Mathematical Analysis
1. Bottle Specifications
This simulation models a standard 500ml PET bottle with the following parameters:
| Height \(L\) |
22 cm |
| Diameter \(2R\) |
7 cm |
| Empty bottle mass \(m_b\) |
25 g |
| Maximum water mass \(m_{w,\max}\) |
500 g |
| Neck height |
4.5 cm |
2. Initial Velocity Components
Given initial speed \(v_0\) and launch angle \(\theta\) (measured from horizontal), the velocity components are:
\[
v_{x0} = v_0 \cos\theta, \quad v_{y0} = v_0 \sin\theta
\]
3. Projectile Motion
The center of mass follows a parabolic trajectory:
\[
\begin{aligned}
x(t) &= x_0 + v_{x0} t \\
y(t) &= y_0 + v_{y0} t - \frac{1}{2}gt^2
\end{aligned}
\]
The flight time (time to return to initial height) is:
\[
T = \frac{2 v_{y0}}{g} = \frac{2 v_0 \sin\theta}{g}
\]
The maximum height reached is:
\[
h_{\max} = \frac{v_{y0}^2}{2g} = \frac{(v_0 \sin\theta)^2}{2g}
\]
4. Rotational Motion (Solid Body Model)
In the solid body approximation, angular velocity remains constant during flight:
\[
\omega(t) = \omega_0 = \text{constant}
\]
The total rotation after flight time \(T\):
\[
\theta(t) = \theta_0 + \omega_0 t
\]
The final angle determines if the bottle lands upright. Starting from \(\theta_0 = \pi\) (cap down):
\[
\theta_f = \pi + \omega_0 T
\]
For upright landing, we need \(\theta_f \approx \pm\pi\) (mod \(2\pi\)), i.e., \(\omega_0 T \approx 2\pi n\) for integer \(n\).
Key Relationship
For a single flip (\(n=1\)) with flight time \(T = \frac{2v_0\sin\theta}{g}\):
\[
\omega_0 = \frac{2\pi}{T} = \frac{\pi g}{v_0 \sin\theta}
\]
5. Moment of Inertia
The moment of inertia about the center of mass depends on fill level. For a solid cylinder of mass \(m\), radius \(r\), and height \(h\), rotating about a perpendicular axis through its center (Wikipedia):
\[
I = \frac{1}{12}m(3r^2 + h^2)
\]
Using the parallel axis theorem, the total moment of inertia is:
\[
I_{\text{total}} = I_{\text{bottle}} + m_b d_b^2 + I_{\text{water}} + m_w d_w^2
\]
where \(d_b\) and \(d_w\) are distances from each component's center to the system COM.
6. Center of Mass Calculation
For a bottle with water fill ratio \(f\) (0 to 1), the center of mass height from the bottom is:
\[
h_{\text{COM}} = \frac{m_w \cdot \frac{h_w}{2} + m_b \cdot \frac{L}{2}}{m_w + m_b}
\]
where \(m_w = f \cdot m_{\text{water,max}}\) is the water mass, \(h_w = f \cdot L_{\text{body}}\) is water height, and \(m_b\) is the empty bottle mass.
7. Landing Stability Criterion
A successful landing requires:
1. Angle condition: \(\theta_f \approx \pm\pi\) (bottle upright, base down)
2. Stability condition: The bottle tips if rotational kinetic energy exceeds the potential energy to raise COM over the base edge:
\[
\frac{1}{2}I\omega^2 < Mg\left(\sqrt{R^2 + h_{\text{COM}}^2} - h_{\text{COM}}\right)
\]
Solving for the maximum stable angular velocity:
\[
\omega_{\text{max}} = \sqrt{\frac{2Mg\left(\sqrt{R^2 + h_{\text{COM}}^2} - h_{\text{COM}}\right)}{I}}
\]
Computed values for this bottle (with \(g = 9.8\) m/s²):
| Fill % |
\(\omega_{\max}\) (rad/s) |
| 10% |
7.9 |
| 20% |
10.4 |
| 30% |
11.7 |
| 40% |
12.0 (peak) |
| 50% |
11.6 |
| 60% |
10.7 |
| 70% |
9.6 |
| 80% |
8.5 |
| 90% |
7.5 |
Model Limitations
- This is a solid body model — water is treated as a rigid mass, not a fluid.
- A more realistic model would require solving the Navier-Stokes equations such as for water sloshing inside the bottle.
- Some reports on the Internet and theoretical results predict optimal fill around 20-40% [Dekker et al., Am. J. Phys. 86, 733 (2018)].