Problem 26: Velocity of Cylinder Jumping Out of Water
1. Identify the Energy Transformation:
We can solve this using the Work-Energy Theorem. The work done by the net force acting on the cylinder results in a change in its kinetic energy. The forces acting on the cylinder are Gravity and Buoyancy.
We can solve this using the Work-Energy Theorem. The work done by the net force acting on the cylinder results in a change in its kinetic energy. The forces acting on the cylinder are Gravity and Buoyancy.
2. Work Done by Gravity ($W_g$):
The cylinder has length $l$. Initially, the top surface is at the water level, so its Center of Mass (COM) is at a depth $l/2$. When it completely leaves the water, its bottom is at the surface, so the COM is at a height $l/2$ above the surface.
The total vertical displacement of the COM is $\Delta h = l$. $$ W_g = -mg\Delta h = -(\rho V) g l $$ Where $V = \pi r^2 l$ is the volume of the cylinder.
The cylinder has length $l$. Initially, the top surface is at the water level, so its Center of Mass (COM) is at a depth $l/2$. When it completely leaves the water, its bottom is at the surface, so the COM is at a height $l/2$ above the surface.
The total vertical displacement of the COM is $\Delta h = l$. $$ W_g = -mg\Delta h = -(\rho V) g l $$ Where $V = \pi r^2 l$ is the volume of the cylinder.
3. Work Done by Buoyancy ($W_B$):
The buoyant force is not constant; it decreases as the cylinder emerges. Let $x$ be the length of the cylinder outside the water. The submerged length is $(l-x)$. $$ F_B(x) = \rho_0 (\text{Submerged Volume}) g = \rho_0 A (l-x) g $$ The work done by buoyancy as the cylinder moves from $x=0$ (fully submerged) to $x=l$ (fully out) is: $$ W_B = \int_{0}^{l} F_B(x) dx = \int_{0}^{l} \rho_0 A g (l-x) dx $$ $$ W_B = \rho_0 A g \left[ lx – \frac{x^2}{2} \right]_0^l = \rho_0 A g \left( l^2 – \frac{l^2}{2} \right) = \frac{1}{2} \rho_0 (Al) g l = \frac{1}{2} \rho_0 V g l $$
The buoyant force is not constant; it decreases as the cylinder emerges. Let $x$ be the length of the cylinder outside the water. The submerged length is $(l-x)$. $$ F_B(x) = \rho_0 (\text{Submerged Volume}) g = \rho_0 A (l-x) g $$ The work done by buoyancy as the cylinder moves from $x=0$ (fully submerged) to $x=l$ (fully out) is: $$ W_B = \int_{0}^{l} F_B(x) dx = \int_{0}^{l} \rho_0 A g (l-x) dx $$ $$ W_B = \rho_0 A g \left[ lx – \frac{x^2}{2} \right]_0^l = \rho_0 A g \left( l^2 – \frac{l^2}{2} \right) = \frac{1}{2} \rho_0 (Al) g l = \frac{1}{2} \rho_0 V g l $$
4. Apply Work-Energy Theorem:
$$ W_{net} = \Delta K.E. $$ $$ W_B + W_g = \frac{1}{2} m v^2 – 0 $$ Substituting the values: $$ \frac{1}{2} \rho_0 V g l – \rho V g l = \frac{1}{2} (\rho V) v^2 $$
$$ W_{net} = \Delta K.E. $$ $$ W_B + W_g = \frac{1}{2} m v^2 – 0 $$ Substituting the values: $$ \frac{1}{2} \rho_0 V g l – \rho V g l = \frac{1}{2} (\rho V) v^2 $$
5. Solve for Velocity ($v$):
Cancel $V$ from all terms and multiply by 2: $$ \rho_0 g l – 2\rho g l = \rho v^2 $$ $$ v^2 = \frac{gl (\rho_0 – 2\rho)}{\rho} $$ $$ v = \sqrt{\frac{gl (\rho_0 – 2\rho)}{\rho}} $$
Cancel $V$ from all terms and multiply by 2: $$ \rho_0 g l – 2\rho g l = \rho v^2 $$ $$ v^2 = \frac{gl (\rho_0 – 2\rho)}{\rho} $$ $$ v = \sqrt{\frac{gl (\rho_0 – 2\rho)}{\rho}} $$
6. Calculation:
Given: $\rho = 250 \, \text{kg/m}^3$, $\rho_0 = 1000 \, \text{kg/m}^3$, $l = 20 \, \text{cm} = 0.2 \, \text{m}$, $g = 10 \, \text{m/s}^2$. $$ v = \sqrt{\frac{10 \times 0.2 \times (1000 – 2(250))}{250}} $$ $$ v = \sqrt{\frac{2 \times (1000 – 500)}{250}} = \sqrt{\frac{2 \times 500}{250}} = \sqrt{4} $$ $$ v = 2 \, \text{m/s} $$
Given: $\rho = 250 \, \text{kg/m}^3$, $\rho_0 = 1000 \, \text{kg/m}^3$, $l = 20 \, \text{cm} = 0.2 \, \text{m}$, $g = 10 \, \text{m/s}^2$. $$ v = \sqrt{\frac{10 \times 0.2 \times (1000 – 2(250))}{250}} $$ $$ v = \sqrt{\frac{2 \times (1000 – 500)}{250}} = \sqrt{\frac{2 \times 500}{250}} = \sqrt{4} $$ $$ v = 2 \, \text{m/s} $$
Answer: The cylinder leaves the water surface with a velocity of 2 m/s.