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Cover image for "The Enterprise" amusement park ride
Paul Cochrane 🇪🇺
Paul Cochrane 🇪🇺

Posted on • Originally published at peateasea.de on

"The Enterprise" amusement park ride

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, vminv_\mathrm{min} , 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 FcF_c . The passenger also experiences a force due to gravity (their weight) which is directed vertically downwards towards the earth; this force we denote by FgF_g . 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 vv when Fc=FgF_c = F_g .

The diagram below describes this situation.

Forces acting on passenger at Ferris wheel's highest point

We know that the centripetal force is

Fc=mac F_c = m a_c

where mm is the passenger’s mass and aca_c is the centripetal acceleration. Note that we’ve re-used the symbol FcF_c because the centripetal force is equal in magnitude to the centrifugal force.

The centripetal acceleration is

ac=v2r a_c = \frac{v^2}{r}

where vv is the speed of the wheel’s rim and rr is the wheel’s radius (see also the derivation of aca_c below).

We now substitute the result for aca_c into the equation for FcF_c , which gives

Fc=mv2r F_c = m \frac{v^2}{r}

for the centrifugal force experienced by the passenger.

The passenger has a weight due to gravity of

Fg=mg F_g = m g

where gg is the acceleration due to gravity and mm 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.

Fc=Fg F_c = F_g

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

mvmin2r=mg m \frac{v_\mathrm{min}^2}{r} = m g

where we’ve called the wheel’s velocity vminv_\mathrm{min} 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

vmin2r=g \frac{v_\mathrm{min}^2}{r} = g

solving this for vminv_\mathrm{min} we get

vmin=gr v_\mathrm{min} = \sqrt{g r}

We know that g=9.81 m/s2g = 9.81\ \mathrm{m/s^2} and that r=11 mr = 11\ \mathrm{m} (the radius is half the diameter which is 22 m), we can calculate that

vmin=9.81×11=10.388 m/s=37.397 km/h v_\mathrm{min} = \sqrt{9.81 \times 11} = 10.388\ \mathrm{m/s} = 37.397\ \mathrm{km/h}

where we’ve used the fact that 1 m/s=3.6 km/h1\ \mathrm{m/s} = 3.6\ \mathrm{km/h} .

Put another way, the speed of a gondola needs to be at least 37.397 km/h37.397\ \mathrm{km/h} for the passenger to remain safely seated in their gondola at its highest point.

Derivation of aca_c

Where did the equation for aca_c 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 ri\mathbf{r}_i and with an initial velocity vi\mathbf{v}_i . Remember, velocity is a vector, hence it has a direction, whereas speed is the magnitude of the velocity, i.e. v=v|\mathbf{v}| = v . After a time Δt\Delta t , this point on the wheel’s rim has moved to a final position denoted by the radius vector rf\mathbf{r}_f , and now has a final velocity of vf\mathbf{v}_f (its speed is still vv ). The radius vector thus rotates in a time Δt\Delta t through an angle Δθ\Delta \theta .

The following diagram describes this situation:

Circle showing initial and final radius and velocity vectors tracing out an angle theta in a time t

Note that the radius vector ri\mathbf{r}_i is perpendicular to its corresponding velocity vector vi\mathbf{v}_i . Also, the radius vector rf\mathbf{r}_f is perpendicular to its corresponding velocity vector vf\mathbf{v}_f .

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 ri\mathbf{r}_i to rf\mathbf{r}_f which we’ll call Δr\Delta \mathbf{r} . These vectors form a triangle (see the diagram below).

Since the radius vector ri\mathbf{r}_i rotates to rf\mathbf{r}_f , making a radius change of Δr\Delta \mathbf{r} in a time Δt\Delta t , this gives us the velocity

v=ΔrΔt \mathbf{v} = \frac{\Delta \mathbf{r}}{\Delta t}

The magnitude of this velocity is the speed of a point on the wheel’s rim:

v=v=ΔrΔt v = |\mathbf{v}| = \frac{|\Delta \mathbf{r}|}{\Delta t}

where we only need to take the magnitude of Δr\Delta \mathbf{r} because Δt\Delta t is a scalar.

We can also draw a similar triangle for the velocity vectors. We join vi\mathbf{v}_i to vf\mathbf{v}_f at their bases and then note that there is a vector from the tip of vi\mathbf{v}_i to the tip of vf\mathbf{v}_f . This is the change in the two vectors which we denote as Δv\Delta \mathbf{v} , forming a triangle.

The radius vector triangle and the velocity vector triangle look like this:

Radius vector triangle and velocity vector triangle

Because vi\mathbf{v}_i is perpendicular to ri\mathbf{r}_i and vf\mathbf{v}_f is perpendicular to rf\mathbf{r}_f , we can see that the angle between ri\mathbf{r}_i and rf\mathbf{r}_f , Δθ\Delta \theta , is also the angle between vi\mathbf{v}_i and vf\mathbf{v}_f . 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

Δrri=Δvvi \frac{|\Delta \mathbf{r}|}{|\mathbf{r}_i|} = \frac{|\Delta \mathbf{v}|}{|\mathbf{v}_i|}

which can be slightly more simply written as

Δrr=Δvv \frac{|\Delta \mathbf{r}|}{r} = \frac{|\Delta \mathbf{v}|}{v}

because ri=r\vert\mathbf{r}_i\vert = r and vi=v\vert\mathbf{v}_i\vert = v .

We can rearrange this equation for Δv\vert\Delta \mathbf{v}\vert like so

Δv=Δrvr |\Delta \mathbf{v}| = \frac{|\Delta \mathbf{r}| v}{r}

We know that acceleration is the change in the velocity vector divided by the change in time

a=ΔvΔt a = \frac{|\Delta \mathbf{v}|}{\Delta t}

hence we can divide both sides of the equation for Δv|\Delta \mathbf{v}| by Δt\Delta t to obtain an equation for the acceleration:

a=ΔrΔtvr a = \frac{|\Delta \mathbf{r}|}{\Delta t} \frac{v}{r}

We also know that the speed vv is the change in the radius vector divided by the change in time, i.e.

v=ΔrΔt v = \frac{|\Delta \mathbf{r}|}{\Delta t}

Substituting this into our equation for the acceleration, aa , we get

a=vvr=v2r a = v \frac{v}{r} = \frac{v^2}{r}

hence we obtain the equation for aca_c mentioned earlier:

ac=v2r a_c = \frac{v^2}{r}

Derivation using calculus

Let’s now consider the situation of a point on the wheel’s rim denoted by the radius vector r\mathbf{r} as described by the following diagram.

Radius vector with x and y components on a circle

The radius vector r\mathbf{r} can be split up into its xx and yy components like so:

r=rxi+ryj \mathbf{r} = r_x \mathbf{i} + r_y \mathbf{j}

where rxr_x is the length of the radius vector along the xx axis, i\mathbf{i} is the unit vector pointing in the xx direction, ryr_y is the magnitude of the radius vector along the yy axis, and j\mathbf{j} is the unit vector pointing in the yy direction.

Let’s denote the angle of the radius vector to the xx axis by the variable θ\theta . We can therefore rewrite the above equation as:

r=rcosθi+rsinθj \mathbf{r} = r \cos \theta \mathbf{i} + r \sin \theta \mathbf{j}

where rr 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 θ\theta within a given time tt . Thus we can write the angular velocity of the wheel like so:

ω=θt \omega = \frac{\theta}{t}

Rearranging this equation for θ\theta we have

θ=ωt \theta = \omega t

Substituting this into our equation for r\mathbf{r} , we have

r=rcos(ωt)i+rsin(ωt)j \mathbf{r} = r \cos (\omega t) \mathbf{i} + r \sin (\omega t) \mathbf{j}

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

v=drdt=rωsin(ωt)i+rωcos(ωt)j \mathbf{v} = \frac{d \mathbf{r}}{d t} = -r \omega \sin (\omega t) \mathbf{i} + r \omega \cos (\omega t) \mathbf{j}

To get this result, we’ve used the property that

ddθsinθ=cosθ \frac{d}{d \theta} \sin \theta = \cos \theta

and

ddθcosθ=sinθ \frac{d}{d \theta} \cos \theta = -\sin \theta

as well as the chain rule.

We can arrive at the acceleration similarly, by differentiating the velocity with respect to time:

a=dvdt=rω2cos(ωt)i+rω2sin(ωt)j \mathbf{a} = \frac{d \mathbf{v}}{d t} = -r \omega^2 \cos (\omega t) \mathbf{i} + r \omega^2 \sin (\omega t) \mathbf{j}

This is our centripetal acceleration.

However, we’re only interested in the magnitude of this vector, a|\mathbf{a}| , which is the square root of the sum of the squares of its components, i.e.

ac=a=r2ω4cos2(ωt)+r2ω4sin2(ωt) a_c = |\mathbf{a}| = \sqrt{r^2 \omega^4 \cos^2 (\omega t) + r^2 \omega^4 \sin^2 (\omega t)}

We can extract the common factor of r2ω4r^2 \omega^4 out of the square root to give

ac=rω2cos2(ωt)+sin2(ωt) a_c = r \omega^2 \sqrt{\cos^2 (\omega t) + \sin^2 (\omega t)}

We know from the trigonometric identity that

cos2θ+sin2θ=1 \cos^2 \theta + \sin^2 \theta = 1

hence the expression under the square root is equal to 1, i.e.:

cos2(ωt)+sin2(ωt)=1 \cos^2 (\omega t) + \sin^2 (\omega t) = 1

which means we can simplify our equation for aca_c further to

ac=rω2 a_c = r \omega^2

We can write the angular velocity in terms of the tangential velocity like so:

ω=vr \omega = \frac{v}{r}

Substituting this value into the above expression for aca_c we get

ac=rv2r2=v2r a_c = r \frac{v^2}{r^2} = \frac{v^2}{r}

which is the result we got by using the geometric argument above.

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