Suppose the surface (radius = r) of the space station in the figure is rotating at 38.6 m/s. What must be the value of r for the astronauts to weigh one-half of their earth-weight?
To determine the value of r for the astronauts to weigh one-half of their earth-weight on the rotating space station, we can use the concept of centrifugal force.
Centrifugal force is the apparent outward force experienced by objects in a rotating frame of reference. It is the force that pushes objects away from the center of rotation.
In this case, we can equate the centrifugal force with the force of gravity to find the value of r.
The formula for centrifugal force is given by:
Fc = m * ω^2 * r
Where Fc is the centrifugal force, m is the mass of the object, ω is the angular velocity (38.6 m/s in this case), and r is the radius of the space station.
The formula for gravitational force is given by:
Fg = m * g
Where Fg is the gravitational force, m is the mass of the object, and g is the acceleration due to gravity on Earth (approximately 9.8 m/s^2).
Since we want the astronauts to weigh one-half of their earth-weight, we can equate the centrifugal force to one-half of the gravitational force:
Fc = (1/2) * Fg
Substituting the formulas for gravitational and centrifugal forces, we get:
m * ω^2 * r = (1/2) * m * g
The mass (m) cancels out from both sides, so we're left with:
ω^2 * r = (1/2) * g
Now, plug in the values:
(38.6 m/s)^2 * r = (1/2) * (9.8 m/s^2)
Simplifying the equation:
r = (1/2) * (9.8 m/s^2) / (38.6 m/s)^2
Calculating the result gives us:
r ≈ 0.125 meters
Therefore, the value of r for the astronauts to weigh one-half of their earth-weight on the rotating space station is approximately 0.125 meters.