Posted by R on .
The roller coaster starts with a velocity of 16 m/s. One of the riders is a small girl of mass 26 kg. Find her apparent weight when the roller coaster is at locations B and C. At these two locations, the track is circular, with the radii of curvature given.
rB = 10 m,
rC = 20 m,
The heights at points A, B, and C are
hA = 22 m,
hB = 32 m,
and
hC = 0.
Assume friction is negligible and ignore the kinetic energy of the wheels. (The figure is not necessarily drawn to scale.)
What is the weight at B?

Physics 
Elena,
Point A: h₁= 22 m, v₁=16 m/s;
Point B: h₂= 32 m, r₂= 10 m, v₂=?;
Δh₁₂ = h₂h₁=3222 = 10 m;
Point C: h₃=0, r₃= 20 m, v₃=?;
Δh₂₃ = h₂h₃ = 32 – 0 = 32 m.
KE₁+PE₁=PE₂ + KE₂,
mv₁²/2 +mgh₁=mgh₂ +mv₂²/2,
mv₁²/2= mv₂²/2+ mgh₂  mgh₁=
=mv₂²/2+ mg Δh₁₂.
v₂ = sqrt{ v₁² 2gΔh₁₂} =
=sqrt{16²  2•9.8•10} =7.75 m/s.
For point B:
ma₂=mgN₂,
N₂=mgma₂=m[g (v₂²/r₂)]=
= 26[ 9.8 – (7.75²/10)]= 98.8 N,
N₂ (normal force) = W₂ (weight)
W₂= 98.8 N.
PE₂ +KE₂ = PE₃+KE₃,
mgh₂ +mv₂²/2 = mgh₃ +mv₃²/2,
mv₃²/2 = mgh₂ +mv₂²/2  mgh₃=
=mv₂²/2 + mg Δh₂₃ ,
v₃ = sqrt{v₂² +2g Δh₂₃}=
=sqrt{ 7.75² + 2•9.8•32) = 26.2 m/s.
For point C:
ma₃= N₃ – mg,
N₃= mg+ma₃=m[g+ (v₃²/r₃)]=
= 26[ 9.8 + (26.2²/20)]= 1148.2 N.
N₃ (normal force) = W₃ (weight)
W₃= 1148.2 N.