Question

A uniform cylinder made of material of density $\rho$ = 250 kg/m³ pressed down and helt at rest in a pool so that its upper circular surface is in level with the water surface. Lenth of the cylinder is l = 20 cm and radius is r = 3.0 cm. When released the cylinder jumps out of water moving vertically. Find velocity with which the cylinder leaves the water surface. Density of water is $\rho_o$ = 1000 kg/m³ and acceleration of free fall is g = 10 m/s².

Answer 1

2 m/s

Answer 2

The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy.

1. Work done by gravity ($W_g$): As the cylinder moves upwards by length $l$, the work done by gravity (acting downwards) is $W_g = -mg \times l = -(\rho \pi r^2 l) g \times l = -\rho \pi r^2 l^2 g$.

2. Work done by buoyancy ($W_B$): The buoyant force varies with the submerged depth. Let $y$ be the upward displacement of the top surface from the water level. The submerged length is $(l-y)$. The buoyant force is $F_B(y) = \rho_o g A (l-y)$, where $A = \pi r^2$. The work done by buoyancy is $W_B = \int_{0}^{l} F_B(y) dy = \int_{0}^{l} \rho_o g \pi r^2 (l-y) dy = \rho_o g \pi r^2 \left[ ly - \frac{y^2}{2} \right]_{0}^{l} = \frac{1}{2} \rho_o g \pi r^2 l^2$.

3. Net work and kinetic energy: The net work is $W_{net} = W_g + W_B = -\rho \pi r^2 l^2 g + \frac{1}{2} \rho_o \pi r^2 l^2 g$. The change in kinetic energy is $\Delta KE = \frac{1}{2} m v_f^2 - 0 = \frac{1}{2} (\rho \pi r^2 l) v_f^2$.

4. Equating work and kinetic energy: $-\rho \pi r^2 l^2 g + \frac{1}{2} \rho_o \pi r^2 l^2 g = \frac{1}{2} \rho \pi r^2 l v_f^2$. Dividing by $\pi r^2 l$: $-\rho l g + \frac{1}{2} \rho_o l g = \frac{1}{2} \rho v_f^2$. $v_f^2 = \frac{lg(\rho_o - 2\rho)}{\rho}$.

5. Calculation: $l = 0.20$ m, $\rho = 250$ kg/m³, $\rho_o = 1000$ kg/m³, $g = 10$ m/s². $v_f^2 = \frac{0.20 \times 10 \times (1000 - 2 \times 250)}{250} = \frac{2 \times (1000 - 500)}{250} = \frac{2 \times 500}{250} = 4$. $v_f = \sqrt{4} = 2$ m/s.