Stability of a Composite Floating Candle
Part (a): Range of Length $l$
1. Condition for Floatation:
For the candle to float, the total weight must be balanced by the buoyant force without the candle sinking completely.
Let $A$ be the cross-sectional area ($\pi r^2$), $l$ be the length of the wax, and $l_{Al} = 1.0 \text{ cm}$ be the length of the aluminium.
Let $h$ be the submerged depth.
$$ \text{Weight} = \text{Buoyancy} $$
$$ (\rho_{wax} l + \rho_{Al} l_{Al}) g A = \rho_{water} h g A $$
Substituting values ($\rho_{wax}=0.8, \rho_{Al}=2.7, \rho_{w}=1$):
$$ 0.8 l + 2.7(1) = 1(h) \implies h = 0.8l + 2.7 $$
To float, the submerged depth $h$ must be less than or equal to the total length $(l + 1)$.
$$ 0.8l + 2.7 \le l + 1 $$
$$ 1.7 \le 0.2l \implies l \ge 8.5 \text{ cm} $$
2. Condition for Stable Vertical Equilibrium:
For a floating body with negligible radius, the center of buoyancy ($B$) must be vertically above the center of mass ($G$).
$$ y_B \ge y_G $$
Taking the bottom of the cylinder as the origin ($y=0$):
Center of Buoyancy ($y_B$): The center of the submerged volume (height $h$) is at $h/2$.
$$ y_B = \frac{h}{2} = \frac{0.8l + 2.7}{2} $$
Center of Mass ($y_G$):
$$ y_G = \frac{m_{Al} y_{Al} + m_{wax} y_{wax}}{m_{total}} $$
$$ y_G = \frac{(2.7A)(0.5) + (0.8lA)(1 + l/2)}{2.7A + 0.8lA} = \frac{1.35 + 0.8l + 0.4l^2}{2.7 + 0.8l} $$
Setting $y_B \ge y_G$:
$$ \frac{0.8l + 2.7}{2} \ge \frac{1.35 + 0.8l + 0.4l^2}{2.7 + 0.8l} $$
Cross-multiplying (since denominators are positive):
$$ (0.8l + 2.7)^2 \ge 2(1.35 + 0.8l + 0.4l^2) $$
$$ 0.64l^2 + 4.32l + 7.29 \ge 2.7 + 1.6l + 0.8l^2 $$
Rearranging to form a quadratic inequality ($Al^2 + Bl + C \le 0$):
$$ 0.16l^2 – 2.72l – 4.59 \le 0 $$
Solving for roots using quadratic formula:
$$ l = \frac{2.72 \pm \sqrt{2.72^2 – 4(0.16)(-4.59)}}{0.32} $$
$$ l = \frac{2.72 \pm \sqrt{7.398 + 2.938}}{0.32} = \frac{2.72 \pm 3.215}{0.32} $$
Taking the positive root:
$$ l_{max} = \frac{5.935}{0.32} \approx 18.54 \text{ cm} $$
Since the inequality is $\le 0$, $l$ must be between the roots.
Combined Range:
$$ 8.5 \text{ cm} \le l \le 18.54 \text{ cm} $$
Part (b): Burning Time
The candle starts at length $l_0 = 12 \text{ cm}$. It remains stable as long as $l \ge 8.5 \text{ cm}$.
Once the length drops below $8.5 \text{ cm}$, the condition $h \le l_{total}$ is violated (the candle sinks).
Change in length allowed:
$$ \Delta l = 12 \text{ cm} – 8.5 \text{ cm} = 3.5 \text{ cm} = 35 \text{ mm} $$
Rate of consumption = $1.0 \text{ mm/min}$.
$$ \text{Time} = \frac{35 \text{ mm}}{1.0 \text{ mm/min}} = 35 \text{ min} $$