Originally published on peateasea.de.
As part of tutoring physics and maths to high school students, I sometimes write up deep-dive explanations of questions arising during lessons. Below I discuss the question of how fast an amusement park ride should rotate so that the passengers don’t fall out when the gondola is at its highest point.
Question
The Enterprise is a ride sometimes found at carnivals and amusement parks. It consists of a large wheel-like structure with gondolas attached to the wheel’s rim, like a Ferris wheel. The wheel initially starts in the horizontal (the plane of the wheel being parallel to the plane of the ground). After the wheel reaches a certain rotational speed, it rotates to the vertical (the plane of the wheel being perpendicular to the plane of the ground). This means that the gondolas at the top of the wheel are upside down (see, for instance, the image above).
Consider such an amusement ride with a diameter of 22m. Calculate both algebraically and numerically the minimum speed, , that this ride must have so that a passenger in a gondola at the wheel’s highest point does not fall onto the gondola’s roof.
Answer
The force keeping the passengers on a circular path is the centripetal force. However, a passenger in a gondola experiences the feeling of being pushed outwards, away from the wheel’s hub, and into the floor of the gondola. The passengers experience the centrifugal force. This is a fictitious force experienced by objects within rotating systems and acts radially away from the axis of rotation (i.e. outwards, away from the wheel’s hub).
Since we wish to consider the experience of a passenger in one of the gondolas (i.e. in one of the rotating frames of reference), it is useful (and relevant) to consider the magnitude of the centrifugal force, which we will denote by . The passenger also experiences a force due to gravity (their weight) which is directed vertically downwards towards the earth; this force we denote by . When the magnitude of the centrifugal force is equal to the weight of the passenger then they will remain in their gondola at the wheel’s highest point and thus won’t fall onto the gondola’s roof. In other words, we need to find the speed when .
The diagram below describes this situation.
We know that the centripetal force is
where is the passenger’s mass and is the centripetal acceleration. Note that we’ve re-used the symbol because the centripetal force is equal in magnitude to the centrifugal force.
The centripetal acceleration is
where is the speed of the wheel’s rim and is the wheel’s radius (see also the derivation of below).
We now substitute the result for into the equation for , which gives
for the centrifugal force experienced by the passenger.
The passenger has a weight due to gravity of
where is the acceleration due to gravity and is the passenger’s mass as before.
For the passenger to remain on the floor of the gondola at the wheel’s highest point, the centrifugal force must be at least equal to the passenger’s weight, i.e.
Substituting the equation for the passenger’s weight as well as the equation for the centrifugal force in terms of the wheel’s velocity into the above equation we get
where we’ve called the wheel’s velocity because this equation represents the situation where the centrifugal force exactly balances the weight force. The force is not too big or too small and hence it’s the minimum velocity required for the passenger not to fall onto the gondola’s roof at the wheel’s highest point.
We can cancel the mass since it appears on both sides of the equation, giving
solving this for we get
We know that and that (the radius is half the diameter which is 22 m), we can calculate that
where we’ve used the fact that .
Put another way, the speed of a gondola needs to be at least for the passenger to remain safely seated in their gondola at its highest point.
Derivation of
Where did the equation for come from? There are two ways to explain this and Physics Ninja explains the calculation particularly well. I’m going to reproduce his explanations here.
Derivation via a geometric argument
A point on the wheel’s rim starts at an initial position denoted by the radius vector and with an initial velocity . Remember, velocity is a vector, hence it has a direction, whereas speed is the magnitude of the velocity, i.e. . After a time , this point on the wheel’s rim has moved to a final position denoted by the radius vector , and now has a final velocity of (its speed is still ). The radius vector thus rotates in a time through an angle .
The following diagram describes this situation:
Note that the radius vector is perpendicular to its corresponding velocity vector . Also, the radius vector is perpendicular to its corresponding velocity vector .
If we draw the radius vectors with their bases starting from the same point, we see that there’s a vector from the tip of to which we’ll call . These vectors form a triangle (see the diagram below).
Since the radius vector rotates to , making a radius change of in a time , this gives us the velocity
The magnitude of this velocity is the speed of a point on the wheel’s rim:
where we only need to take the magnitude of because is a scalar.
We can also draw a similar triangle for the velocity vectors. We join to at their bases and then note that there is a vector from the tip of to the tip of . This is the change in the two vectors which we denote as , forming a triangle.
The radius vector triangle and the velocity vector triangle look like this:
Because is perpendicular to and is perpendicular to , we can see that the angle between and , , is also the angle between and . Hence the triangle for the velocity vectors is similar to the triangle for the radius vectors. Because these two triangles are similar (and isosceles), the ratio of the shortest side to one of the long sides for the radius vector triangle will be equal to the ratio of the shortest side to one of the long sides for the velocity vector triangle. In other words
which can be slightly more simply written as
because and .
We can rearrange this equation for like so
We know that acceleration is the change in the velocity vector divided by the change in time
hence we can divide both sides of the equation for by to obtain an equation for the acceleration:
We also know that the speed is the change in the radius vector divided by the change in time, i.e.
Substituting this into our equation for the acceleration, , we get
hence we obtain the equation for mentioned earlier:
Derivation using calculus
Let’s now consider the situation of a point on the wheel’s rim denoted by the radius vector as described by the following diagram.
The radius vector can be split up into its and components like so:
where is the length of the radius vector along the axis, is the unit vector pointing in the direction, is the magnitude of the radius vector along the axis, and is the unit vector pointing in the direction.
Let’s denote the angle of the radius vector to the axis by the variable . We can therefore rewrite the above equation as:
where is the radius of the wheel and the magnitude of the radius vector.
We know that the wheel is rotating at a constant speed, hence the radius vector will rotate through a constant angle within a given time . Thus we can write the angular velocity of the wheel like so:
Rearranging this equation for we have
Substituting this into our equation for , we have
We know that velocity is the time derivative of displacement, hence we can find the velocity vector by differentiating the equation for the radius vector
To get this result, we’ve used the property that
and
as well as the chain rule.
We can arrive at the acceleration similarly, by differentiating the velocity with respect to time:
This is our centripetal acceleration.
However, we’re only interested in the magnitude of this vector, , which is the square root of the sum of the squares of its components, i.e.
We can extract the common factor of out of the square root to give
We know from the trigonometric identity that
hence the expression under the square root is equal to 1, i.e.:
which means we can simplify our equation for further to
We can write the angular velocity in terms of the tangential velocity like so:
Substituting this value into the above expression for we get
which is the result we got by using the geometric argument above.
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