Question

A cylinder with a diameter 1.0 m is stands in water with its axis vertical. When depressed
slightly and released, it oscillates with a time period of 2 s. Determine the mass of the
cylinder.

Answer 1

Answer on Question#38888 - Math - Other

A cylinder with a diameter $1.0\mathrm{m}$ is stands in water with its axis vertical. When depressed slightly and released, it oscillates with a time period of 2 s. Determine the mass of the cylinder.

Solution

We denote the mass of the cylinder by $m$ then its weight is equal to $P = mg$ , where $g = 9.8 \, m/s^2$ is the gravitational acceleration. While being in the position of equilibrium the cylinder is immersed in water to a depth equal to $a$ (See the figure).

In this position the force of gravity and the force of Archimedes are in equilibrium. Let $\gamma$ denotes the specific weight of the water. Because the volume of the immersed part of the cylinder equals $V = \pi r^2 a$ , where the cylinder radius $r = 0,5m$ , the force of Archimedes is equal to $F_{ar} = \gamma V = \pi \gamma r^2 a$ . Hence, we have from the equation $P = F_{ar}$ , i.e.

$$m g = \pi \gamma r ^ {2} a.$$

To determine the motion of the cylinder with respect to the position of equilibrium we consider the action of water as an additional force of Archimedes. We direct downward the $Ox$ - axis, putting its origin in the placement of equilibrium of the gravity center $C$ .

The force of Archimedes is equal to

$$F _ {a r} = \pi \gamma r ^ {2} (a + x).$$

From Newton's second law of motion of the center of gravity with using (1) and (2) we have

$$\begin{array}{l} m \ddot {x} = P - F _ {a r} \\ = m g - \pi \gamma r ^ {2} a - \pi \gamma r ^ {2} x \\ = - \pi \gamma r ^ {2} x. \\ \end{array}$$

Hence, the position $x(t)$ of the mass center satisfies the second-order linear homogeneous differential equation

$$m \ddot {x} + \pi \gamma r ^ {2} x = 0.$$

Dividing the both sides of the equation by $m \neq 0$ we obtain

$$\ddot{x} + \frac{\pi \gamma r^{2}}{m} x = 0.$$

Now if we denote $k^2 = \pi \gamma r^2 / m$ then the last equation is rewritten

$$\ddot{x} + k^{2} x = 0.$$

We expect oscillatory motion. If we attempt a solution of (3) of the form $x(t) = A \cos(\omega t + \varphi)$ for some frequency $\omega$ and amplitude $A$, we find upon substitution that $\omega = k$. Therefore the displacement of the cylinder is given by

$$x(t) = A \cos(kt + \varphi).$$

These solutions represent an oscillations of frequency $k$, and period $T = 2\pi / k$, so $k = 2\pi / T$. Hence

$$m = \frac{\pi \gamma r^{2}}{k^{2}} = \frac{\gamma r^{2} T^{2}}{4\pi}.$$

Thus substituting all known values into the formula (4) we evaluate

$$m = \frac{1000 \, kg/m^3 \, 9,8 \, m/s^2 \, 0,25 \, m^2 \, 4 \, s^2}{4\pi} \approx 779.8692209 \, kg \approx 780 \, kg.$$

Answer:

780 kg