I am pretty sure that you know about the bottle flip challenge that has gone viral last year. I am trying to take it further to challenge my high school friends that it can be calculated, even theoretically, if the bottle flip will be successful or not. So, I am asking you to help me find the necessary calculations. I thought of projectile motion as if the initial velocity and the launching angle are known, we can't calculate where the bottle will land. But I am pretty sure the rotation of the bottle affects the calculations as well; I don't know how though. May be calculating average revolutions made by the bottle per second can help if it will land on its bottom.

My question is: what are the calculations needed to know if a bottle flip is successful?

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    $\begingroup$ Is the bottle completely full, completely empty, or somewhere in between? The latter case is much more complicated than the first two. $\endgroup$ – probably_someone Feb 5 '18 at 18:46
  • $\begingroup$ Let's assume it's completely full $\endgroup$ – Petro885 Feb 5 '18 at 19:00
  • $\begingroup$ Possible duplicate of Flipping a bottle, or The physics of bottle flipping $\endgroup$ – sammy gerbil Feb 5 '18 at 19:53
  • $\begingroup$ The easiest flip to do is with the bottle about 1/3 full of water. The sloshing of water is key to success. Flipping a full bottle is likely to fail. $\endgroup$ – sammy gerbil Feb 5 '18 at 20:01

Let's assume the bottle is completely full (i.e. it acts as a rigid body and essentially ignores air resistance), and that it is initially held horizontally, by the cap, and thrown straight upward. Then you'll need four things:

  • The initial height $h$ above the landing surface,

  • The distance $d$ between the bottom of the bottle and its center of mass,

  • The initial vertical velocity $v_0$, and

  • The initial angular velocity $\omega$ about the bottle's center of mass.

Since air resistance is ignored, there is no torque on the bottle while in the air; as such, it rotates with constant angular velocity $\omega$ the whole time. Requiring that the bottle lands upright is equivalent to saying that the total angle rotated in flight is equivalent to $3\pi/2$ (since the bottle, when flipped, rotates in a manner such that the cap is the first end to face the ground). This constraint is equivalent to:

$$\omega t=2\pi(n+3/4)$$

for flight time $t$ and some integer $n\geq0$. Now, since the bottle is a rigid body, gravity acts as if it were a point mass. As such, the flight time can be calculated from the normal kinematic formula:

$$h+v_0 t-\frac{1}{2}gt^2=d$$

taking the height of the landing surface itself to be 0. Using the quadratic formula,


Since we require a positive time, we must choose the minus sign in the $\pm$, making it


Plugging this into our earlier constraint, we see that for a given $v_0$ and $h$,

$$\omega=\frac{2\pi g}{\sqrt{v_0^2+2g(h-d)}-v_0}(n+3/4)$$

so there are a multitude of possible angular velocities that will guarantee a landing. For the minimum angular velocity, set $n=0$.

  • $\begingroup$ In reality, the bottle may land correctly for a range of $\omega$ around these values, as small uncertainties will correct themselves as the bottle naturally rights itself. The particular range of $\omega$ for which this is true depends on the particular shape and size of the bottle; namely, each bottle has a tipping angle $\theta_t$, which is the angle at which its center of mass passes over its corner. Then, the constraint becomes $|\omega t-2\pi(n+3/4)|<\theta_t$. $\endgroup$ – probably_someone Feb 5 '18 at 19:23
  • $\begingroup$ Knowing whether the base is (close to) horizontal on landing is only part of the solution. You also need to know if the bottle will topple when it lands, due to its angular momentum. $\endgroup$ – sammy gerbil Feb 5 '18 at 20:07
  • $\begingroup$ @sammygerbil Nevertheless, this is accurate for low $\omega$, which is usually the case in bottle flips (in most of the YouTube videos, you rarely see them rotate around more than $3\pi/2$). $\endgroup$ – probably_someone Feb 5 '18 at 20:19
  • $\begingroup$ Low speed of rotation seems obvious to me as a condition for success. Also the CM should have low speed on landing, to minimise the rebound. This suggests throwing up onto a ledge above launch point is best. Then $v_0$ can be calculated from $h$. $\endgroup$ – sammy gerbil Feb 5 '18 at 20:40

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