The Teardrop Shaped Roller Coaster Loop

At the recent CAPT "Unseasoned" Physics Teacher Workshop Alan Haught mentioned that he had scanned a web page for the England Amusement Park Society (or some such organization) and had found out that at least some roller coaster loops have a radius of curvature for the bottom of the loop that is double that for the top half. Most of us probably show that if the entire loop is circular and the velocity at the top is critical, the normal force N at the bottom is six times the weight of the rider (six "gees"). The general result is that the difference between the apparent weights at the bottom and top is six times the weight of the rider, independent of the speed at the top. What if the loop is as the society claims?

Shown at right is such a loop. Applying conservation of energy between points A and B, assuming that the speed at A is critical, or , yields

(1/2)mvB2 = mg(3R) + (1/2)m()2.

Thus vB2 = 7Rg. Applying Fnet = ma at B,

N - mg =

= , or

N = 4.5mg.

Not a great improvement from N = 6mg for a circular loop, but a force that is at least slightly more tolerable. If the radius of the bottom portion is n times the radius of the top portion and the speed at the top is critical, can you show that N = ? It's interesting to note that as n increases, N approaches 3mg, so at B the rider must feel at least 3 gees.

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Are U = mgh and U = - Compatible?

Students are at the very least troubled by if not skeptical of the expression U = - for gravitational potential energy, even if one carefully derives the expression. The following example can be used to show that application of this and U = mgh yields the same answer, hopefully allaying some of the anxiety.

Assuming that a small object is dropped to earth (or any planet) from a height h that is small compared to the earth's radius R (h << R), what is the speed just before impact with the ground? Since h << R, g is essentially constant and conservation of energy yields mgh = (1/2) mv2, or v2 = 2gh. Let's try the same problem using the other expression for U, with M the mass of the earth and m the mass of the object. Recalling that r must be measured from the center of the earth,

- = -

Dividing by m and rearranging,

= .

If at this point h is neglected in comparison to R, v = 0, clearly a nonsense result. However, if a little more algebra is done everything works out. Multiplying by 2 and finding a common denominator,

v2 = .

Now if R + h @ R, this becomes

v2 = .

However, the term in parentheses is simply that for the acceleration due to gravity near the earth's surface, or v2 = 2gh!