Why Does Teardrop Shape Matter in Roller Coaster Loop Design?

[brooke89]

Roller Coaster problem!

Homework Statement

A roller coaster at 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, with a maximum speed of 31.0 m/s (nearly 70mi/h) at the bottom. Suppose the speed at the top is 13.0 m/s and the corresponding centripetal acceleration is 2g.
(a)What is the radius of the arc of the teardrop at the top?
(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 20.0 m. If the cars have the same speed, 13.0 m/s at the top, what is the centripetal acceleration at the top? Comment on the normal force at the top in this situation.

Homework Equations

F= ma_c
Ac=v²/r
F=n_top +mg = mv²/r

The Attempt at a Solution

Ac=v^r/r
2=(13.0m/s)/r
r=6.5m

F=n_top +mg = mv²/r
F=n_top +(9.8m/s²)M=mv²/r

That's the farthest that I got...I don't know if I'm starting this right or even using the right equations. I have no idea how to do the rest.

 

[Doc Al]

Homework Equations

F= ma_c
Ac=v²/r
F=n_top +mg = mv²/r

All good.

The Attempt at a Solution

Ac=v^r/r
2=(13.0m/s)/r
r=6.5m

Correct this. You missed the v². Also, the acceleration is not 2, it's 2g.

F=n_top +mg = mv²/r
F=n_top +(9.8m/s²)M=mv²/r

Looks good to me. Just solve for that normal force.