Floating Cylinder with Frog Departure & Simple Harmonic Motion
a) What is the total mass of the wood along with the frog?
b) After the frog slowly leaves the wood, only one-third of the wood remains immersed. Calculate the mass of the frog.
c) Calculate $x$, the distance between the water level and the center of the circular end of the wooden piece when the frog is aboard, and after the frog departs.
d) Describe the motion of the wood after the instant the frog departs into the water, and derive the equation for $x(t)$ along with its oscillation frequency.
Step 1: System Geometry & Submerged Volume Calculation
$$A = \pi R^2 = \pi (2)^2 = 4\pi \approx 12.566\text{ cm}^2$$
The total volume of the wooden cylinder is:
$$V_{\text{total}} = A \cdot L = 4\pi \times 20 = 80\pi \approx 251.327\text{ cm}^3$$
When half of the wood is immersed in water, the submerged volume is:
$$V_{\text{sub1}} = \frac{1}{2} V_{\text{total}} = 40\pi \approx 125.664\text{ cm}^3$$
Step 2: Total Mass of Wood and Frog (Part a)
$$F_b = (m_{\text{wood}} + m_{\text{frog}}) g = \rho_w V_{\text{sub1}} g$$
Dividing through by $g$ and substituting $\rho_w = 1.0\text{ g/cm}^3$:
$$m_{\text{total}} = m_{\text{wood}} + m_{\text{frog}} = \rho_w V_{\text{sub1}} = 1.0 \times 40\pi = 40\pi\text{ g} \approx 125.66\text{ g}$$
Thus, the combined mass of the wood and the frog is exactly $40\pi\text{ g} \approx 125.66\text{ g}$.
Step 3: Mass of Wood and Isolated Mass of Frog (Part b)
$$V_{\text{sub2}} = \frac{1}{3} V_{\text{total}} = \frac{80\pi}{3}\text{ cm}^3$$
The wood floats alone in static equilibrium, so its buoyant force balances its own weight:
$$m_{\text{wood}} = \rho_w V_{\text{sub2}} = 1.0 \times \frac{80\pi}{3} = \frac{80\pi}{3}\text{ g} \approx 83.78\text{ g}$$
The density of the wooden piece is:
$$\rho_{\text{wood}} = \frac{m_{\text{wood}}}{V_{\text{total}}} = \frac{1}{3}\text{ g/cm}^3 \approx 0.333\text{ g/cm}^3$$
The mass of the frog is the difference between total initial mass and isolated wood mass:
$$m_{\text{frog}} = m_{\text{total}} - m_{\text{wood}} = 40\pi - \frac{80\pi}{3} = \frac{40\pi}{3}\text{ g} \approx 41.89\text{ g}$$
Step 4: Distance $x$ from Waterline to Cylinder Axis (Part c)
$$x_1 = 0\text{ cm}$$
Case 2 (Without frog): One-third of the cross-sectional area is submerged. Let $\theta$ be the central angle subtended by the submerged circular arc at the cylinder axis. The submerged segment area is given by the formula:
$$A_{\text{sub}} = \frac{1}{2} R^2 (\theta - \sin\theta) = \frac{1}{3} \pi R^2 \implies \theta - \sin\theta = \frac{2\pi}{3} \approx 2.0944\text{ rad}$$
Solving $\theta - \sin\theta = 2.0944$ iteratively gives $\theta \approx 2.6053\text{ rad}$ ($149.28^\circ$).
The center of the circular end lies above the waterline by distance:
$$x_2 = R \cos\left(\frac{\theta}{2}\right) = 2.0 \times \cos(1.3027\text{ rad}) = 2.0 \times \cos(74.64^\circ) = 2.0 \times 0.2649 = 0.53\text{ cm}$$
Thus, the cylinder axis rests $0.53\text{ cm}$ above the water surface.
Step 5: Dynamic Oscillation Analysis & Differential Equation (Part d)
Near equilibrium, the waterline width is $w = 2\sqrt{R^2 - x_2^2} = 2\sqrt{2^2 - 0.53^2} = 2\sqrt{3.719} \approx 3.857\text{ cm}$.
For small vertical displacement $y$, the restoring buoyant force is:
$$F_{\text{net}} = -\rho_w g (w L) y$$
Applying Newton's second law for the wood of mass $m_{\text{wood}}$:
$$m_{\text{wood}} \frac{d^2 y}{dt^2} + (\rho_w g w L) y = 0 \implies \frac{d^2 y}{dt^2} + \omega^2 y = 0$$
where the angular frequency of vertical oscillation is:
$$\omega = \sqrt{\frac{\rho_w g w L}{m_{\text{wood}}}} = \sqrt{\frac{1.0 \times 980 \times 3.857 \times 20}{83.78}} = \sqrt{\frac{75597.2}{83.78}} = \sqrt{902.33} \approx 30.04\text{ rad/s}$$
The natural period of oscillation is:
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{30.04} \approx 0.209\text{ s}$$
With initial conditions $y(0) = -0.53\text{ cm}$ and $\dot{y}(0) = 0$, the displacement as a function of time is:
$$x(t) = 0.53 - 0.53 \cos(30.04 t)\text{ cm}$$
| Volume and total mass derivation with Archimedes equilibrium (Part a) | 6 Marks |
| Isolated wood mass, density calculation, and frog mass calculation (Part b) | 6 Marks |
| Geometric circular segment area transcendental equation and $x$ derivation (Part c) | 8 Marks |
| Equation of motion, restoring stiffness $\rho_w g w L$, and frequency $\omega$ (Part d) | 10 Marks |
In buoyancy oscillations, the restoring force is proportional to the waterline cross-section $w L$. For a cylinder, $w$ depends non-linearly on immersion depth, but for small perturbations, linearization around the static equilibrium position yields clean simple harmonic motion.
m_{\text{total}} = \rho_w \left(\frac{1}{2}\pi R^2 L\right) = 40\pi\text{ g} \approx 125.66\text{ g}\\
m_{\text{frog}} = \frac{40\pi}{3}\text{ g} \approx 41.89\text{ g}\\
\theta - \sin\theta = \frac{2\pi}{3} \implies x_2 = R\cos(\theta/2) \approx 0.53\text{ cm}\\
\omega = \sqrt{\frac{\rho_w g w L}{m_{\text{wood}}}} \approx 30.04\text{ rad/s}, \quad T \approx 0.21\text{ s}