Problem Solution: Stability Range for a Floating Glass Tube
Physical Analysis: We need to determine the range of water height $h$ inside a thin glass tube such that it floats vertically in stable equilibrium. This involves two conditions: 1. Floatation: The tube must not sink. 2. Stability: The system must not tip over. The problem explicitly hints that for stability, the center of mass of the system (tube + water inside) must be lower than the center of buoyancy (center of displaced water).
Part 1: The Floatation Condition (Upper Limit)
Let the length of the tube be $l$, mass $m$, and cross-section $S$. The water density is $\rho_0$. The height of water inside is $h$.
The total mass of the system is: $$ M_{\text{sys}} = m_{\text{tube}} + m_{\text{water}} = m + \rho_0 S h $$ For the tube to float, the buoyant force must balance the weight. Let $x$ be the submerged depth of the tube. $$ \rho_0 S x g = (m + \rho_0 S h) g $$ $$ x = \frac{m}{\rho_0 S} + h $$ For the tube not to sink completely, the submerged depth $x$ must be less than the length of the tube $l$ (assuming negligible wall thickness/bottom curvature for height limit). $$ x < l \implies \frac{m}{\rho_0 S} + h < l $$
Part 2: The Stability Condition (Lower Limit)
For stable equilibrium, the Center of Gravity (G) of the loaded tube must be below the Center of Buoyancy (B) (the center of mass of the displaced water).
1. Position of Center of Gravity ($Y_G$):
Measuring from the bottom of the tube:
– CM of tube (mass $m$) is at $l/2$.
– CM of water inside (mass $\rho_0 S h$) is at $h/2$.
$$ Y_G = \frac{m(l/2) + (\rho_0 S h)(h/2)}{m + \rho_0 S h} = \frac{ml + \rho_0 S h^2}{2(m + \rho_0 S h)} $$
2. Position of Center of Buoyancy ($Y_B$): The center of buoyancy is the centroid of the submerged volume. Since the tube is cylindrical and submerged to depth $x$: $$ Y_B = \frac{x}{2} $$ Substituting $x = \frac{m + \rho_0 S h}{\rho_0 S}$: $$ Y_B = \frac{m + \rho_0 S h}{2 \rho_0 S} $$
3. Stability Inequality ($Y_G < Y_B$): $$ \frac{ml + \rho_0 S h^2}{2(m + \rho_0 S h)} < \frac{m + \rho_0 S h}{2 \rho_0 S} $$ Canceling the factor of 2 and cross-multiplying (note all terms are positive): $$ \rho_0 S (ml + \rho_0 S h^2) < (m + \rho_0 S h)^2 $$ $$ \rho_0 S m l + (\rho_0 S h)^2 < m^2 + (\rho_0 S h)^2 + 2 m \rho_0 S h $$ Subtract $(\rho_0 S h)^2$ from both sides: $$ \rho_0 S m l < m^2 + 2 m \rho_0 S h $$ Divide by $m$: $$ \rho_0 S l < m + 2 \rho_0 S h $$ $$ \rho_0 S l - m < 2 \rho_0 S h $$ $$ h > \frac{\rho_0 S l – m}{2 \rho_0 S} = \frac{1}{2} \left( l – \frac{m}{\rho_0 S} \right) $$