Find a differential equation for this problem A cylinder of mass $m$ and radius $r$ is sinking in another cylinder filled with water, with radius $R$, where both of their axis are vertically positioned. Let $x$ be the height of the water respect with the level it would occupy in the absence of the small cylinder. Write the differential equation for $x$.Consider the specific weight of the water equal to 1.
I've tried applying the Newton's second law for finding $dx^2/dt^2$, but this $x$ would not be the acceleration of the rising water, but the one of the position of the bottom of the small cylinder respect with the top of the water. The solution given is the next one, $$ \frac{d^2x}{dt^2}= \frac{r^2}{R^2-r^2}\left(g-\frac{\pi R^2}{m}x\right)$$
I would really appreciate if you can help me do the procedure of this. Thanks.
 A: Suppose the cylinder has moved to a depth $y$ below the original water level, and the water level has risen by $x$. Since the cylinder is submerged to a depth $x + y$, the upward buoyant force on it is the weight of the displaced cylinder, $\pi r^2 (x+y)\rho_w g$, where $\rho_w$ is the density of water. The downward force is just its weight $mg$. Since $y$ increases as the cylinder moves downward, we have from Newton's second law
$$
m\frac{d^2y}{dt^2} = mg - \pi r^2 (x+y)\rho_w g
$$
Now, the volume of the water above the original water level is equal to the volume that the cylinder pushed out of the way, so we have $\pi r^2 y = \pi (R^2 - r^2) x$, or $y = x(R^2 -r^2)/r^2$. Then $x + y = xR^2/r^2$ and we have
$$
m\frac{R^2 - r^2}{r^2}\frac{d^2x}{dt^2} = mg - \pi r^2\rho_wgx\frac{R^2}{r^2},
$$
and simplifying gives
$$
\frac{d^2x}{dt^2} = \frac{r^2}{R^2-r^2}\left(g - \frac{\rho_wg\pi R^2}{m} x\right).
$$
$\rho_W g$ is the specific weight of water, so setting it equal to $1$ gives the equation in the book.
