6 pts You are designing a rollercoaster One section of the t

[6 pts] You are designing a roller-coaster. One section of the track will be advertised as

“The Loop.” The shape of the track is defined parametrically by the equations

x = [u/2 + sin(u)]H/2

y = [1 cos(u)]H/2

z = 0

where y is the height above ground and 0 < u < 2. Find the force due to the track on

the car of mass M when the car reaches its maximum height at u = , ignoring friction.

Assume the car starts at u = 0 with speed v0.

Solution

x = [u/2 + sin(u)]H/2
y = [1-cos(u)]H/2
z = 0

initial KEi = 0.5*m*(Vo)^2
xi = 0
yi = 0
zi = 0

final KEf = 0.5m(Vo)^2 - mgyf
xf = pi*H/4
yf = H
zf = 0

KEf = 0.5m(Vo)^2 - mgH = 0.5mv^2
v = sqroot([Vo]^2 - 2gH)

Dx/Dy = [0.5 + cos(u)]/[sin(u)]

At last point, Dy/Dx = 0
Dy/Dx = sin(u)/(0.5 + cos(u))
D2y/Dx2 = [[0.5 + cos(u)]cos(u) - sin(u)*(-sin(u))][Du/Dx]/[0.5 + cos(u)]^2
Dx = [0.5 + cos(u)]HDu/2
Du/Dx = 2/H[0.5+cos(u)]
D2y/Dx2 = [[0.5 + cos(u)]cos(u) - sin(u)*(-sin(u))][2/H[0.5+cos(u)]]/[0.5 + cos(u)]^2
at last point
D2y/Dx2 = [[0.5 + 1][2/H[-0.5]]/[-0.5]^2 = -1.5*2/H*(0.125) = -3/H

radius of curvature = (H/[3]) = H/3

FBD -> mg - N = mv^2/R
N = mg - 3m[(Vo)^2 - 2gH]/H