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Equation
F_{c} = ma_{c} = mv^{2}/r
2.7
Equation
h ≥ 2.5r
2.8
For more information, check out the following web sites:
How roller coasters work
Coaster Creator game
Centripetal Force
Now let’s look at the roller coaster loop-d-loop. In order for the coaster to successfully travel along the loop without falling, it must have sufficient velocity. This critical velocity is a function of the radius of the loop as well as the force of gravity. Here, we encounter a new force beyond the force of gravity, “centripetal force”. Literally translated from the Greek, the word centripetal means center-seeking. Centripetal force is the force, directed along the radius of a curved path that keeps an object on the path moving at a uniform velocity. It is always perpendicular to the velocity vector.
During a loop-d-loop the coaster and people in it experience an inertial force that moves them forward, while simultaneously experiencing a centripetal force at right angles to the forward force that pushes them against the track or the backs of the seats. They are being accelerated toward the center of the loop by the track. This acceleration toward the center is felt by the passengers as a force pulling them down in their seats, which is where riders want to be.
Centripetal force, F_{c}, is defined as:
centripetal force = mass x velocity squared ÷ radius F_{c} = ma_{c} = mv^{2}/r
The coaster can make it successfully all the way around the loop by ensuring that the centripetal acceleration is at least equal to the gravitational force at the highest point on the loop. To make sure the coaster doesn’t fall from point B, the roller coaster designer makes this calculation for a loop with a radius of r and a diameter of 2r at point B, the highest point on the loop.
We can now write that:
mv^{2}/r ≥ mg = v^{2}/r ≥ g and thus v^{2} ≥ gr
where ≥ means greater than or equal to
Combining equations 2.5 and v^{2} ≥ gr, we get
v^{2} = 2g(h-y)
2g(h-2r) ≥ gr
2h≥ 5r
height is greater than or equals 2.5 x radius h ≥ 2.5r
We have proved that the roller coaster start height must be at least 2.5 times as high as the radius of the circle in order for the coaster to safely make it around the loop (of course, remember that was assuming there would be zero friction; taking that into account requires even more initial height). And just to be safe, the wheels of the coaster are mechanically trapped so the coaster can’t fall from the tracks (just in case the roller coaster designer got a grade of D in physics).
Answer: 50 m
Solution:
h ≥ 2.5 r
h ≥ 2.5 • 20m ≥ 50m
Answer: 50 m
Gravity cancels out so it is not in Equ. 2.7. The height for the Moon is the same as for the Earth.
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