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?
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.
To find the angular displacement of the ball while it is in the air, we can use the relationship between linear and angular motion.
The linear speed, v, is given as 3.41 m/s, and the radius of the ball, r, is given as 0.211 m. The linear speed is related to the angular speed, ω, by the equation v = rω.
We can find the angular speed, ω, by rearranging the equation as ω = v/r. Substituting the given values, we get ω = 3.41 m/s / 0.211 m.
Now, we need to find the time, t, that the ball is in the air. We can use the equation s = ut + 0.5at^2, where s is the vertical distance, u is the initial vertical velocity (which is 0 since the ball rolls off horizontally), and a is the acceleration due to gravity (which is approximately 9.8 m/s^2). Solving for t, we get t = sqrt(2s/a).
Substituting the given values, we get t = sqrt(2 * 2.27 m / 9.8 m/s^2).
Finally, we can find the angular displacement, θ, by multiplying the angular speed, ω, by the time, t. θ = ω * t.
Substituting the calculated values, we get θ = (3.41 m/s / 0.211 m) * sqrt(2 * 2.27 m / 9.8 m/s^2).
Evaluating this expression gives us the final answer for the angular displacement of the ball while it is in the air.