Consider a circular roller coaster loop of radius 15.1 m. You already know that

Question

Consider a circular roller coaster loop of radius 15.1 m. You already know that if the car starts from a height of 2.5 r at rest, the normal force acting on the car at the top will vanish.


Now assume the car starts from a level track, a height of 15.1 m above the ground. The track first guides the car to ground level and then into the loop. At which initial speed must the car move on the elevated track (height 15.1 m) such that it still has the required speed at the top of the loop for the normal force to vanish?

Explanation / Answer

From first part 1/2(mv^2)=mg(H-h)

                                           = mg (2.5r-r)

                                           = 1.5(mgr)

                                   v^2=3gr

                                    v=sqrt(3gr)

Now The normal force N= centripetal force= mv^2/r= 3mg

Now in the second part

                                 mgh=(1/2)mv^2 + 3mg

                                   2(gh-3g)=v^2

                                    2*9.8*(15.1-3)=v^2

                                      v=15.4 m/s

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