A Ferris wheel has a radius of 14 m. What period of rotation would give the occupants a feeling of "weightlessness" at the top?
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Require that R*w^2 = g
Solve for the required anguloar velocity w, in radians/s
w = sqrt(g/R) = 0.8367 rad/s
Period = 2*pi/w = 5.26 s
To determine the period of rotation that would give the occupants a feeling of "weightlessness" at the top of the Ferris wheel, we can use the concept of centripetal acceleration.
The centripetal acceleration is given by the formula:
a = v^2 / r
Where:
a is the centripetal acceleration
v is the linear velocity
r is the radius of the Ferris wheel
At the top of the Ferris wheel, the occupants would experience a feeling of "weightlessness" when the centripetal acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.
Since weightlessness occurs when a = g, we can set up the equation:
v^2 / r = g
Rearranging the equation to solve for v, we get:
v = √(r * g)
We can substitute the given values into the equation:
v = √(14 * 9.8)
v ≈ 13.89 m/s
The period of rotation (T) is related to the linear velocity by the equation:
v = 2 * π * r / T
Rearranging the equation to solve for T, we get:
T = 2 * π * r / v
Substituting the values, we have:
T = 2 * π * 14 / 13.89
Calculating the value, we find:
T ≈ 6.37 seconds
Therefore, a period of rotation of approximately 6.37 seconds would give the occupants a feeling of "weightlessness" at the top of the Ferris wheel.