You are designing the section of a roller coaster ride shown in the figure. Prev

Question

You are designing the section of a roller coaster ride shown in the figure. Previous sections of the ride give the train a speed of 15.5 m/s at the top of the incline, which is 35.9 m above the ground. As any good engineer would, you begin your design with safety in mind. Your local government's safety regulations state that the riders' centripetal acceleration should be no more than n = 189 g at the top of the hump and no more than N = 5.29 g at the bottom of the loop. For this initial phase of your design, you decide to ignore the effects of friction and air resistance. What is the minimum radius you can use for the semi-circular hump? What is the minimum radius you can use for the vertical loop?

Explanation / Answer

  total mechanical energy at start
TME = mgh + ½mv² = M*(gh + v²/2)
TME = M*(9.8m/s² * 35.9m + (15.5m/s)²/2) = M * 471.945m²/s²

At the top of the hump of radius r,
TME = mgr + ½mv² = M * 471.945m²/s² mass m cancels
g*r + v²/2 = 471.945 m²/s²

We can't allow the centripetal acceleration to exceed 1.89g:
v²/r = 1.89g, or
v² = 1.89gr

so
g*r + 1.89g*r/2 = 471.945 m²/s²
10.745m/s² * r = 471.945m²/s²
r = 43.9 m min radius of hump

At the bottom of the loop,
½mv² = M * 471.945m²/s² mass m cancels
v²/2 = 471.945 m²/s²

and now v²/r = 5.29g, or
v² = 5.29*9.8m/s² * r= 51.84m/s² * r
so
51.84m/s² * r / 2 = 471.945 m²/s²
r = 18.2 m min radius of loop

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