In an exhibit at the NASA museum in Huntsville, AL, there is a “human centrifuge”. Guests
stand upright against the walls of a circular room with a radius of 8 meters which then spins
at an angular velocity ω about an axis through its center. Then, once the room is spinning,
the floor drops away from the guests’ feet. They feel themselves pressed against the walls,
but do not fall.
a) Draw a force diagram for the person in this exhibit. Make sure you label the axes in your
force diagram to indicate which direction is “up”, and which direction points toward the
center of the circular room.
b) Suppose the coefficient of static friction between the guests’ clothing and the wall is 0.4.
At what angular velocity ω must the room spin to ensure that the guests do not fall?
c) Guests in this exhibit feel themselves “pressed against the walls”, and find it very difficult
to lift their arms away from the wall; for instance, it may require a force of 25 N to hold
a one-kilogram object in place away from the wall. Is there a force pressing the guests and
their hands against the wall? If so, what is that force? If not, how do you explain the
difficulty lifting objects away from the wall?
a) To draw a force diagram for the person in this exhibit, we need to consider the forces acting on the person.
1. The weight of the person acts vertically downward towards the center of the Earth.
2. The normal force acts perpendicular to the surface of the wall and opposes the weight.
3. The static friction force acts horizontally towards the center of the circular room and opposes the tendency of the person to slide down.
In the force diagram, the vertical axis can be labeled as "up" and the axis pointing towards the center of the circular room can be labeled as "center".
b) To determine the angular velocity ω at which the guests do not fall, we need to consider the maximum static friction force that can prevent the person from sliding down. The maximum static friction force is given by the equation:
friction = coefficient of static friction * normal force
In this case, the normal force is equal to the weight of the person. So, the equation becomes:
friction = coefficient of static friction * weight
Since the static friction force is responsible for providing the centripetal force required to keep the person moving in a circular path, we can equate it to the centripetal force:
friction = m * (ω^2 * r)
Here, m is the mass of the person and r is the radius of the circular room.
Setting the two equations equal to each other gives us:
coefficient of static friction * weight = m * (ω^2 * r)
We can solve this equation for ω by rearranging:
ω^2 = (coefficient of static friction * weight) / (m * r)
ω = sqrt((coefficient of static friction * weight) / (m * r))
Substituting the given values of the coefficient of static friction (0.4) and the radius (8 meters), and assuming an average weight of a person, we can calculate the value of ω.
c) The difficulty in lifting objects away from the wall is due to the centripetal force acting on the objects. When the room is spinning, there is a centripetal force that pushes everything towards the center of the circular room, including the guests and any objects they are holding.
The force that presses the guests and their hands against the wall is the static friction force. This force provides the necessary centripetal force to keep the guests in a circular path and prevents them from falling.
The force needed to hold a one-kilogram object in place away from the wall (25 N in this case) is also a result of the static friction force. If the object is not held firmly, the static friction force will not be able to provide enough centripetal force and the object will slide down towards the center of the circular room.
Therefore, the difficulty in lifting objects away from the wall is due to the opposing force provided by static friction, which is required to maintain the circular motion of the guests.