A Ferris wheel (a vertical circle for the purposes of this question) with a radius 42.2 m is initially rotating at a constant rate, completing one revolution every 33.2 seconds. Suppose the Ferris wheel begins to decelerate at the rate of 0.0542 rad/s2.

Question

[Note: If it helps you, for sake of definiteness, you may take the wheel to be rotating clockwise… though this is immaterial.]
 
a) Find the magnitude of the passenger's acceleration at that instant in m/s2.

b) What is the angle between the passenger's velocity vector and acceleration vector at that instant (in degrees)?

Answer 1

ω₀=2πn₀=2π/33.2=0.19 rad/s
centripetal acceleration is a(cen) = ω₀²R=0.19²•42.2=1.52 m/s²
tangential acceleration is a(tan) =ε•R =0.0542•42.2=0.12 m/s²
acceleration is a=sqrt{ a(cen)²+a(tan)²}= sqrt{1.52²+0.12²}=1.525
α=arctan(a(tan)/a(cen)} =4.51°
the angle between the passenger's velocity vector and acceleration vector =90+4.51 =94.51°

Answer 2

Thank you very much!