Computation of Phase-Transition Driving Velocity Resulting from Catastrophic Instability of Mountain Dew Due to Centripetal Force

Clifford B. Miller

28 August 1998

We compute the ``phase-transition driving velocity" v, that is, the velocity at which the driver throws a hairy fit because his can of Mountain Dew $^{\mbox{\tiny TM}}$ has tipped over, while rounding a curve of radius r_d. We assume that the can rests on a flat surface, inside the car (otherwise wind shear would have to be accounted for), and that the frictional force between the can and the surface is sufficient to keep the can from sliding, which will in general be the case as numerous Mountain Dews will have spilled there a priori before because the driver did not know how to accurately compute the phase-transition driving velocity.

We make the general observation that $\tau_+$, the torque of gravity acting on the can to keep it from tipping, must always be greater than $\tau_-$, the torque of centripetal force that occurs while rounding a curve. The critical point, occurring just before a spill, is expressed as

$$\begin{eqnarray*}\tau_+ &=& \tau_- \\ x F_z &=& z F_x \end{eqnarray*}$$

where the torque fulcrum lies at the lower corner of the can, and

• z is the vertical distance from the surface to the can/soda system's center of mass (the ``moment arm" with which centripetal force generates $\tau_-$),
• x is the horizontal distance from the edge of the can to that same center of mass (the moment arm with which gravity generates $\tau_+$),
• F_z is the force of gravity, and
• F_x is the centripetal force occurring as the car turns.

From basic physics we know that F_z = mg and F_x = mv²/r_d, where

• m is the mass of the (can + soda) system,
• g is the gravitational constant,
• v is the linear velocity of the car, and
• r_d is the radius of curvature of the car's path.

The complex part of the calculation is the length of the two moment arms x and z. A naïve approach might lead us to substitute

$$\begin{eqnarray*}x &=& r_c, \\ z &=& \hat{h}/2 \end{eqnarray*}$$

where r_c is the radius of the can, and $\hat{h} \equiv fh$ is the level of the soda in the can (remembering that the soda may be partially drunk and therefore not reach the full height of the can). Here h is the can's height and f is its fractional fullness (f=0 is an empty can, and f=1 is a full can).

However, we need to consider that the can itself has mass independent of the soda within, and therefore that the height of the center of mass is a more complex expression (we assume for now that the can is radially symmetric, as is the soda -- otherwise ugh!).

The general expression for the center of mass is

$${\bf r} = \frac {\sum_i m_i {\bf r}_i} {\sum_i m_i}$$

where the sum is over objects (in this case, can and soda), m_i is the mass of object i, and ${\bf r}_i$ is the location of the center of mass of object i. With our assumption of radial symmetry we can state that the centers of mass both lie on the central axis of the can/soda system, and thus express it as ${\bf r}_i = z_i$. Now we know that

$$\begin{eqnarray*}m_1 &=& c \\ z_1 &=& h/2 \\ m_2 &=& fs \\ z_2 &=& fh/2 \end{eqnarray*}$$

where c is the mass of the can (empty) and s is the mass (tare weight) of the (full) soda. Thus we have

$$z(f) = {h \over 2} \left( \frac {c + f^2 s} {c + fs} \right)$$

which shows that the height of the center of mass of the can/soda system is a function of how full the can is.

Substituting these values into the original equation, we have

$$\begin{eqnarray*}x F_z &=& z F_x \\ r_c \cdot m g &=& z(f) \cdot {mv^2 \over r_d} \end{eqnarray*}$$

or, rearranging,

$$\begin{eqnarray*}v &=& \sqrt{g r_c r_d \over z(f)} \\ &=& \sqrt{g r_c r_d \, {2 \over h} \frac{c+fs}{c+f^2s} } \end{eqnarray*}$$

Now this is one of those nice, neat equations that we love to see from our work in physics. And we could let the matter rest there, substituting in actual values from the actual world to arrive at some hard numbers, and then go into the wet lab (so to speak) with confidence. But alas, our experiments would go awry, because we neglected a very important subtlety of the problem.

As the car is going around the curve, the centripetal force acts on the can and soda. However, the soda does not stay fixed in place as this happens -- it is a fluid, and so it is pushed toward one side of the can. This results in a shift in the center of mass -- not only horizontally, but vertically as well, as shown by the following calculations.

The shape of the soda is now no longer a cylinder, but a cylinder with an angled top. Since a contained liquid seeks its own level perpendicular to the direction of force acting on it, we can determine the exact geometry of the soda. If the soda is in equilibrium (for now assume the car is not changing velocity, and is turning at a constant rate), the top of the soda will be a planar surface, tilted at an angle $\alpha$ by centripetal force such that

$$\tan \alpha = \frac{F_x}{F_z} = \frac{v^2/r_d}{g}$$

This raises the level of the liquid (at the edge of the can furthest from the turn's center) by an amount

$$a = r \tan \alpha$$

above the equilibrium (non-turning) height of the liquid, fh. A similar but opposite displacement occurs on the other side of the can. Thus an expression for the height of the liquid as a function of position in the can is derived as:

$$z(r,\theta) = fh + a \,{r \over r_c} \cos \theta$$

where we position the can's bottom center at the origin of a cylindrical coordinate system $(r,\theta,z)$ and $\theta = 0$ represents the edge of the can where the liquid is highest (z=fh+a).

As an exercise we'll show the actual computation of the center of mass of the can/soda system with this new geometry.

We can turn the discrete center-of-mass equation into a continuous one, in order to compute the contribution of the soda to the center of mass, with the equation

$${\bf r}_s = \frac {\int \rho {\bf r} \; dV} {\int \rho \; dV}$$

where the integrals are over the volume V of the soda, and $\rho$ represents the density of the soda at position r. Luckily the soda is uniform (bleah!) and so $\rho$ is a constant that can be factored out of the equation altogether:

$${\bf r}_s = \frac {\int {\bf r} \; dV} {\int \; dV}$$

The denominator of this equation is simply the volume of the soda itself, which we already know to be $\pi r_c^2 fh$ (since sloshing of the soda does not change its volume). The numerator is more complicated:

$$\begin{eqnarray*}& & \int {\bf r} \; dV \\ &=& \int_0^{r_c} \int_0^{2\pi} \int... ... r\cos\theta, r\sin\theta ,z \rangle \; r \; dr \; d\theta \; dz \end{eqnarray*}$$

where we use the $\langle x,y,z \rangle$ notation to denote the decomposition of the vector r into its cartesian components (expressed in cylindrical coordinates). This is in effect three volume integrals, one for each component of the center-of-mass vector. Now, we can realize through symmetry that the y component of the center of mass vector is 0, since gravity and centripetal force are acting in the xz plane only. Thus the remaining components are computed as follows.

$$\begin{eqnarray*}X &=& \int_0^{r_c} \int_0^{2\pi} \int_0^{z(r,\theta)} (r \cos\... ...right\vert _0^{r_c} \cdot \pi \\ &=& {a \pi \over 4} \; r_c^3 \end{eqnarray*}$$

$$\begin{eqnarray*}Z &=& \int_0^{r_c} \int_0^{2\pi} \int_0^{z(r,\theta)} (z) \; r... ...=& \left( {f^2 h^2 \over 2} + {a^2 \over 8} \right) \; \pi r_c^2 \end{eqnarray*}$$

Thus we may express the combined vector for the soda's center of mass as

$$\begin{eqnarray*}{\bf r}_s &=& \frac {\int {\bf r} \; dV} {\int \; dV} \\ &=& \... ...t< {ar_c \over 4fh} , 0 , {fh \over 2} + {a^2 \over 8fh} \right> \end{eqnarray*}$$

This equation shows that the ``sloshing effect" causes the center-of-mass of the soda to move up slightly (a²/8fh) from its equilibrium position fh/2, and also to move slightly off-axis (the non-zero x component of ar_c/4fh). We see that if a=0, i.e. no centripetal force is acting on the system, these perturbations vanish and the center of mass returns to its cylindrically-centered position of z=fh/2.

Now we may compute the center-of-mass of the entire can/soda system:

$$\begin{eqnarray*}{\bf r}_{cs} &=& \frac { m_c {\bf r}_c + m_s {\bf r}_s } { m_... ...ver 2h^2} , 0 , c + f^2 s + {sa^2 \over 4h^2} \right> \\ &=& \end{eqnarray*}$$

We can use this equation to directly express the moment arms x and z (note that the moment arm x is computed from the edge of the can, not the center, hence its form differs slightly from that of the x component of the can/soda system center-of-mass):

$$\begin{eqnarray*}x &=& r_c - {1 \over c+fs} {h \over 2} {sar_c \over 2h^2} = {1... ...c+fs} {h \over 2} \left( c + f^2 s + {sa^2 \over 4h^2} \right) \end{eqnarray*}$$

And finally, rearranging our original equation for the can's equilibrium,

$$\begin{eqnarray*}x F_z &=& z F_x \\ {F_z \over F_x} &=& {z \over x} \end{eqnarray*}$$

Substituting in our derived values for F_z, F_x, z, and x, and factoring 1/(c+fs) out of both z and x, we get

$${g \over v^2/ r_d} = {h \over 2r_c} \frac { c + f^2 s + {sa^2 \over 4h^2} } { c+fs - {sa \over 4h} }$$

Inverting,

$$\begin{eqnarray*}{v^2/r_d \over g} &=& {2r_c \over h} \frac { c+fs - {sa \ove... ... { c+fs - {sa \over 4h} } { c + f^2 s + {sa^2 \over 4h^2} } } \end{eqnarray*}$$

We now leave it as an exercise to the reader to realize that a, the ``extra height" attained by the soda due to the ``sloshing effect", is actually dependent on v²/r_d, and thus that the above equation becomes a cubic in v. Derive a closed-form expression for v. (But before you begin, drink another can of Mountain Dew.)

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