A ball of radius 0.2m rolls along a horizontal table top with a constant linear speed of 3.04 m/s. The ball rolls off the edge and falls a vertical distance of 2.02m before hitting the floor. What is the angular displacement of the ball while it is in the air.
Angular spin rate (w) remains the same after falling off the edge. Multiply w by the time it takes to fall 2.02 m.
w = V/R
Time to fall H = sqrt(2H/g)
= 0.642 s when H = 2.02 m
To find the angular displacement of the ball while it is in the air, we need to consider the relationship between linear motion and rotational motion.
The linear speed of the ball can be related to its angular speed by the formula:
v = ω * r
where:
v is the linear speed of the ball,
ω (omega) is the angular speed of the ball, and
r is the radius of the ball.
In this case, we know the linear speed (v = 3.04 m/s) and the radius of the ball (r = 0.2 m). We can rearrange the formula to solve for the angular speed:
ω = v / r
ω = 3.04 m/s / 0.2 m
ω ≈ 15.2 rad/s
Now we can find the time it takes for the ball to fall a distance of 2.02 m. In free fall, the vertical displacement of an object can be related to time using the formula:
y = (1/2) * g * t^2
where:
y is the vertical displacement,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time.
In this case, we know the vertical displacement (y = 2.02 m) and the acceleration due to gravity (g = 9.8 m/s^2). We can rearrange the formula to solve for the time:
t^2 = (2 * y) / g
t^2 = (2 * 2.02 m) / 9.8 m/s^2
t^2 ≈ 0.4122 s^2
Taking the square root of both sides, we find:
t ≈ 0.641 s
Now we can find the angular displacement of the ball in the air. The formula relating angular displacement, angular speed, and time is:
θ = ω * t
where:
θ is the angular displacement,
ω (omega) is the angular speed, and
t is the time.
Plugging in the values we found:
θ = 15.2 rad/s * 0.641 s
θ ≈ 9.75 radians
Therefore, the angular displacement of the ball while it is in the air is approximately 9.75 radians.