A Ferris wheel rotating at 20 rad/s decelerates with a constant angular acceleration of -5.0 rad/s2.?

2 Answers

• ωf = ωi + at

  t = (ωf – ωi)/a = (0-20)/-5 = 4s

  θ = ωt + 0.5at^2

  θ = (20*4) + (0.5 * -5 * 4^2)

  = 80 + (-40)

  = 40 radians

  2π rad = 1 rev

  40 rad = 40/2π rev = 6.36 rev.

  So the ferris wheel will have 6.36 rev.

• w(1)=20 w(2)=0 a=-50 0^2-20^2=2(-50)x x=4 revolutions