Question

A uniform vertical cylinder (density = $\sigma$) is released from rest when its lower end just touches the liquid (density = $\rho$) surface of a deep lake. Calculate maximum displacement of cylinder (in meter) Take ($\ell = 8m$ and $\frac{\sigma}{\rho} = \frac{1}{2}$)

Answer 1

8

Answer 2

Let the cylinder have cross-sectional area $A$ and length $\ell$. The density of the cylinder is $\sigma$ and the density of the liquid is $\rho$. The mass of the cylinder is $m = \sigma A \ell$. The weight of the cylinder is $W = mg = \sigma A \ell g$.

The cylinder is released from rest when its lower end just touches the liquid surface, so the initial position is $x=0$ (taking the downward direction as positive and the liquid surface as the origin) and the initial velocity is $v(0) = 0$.

When the cylinder is submerged in the liquid by a depth $x$, the volume of the submerged part is $V_{submerged} = A x$ if $0 \le x \le \ell$. The buoyant force is $F_B = \rho V_{submerged} g = \rho A x g$. This force acts upwards.

The net downward force on the cylinder when $0 \le x \le \ell$ is $F_{net}(x) = W - F_B(x) = \sigma A \ell g - \rho A x g$. The equation of motion is $m a = F_{net}$, so $\sigma A \ell a = \sigma A \ell g - \rho A x g$. $a = \frac{d^2 x}{dt^2} = g - \frac{\rho}{\sigma \ell} g x$.

The maximum displacement occurs when the velocity of the cylinder becomes zero. We can use the work-energy theorem. The change in kinetic energy is $\Delta KE = KE_f - KE_i = 0 - 0 = 0$. The work done by the net force is equal to the change in kinetic energy. The work done by gravity as the cylinder moves from $x=0$ to $x_{max}$ is $W_g = \int_0^{x_{max}} W dx = \int_0^{x_{max}} \sigma A \ell g dx = \sigma A \ell g x_{max}$. The work done by the buoyant force as the cylinder moves from $x=0$ to $x_{max}$ is $W_B = \int_0^{x_{max}} (-F_B(x)) dx$. Since the buoyant force is upwards and displacement is downwards, the work done is negative.

We need to consider two cases for the buoyant force depending on whether $x_{max} \le \ell$ or $x_{max} > \ell$.

Case 1: Maximum displacement $x_{max} \le \ell$. In this case, the buoyant force is $F_B(x) = \rho A x g$ for $0 \le x \le x_{max}$. $W_B = \int_0^{x_{max}} -\rho A x g dx = -\rho A g \int_0^{x_{max}} x dx = -\frac{1}{2} \rho A g x_{max}^2$. By the work-energy theorem, $W_g + W_B = 0$. $\sigma A \ell g x_{max} - \frac{1}{2} \rho A g x_{max}^2 = 0$. Since $x_{max} \neq 0$ (as $\sigma < \rho$, the cylinder will sink), we can divide by $A g x_{max}$: $\sigma \ell - \frac{1}{2} \rho x_{max} = 0$. $x_{max} = \frac{2 \sigma \ell}{\rho} = 2 \ell \frac{\sigma}{\rho}$. Given $\ell = 8m$ and $\frac{\sigma}{\rho} = \frac{1}{2}$. $x_{max} = 2 \times 8m \times \frac{1}{2} = 8m$. We assumed $x_{max} \le \ell$. Since $\ell = 8m$, $x_{max} = 8m$ satisfies this condition. So this is the correct maximum displacement.

Case 2: Maximum displacement $x_{max} > \ell$. In this case, the cylinder is fully submerged for part of the motion. Let's split the work integral into two parts: from $x=0$ to $x=\ell$ and from $x=\ell$ to $x=x_{max}$. For $0 \le x \le \ell$, $F_B(x) = \rho A x g$. The work done by buoyant force is $\int_0^{\ell} -\rho A x g dx = -\frac{1}{2} \rho A g \ell^2$. For $x > \ell$, the cylinder is fully submerged, so the buoyant force is constant, $F_B = \rho A \ell g$. The work done by buoyant force from $x=\ell$ to $x=x_{max}$ is $\int_{\ell}^{x_{max}} -\rho A \ell g dx = -\rho A \ell g (x_{max} - \ell)$. The total work done by buoyant force is $W_B = -\frac{1}{2} \rho A g \ell^2 - \rho A \ell g (x_{max} - \ell)$. The work done by gravity is $W_g = \sigma A \ell g x_{max}$. By the work-energy theorem, $W_g + W_B = 0$. $\sigma A \ell g x_{max} - \frac{1}{2} \rho A g \ell^2 - \rho A \ell g (x_{max} - \ell) = 0$. Divide by $A g$: $\sigma \ell x_{max} - \frac{1}{2} \rho \ell^2 - \rho \ell x_{max} + \rho \ell^2 = 0$. $x_{max} (\sigma \ell - \rho \ell) + \frac{1}{2} \rho \ell^2 = 0$. $x_{max} \ell (\sigma - \rho) + \frac{1}{2} \rho \ell^2 = 0$. Since $\ell \neq 0$, $x_{max} (\sigma - \rho) + \frac{1}{2} \rho \ell = 0$. $x_{max} (\rho - \sigma) = \frac{1}{2} \rho \ell$. $x_{max} = \frac{\frac{1}{2} \rho \ell}{\rho - \sigma} = \frac{\frac{1}{2} \ell}{1 - \frac{\sigma}{\rho}}$. Using the given values, $\ell = 8m$ and $\frac{\sigma}{\rho} = \frac{1}{2}$. $x_{max} = \frac{\frac{1}{2} \times 8m}{1 - \frac{1}{2}} = \frac{4m}{\frac{1}{2}} = 8m$. This result $x_{max} = 8m$ is consistent with the condition $x_{max} > \ell$ only if we consider the boundary case $x_{max} = \ell$. The result from Case 1 gives $x_{max} = 8m$, which is equal to $\ell$. So the maximum displacement is exactly the length of the cylinder.

The maximum displacement of the cylinder is 8 meters.