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Indira_Clueless
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Homework Statement
Task: [/B]
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.
Homework Equations
* Fg = m * g
* FZP = (m * v2) / r
* Ekin = 1/2 * m * v2
* Epot = m * g * h
* cos(α) = adjacent / hypotenuse
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: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)
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)
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