Question

A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its lower end. The cylinder is held so that the spring is neither stretched nor compressed, then a tank of water is placed underneath with the surface of the water just touching the bottom of the cylinder. When released, the cylinder will oscillate a few times but, damped by the water, quickly reach an equilibrium position.When in equilibrium, what length of the cylinder is submerged?y=?m

Answer 1

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be $\who_w = 1000 kg/m^3$

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

$F_s + F_b = W$

Where $F_s = kx$N is the spring force, $F_b = W_w = m_wg = \rho_w V_s g$ is the buoyancy force, which equals to the weight $W_w$ of the water displaced by the submerged portion of the cylinder, which is the product of water density $\rho_w$, submerged volume $V_s$ and gravitational constant g. W = mg is the weight of the metal cylinder.

$kx + \rho_w V_s g = mg$

The submerged volume would be the product of cross-section area and the submerged length x

$V_s = Ax = \pi(d/2)^2x$

Plug that into our force equation and we have

$kx + \rho_w \pi(d/2)^2x g = mg$

$x(k + \rho_w g \pi d^2/4) = mg$

$x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m$