length of the segment only gives me a part of the height so I used trigonometry to calculate the green height using the angle β. I thought since the triangle the radius and s is a isosceles triangle, the other two angles besides α are ought to be the same. If they are, one of them could be calculated by:
β = (180° - α) / 2​
By using trigonometry, I found out that the height of the object should generally be:
h = cos(β) * s​

Putting it all together, the formula for my height would be:
h = cos((180°-α)/2) * 2 * r * sin(α/2)
And now all there is left is putting it in the formula for the velocity I derived above, which results in:
v(α) = √( vu2 - 2 * g * cos((180°-α)/2) * 2 * r * sin(α/2) )
That's a very long formula if you ask me and somehow I don't believe it to be right.

In order to calculate whether or not the object passes through the highest point of the loop (at α=180°), Fzp ≥ Fg which means v(180°) ≥ √(r * g)

If I put everything in the formula, v(180°) = 22.81 m/s which seems legit enough.
√(15 m * 9.81 m/s2 ) = 12.13 m/s which means that the object should pass through the highest position.

Could someone please help me solve the question and correct me where I got wrong. I'd be so grateful for every help.

With regards,
Indira

(P.S.: Hopefully my translation of the task doesn't suck that badly, translating physical terms to english is still quite new for me)

Last edited: Nov 11, 2015
2. Nov 11, 2015

### Khashishi

Your expression for the height is too complicated and wrong. Since height is arbitrary, you can just set the zero of height to the be center of the circle.

3. Nov 11, 2015

### Indira_Clueless

But how do I derive an equation for v(α) if not through trying to derive the height by α?

Last edited: Nov 11, 2015
4. Nov 11, 2015

### Khashishi

I didn't say you shouldn't calculate the height! You need to calculate the height correctly.

5. Nov 11, 2015

### Indira_Clueless

Could you maybe help me calculate it correctly? I have been stuck for a while and probably miss the obvious (which I apologize for)

/Edit: I can't really set the height of the middle of the loop to zero as it would alter the calculation for potential energy and by that alter the whole formula right?

6. Nov 11, 2015

### haruspex

No, it's correct, but it is much more complicated than it needs to be.
@Indira_Clueless , go back to the original diagram and draw a horizontal line from the position of the cage to the vertical dashed line. How far is it from that intersection to the circle's centre?

7. Nov 11, 2015

### Khashishi

Potential energy has an arbitrary zero point. You can set the zero of height wherever you want and it should not affect your final answer. Also, if you know how to calculate the height from the center of the circle, you can just add the height of the center.

8. Nov 11, 2015

### Khashishi

Well I'll be damned. . I thought it was wrong because I would never have thought to calculate it that way.

9. Nov 11, 2015

### Indira_Clueless

@haruspex
To be honest, I don't know what you mean .... I'm sorry for being so supid at the moment, I'm just really feeling helpless and am apparently not able to comprehend easy questions.

@Khashishi believe me, it took forever to come up with this bullshit (sorry)

10. Nov 11, 2015

### haruspex

Let me label some points. C is centre of circle, O is bottom of circle, A is current position of cage.
Horizontal line from A intersects OC at B.
OC=AC=r.
You know angle BCA. Study triangle ABC.
How long is BC?

11. Nov 11, 2015

### Khashishi

What do you get for height if you set the zero point in the center of the circle?
Now just take that result and add the height of the center of the circle.

12. Nov 11, 2015

### Indira_Clueless

@haruspex

well I guess the length of BC would be √(CA2 - BA2) ?
I mean, if I got you correctly ...

/Edit: Which means the height would be r - BC ... right?

13. Nov 11, 2015

### Indira_Clueless

@Khashishi if I set the general height of the circle to zero in the center, then the max height would be r and the starting height would be -r ?

14. Nov 11, 2015

### Khashishi

yeah

15. Nov 11, 2015

### Indira_Clueless

I could calculate the upper part (positive part) of the height with sin(α) but I'm not sure what to do with the negative part

16. Nov 11, 2015

### Khashishi

Who said height can't be negative?

17. Nov 11, 2015

### Indira_Clueless

I do know it can be, it depends on where you set the start point (for example of a movement. If I throw something down and I set the startpoint of the motion to where the fall started, then the height the object is gain would be negative). But my α would be negative too right? Since my startpoint is set not at the lowest point of the loop but the middle point. Which means the angle for the first half of the movement (the upmoving along the loop) is -90°≤α≤0° and 0°≤α≤90° for the second half...

18. Nov 11, 2015

### Khashishi

From the diagram, it is clear that $\alpha$ is 0 at the bottom and is increasing as you go counterclockwise. There's no need for a negative $\alpha$.

19. Nov 11, 2015

### Indira_Clueless

Ah and then I can use pretty simple trigonometry ... Yeah I think I get how this is a way.

20. Nov 11, 2015

### haruspex

Yes, but use the known angle BCA (=alpha) and the known distance AC=r.