A ball of radius 0.300 m rolls along a horizontal table top with a constant linear speed of 3.30 m/s. The ball rolls off the edge and falls a vertical distance of 2.50 m before hitting the floor. What is the angular displacement of the ball while it is in the air? (All problems in this section assume that there is no slipping of the surfaces in contact during the rolling motion.)
Multiply the rotation rate of the ball, which is (Linear speed)/R, by the time "t" that it takes the ball to hit the ground. Use Y = (1/2) g t^2 for that, and sove for t. Your answer will be in radians.
To find the angular displacement of the ball while it is in the air, we need to calculate the time it takes for the ball to hit the ground.
Step 1: Calculate the time "t" using the equation Y = (1/2) g t^2 for vertical motion, where Y is the vertical distance fallen, g is the acceleration due to gravity, and t is the time.
Given:
Vertical distance fallen, Y = 2.50 m
Acceleration due to gravity, g = 9.8 m/s^2
Using the equation, we can rearrange it to solve for time:
2.50 m = (1/2) * (9.8 m/s^2) * t^2
5.00 = 4.9 t^2
t^2 = 5.00 / 4.9
t^2 ≈ 1.02
t ≈ √(1.02)
t ≈ 1.01 s
Step 2: Calculate the angular displacement using the rotation rate of the ball.
The rotation rate of the ball is given by (Linear speed) / R, where R is the radius of the ball.
Given:
Linear speed of the ball, v = 3.30 m/s
Radius of the ball, R = 0.300 m
Rotation rate = v / R = (3.30 m/s) / (0.300 m) = 11.00 rad/s
Step 3: Multiply the rotation rate by the time "t" to find the angular displacement.
Angular displacement = rotation rate * t = 11.00 rad/s * 1.01 s = 11.11 radians
Therefore, the angular displacement of the ball while it is in the air is approximately 11.11 radians.