Question

PLEASE ANSWER CORRECTLY THE CORRECT ANSWER IS 46

A roller coaster cart starts at the top of a quarter circular hill (no friction) of some radius at a speed of 11.8 m/s.   At the bottom of the hill it encounters a horizontal surface with friction and the coefficient of kinetic friction is 0.55. It travels over that horizontal surface for 8 meters and encounters a frictionless loop with a radius of 23 meters. If it is to just barely make it around the top of the loop (barely meaning that it's not going to fall of the track at the top), what is the minimum radius, in meters, for the quarter circular hill? $hw6_roller.JPG$

Selected Answer: 34.5 Correct Answer: 46.00 ± 10%

Answer 1

here,

initial speed , vi = 11.8 m/s

let the radius of the quater circle be r1

let the speed at the bottom of the circle be v2

using conservation of energy

0.5 * m * v2^2 = m*g*r1 + 0.5 * m * vi^2

v2 = sqrt( 2* g*r1 + 139.24)

deaccelration due to friction , a = u*g

a = 5.49 m/s^2

let the velocity after the frictional track be v3

using third equation of motion

v3^2 - v2^2 = - 2 * a*s

v3^2 - (sqrt( 2* g*r1 + 139.24))^2 = - 2 * 5.49 * 8

V3 = sqrt( 2* g*r1 + 51.4)

velocity at the top of the circle be v4 ,

0.5 * m*v4^2 = 0.5 * m * v3^2 - m*g*(2*r2)
v4 = sqrt( 2* g*r1 - 850.2) m/s

If it is to just barely make it around the top of the loop

then , m*g = m*v4^2/r2

9.8 = ( 2* g*r1 - 850.2) / 23

r1 = 54.88 m

the radius of the quater circle is 54.88 m