A roller coaster at the Six Flags Great America amusement park in Gurnee, Illino

Question

A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop. The cars ride on the inside of the loop at the top, and the speeds are fast enough to ensure that the cars remain on the track. The biggest loop is 40.0 m high. Suppose the speed at the top is 14.1 m/s and the corresponding centripetal acceleration is 2g.

(a) What is the radius of the arc of the teardrop at the top?

                           m

(b) If the total mass of a car plus the riders is M, what force does the rail exert on the car at the top?

(c) Suppose the roller coaster had a circular loop of radius 21.1 m. If the cars have the same speed, 14.1 m/s at the top, what is the centripetal acceleration at the top?

(d) Comment on the normal force at the top in the situation described in part (c) and on the advantages of having teardrop-shaped loops.

Explanation / Answer

Centripetal acceleration of an object in circular motion of radius "R" and speed "v" is; v^2/R

Even though your track is not a complete circle with the same radius, the car's at the top don't know that. So their centripetal acceleration uses a small section of arc of circular radius "R"; even though that radius will change.

So at the top;

(a)

v^2/R = 2g

R = v^2/2g = 10 m (answer)

(b)
At the top the forces on the car producing the centripetal accel. are its weight "mg" down and the normal force "N" of the track pushing down (Normal forces can not pull). From 2nd law;

Mg + N = M(2g)

N = Mg (answer)

(c) Use the centripetal accel eq.; v^2/R

So, Centripetal accel = 14.1^2/21.1 = 9.42 (answer)

(d) At the top the forces on the car producing the centripetal accel. are its weight "mg" down and the normal force "N" of the track pushing down (answer)

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