A ball of radius 0.211 m rolls along a horizontal table top with a constant linear speed of 3.41 m/s. The ball rolls off the edge and falls a vertical distance of 2.27 m before hitting the floor. What is the angular displacement of the ball while the ball is in the air?

Answer 1

An angular velocity of the ball is
ω = v/R=3.41/0.211=16.16 rad/s.
The ball is falling down, therefore,
h=gt^2/2.
Then
t=sqr(2h/g)=sqr(2•2.27/9.8)=0.68 s.
The angular displacement of the ball is
φ = ω •t =16.16•0.68 = 11 rad.