Question

You are testing a new roller coaster ride in which a car of mass mm moves around a vertical circle of radius RR. In one test, the car starts at the bottom of the circle (point A) with initial kinetic energy KiKi. When the car reaches the top of the circle (point B), its kinetic energy is Ki/4Ki/4, and its gravitational potential energy has increased by Ki/2Ki/2. What was the speed of the car at point A, in terms of gg and RR? Express your answer in terms of the variable RRR and constant ggg.

Answer 1

Answer:

vi = 2.83 √gR

Explanation:

For this exercise we can use the law of conservation of energy

Let's take a reference system that is at point A, the lowest

Starting point. Lower, point A

           Em₀ = Ki = ½ m vi²

Final point. Higher, point B

            $Em_{f}$ = K + U

             

 It indicates that at this point the kinetic energy is ki / 2 and the potential energy is ki / 2

             K = ki / 2

             U = m g (2R)

Energy is conserved so

             Em₀ = Em_{f}

             ½ m vi² = ½ (1/2 m vi²) + m g 2R

             ½ m vi² (1- ½) = m g 2R

              vi² = 4 g 2 R

              vi = √ 8gR = 2 √2gR

              vi = 2.83 √gR