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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?

Petro885
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1 Answers1

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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,

$$t=-\frac{v_0\pm\sqrt{v_0^2+2g(h-d)}}{g}$$

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

$$t=\frac{\sqrt{v_0^2+2g(h-d)}-v_0}{g}$$

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 $n=0$.

probably_someone
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