Question

In A uniform solid sphere of density$\frac{\mathbf{1}}{\mathbf{2}}$is floating in water. (Compare Chapter 8, Problem (5.37).) It is pushed down just under water and released. Write the differential equation of motion (neglecting friction) and solve it to obtain the period in terms of $\mathit{K}\mathbf{\left(}{\mathbf{5}}^{\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}}\mathbf{\right)}$. Show that this period is approximately 1.16 times the period for small oscillations.

Short answer

The period is approximately 1.16 times the period for small oscillations.

Step-by-step solution

Given Information

A uniform solid sphere of density$\frac{1}{2}$ is floating in water.

Definition of the second law of Newton.

TheNewton be second law of motion states that ${\mathbf{m}}_{\mathbf{s}}\stackrel{\mathbf{.}\mathbf{.}}{\mathbf{z}}\mathbf{=}{\mathbf{m}}_{\mathbf{s}}\mathbf{g}\mathbf{-}{\mathbf{\rho }}_{\mathbf{\omega }}\mathbf{aV}\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{g}$.

Prove the statement.

A uniform solid sphere of density$\frac{1}{2}$is floating in the water.

Let the radius of the sphere be R.

Newton be second law of motion states thatlocalid="1664323424703" $m_s\stackrel{..}{z}=m_{s}g-{\rho }_{\omega }aV\left(z\right)g$

$m_s$is sphere mass.

${\rho }_{\omega }$is water density.

$V\left(z\right)$is the volume of the sphere beneath surface.

Slice the sphere into a bunch of disks of thickness dz

Radii be $\sqrt{R^2-z^2}$.

$$\begin{gathered} V=\sqrt{{\left(R^2-z^{\text{'}2}\right)}^2}\pi dz\text{'}+\frac{1}{2}\frac{4\pi }{3}{R}^3 \\ V=\left(R^{2}z-\frac{z^{\text{'}3}}{3}\right)\pi +\frac{2\pi }{3}{R}^3 \end{gathered}$$

Since $m_s={\rho }_s\frac{4\pi }{3}{R}^3$

Let${\rho }_s=\frac{1}{2}$

Substitute values mentioned above in the equation mentioned below.

$$\begin{gathered} m_s\stackrel{..}{z}=m_{s}g-{\rho }_{\omega }aV\left(z\right)g \\ \stackrel{..}{z}=\left(1-\frac{V\left(z\right)}{m_s}\right)g \\ \stackrel{..}{z}=\left(\frac{-3z}{2R}+\frac{1}{2}{\left(\frac{z}{R}\right)}^3\right) \\ \stackrel{..}{z}=\frac{-g}{R}\left(\frac{3z}{2R}-\frac{1}{2}{\left(\frac{z}{R}\right)}^3\right) \end{gathered}$$

Substitute values mentioned below in above equation.

$$\begin{gathered} \xi =\frac{z}{R} \\ \tau =\sqrt{\frac{g}{R}}t \end{gathered}$$

The Equation becomes as follows.

$\xi \text{'}\text{'}\left(\tau \right)=\left(\frac{-3}{2}\xi +\frac{1}{2}{\left(\xi \right)}^3\right)$

Initially the sphere is not moving and is just beneath the water surface, the initial condition for$\xi \left(\tau \right)$are mentioned below.

$$\begin{gathered} z\left(0\right)=R \\ \stackrel{.}{z}\left(0\right)=0 \\ \xi \left(0\right)=1 \\ \xi \text{'}\left(0\right)=0 \end{gathered}$$

Substitute values mentioned above in the equation mentioned below.

The equation becomes as follows

$$\begin{gathered} \frac{1}{2}\xi {\text{'}}^2-\frac{1}{2}\xi \text{'}\left(0\right)=\left(\frac{-3}{4}\xi +\frac{1}{8}{\left(\xi \right)}^3\right)-\left(\frac{-3}{4}{\xi }^2\left(0\right)+\frac{1}{8}{\left(\xi \right)}^4\left(0\right)\right) \\ \xi {\text{'}}^2=\left(\frac{-3}{2}{\xi }^2+\frac{1}{4}{\left(\xi \right)}^4+\frac{5}{4}\right) \\ \xi {\text{'}}^2=\frac{1}{4}\left(-6{\xi }^2+{\left(\xi \right)}^4+5\right) \\ \xi {\text{'}}^2=\frac{5}{4}(1-{\xi }^2)(1-\frac{1}{5}{\xi }^2) \end{gathered}$$

Integrate for period oscillation.

$$\begin{gathered} \underset{1}{\overset{0}{\int }}\frac{d\xi }{\sqrt{(1-{\xi }^2)(1-k^2{\xi }^2})}=-\sqrt{\frac{5}{4}}\underset{0}{\overset{\frac{T}{4}}{\int }}dr \\ T=\frac{8}{\sqrt{5}}K\left(\frac{1}{\sqrt{5}}\right) \\ T\approx 5.94 \end{gathered}$$

The statement mentioned above states that$\xi \text{'}\text{'}\left(\tau \right)=\left(\frac{-3}{2}\xi \right)$.

Substitute the values in the equation mentioned above.

$$\begin{gathered} T_{80}=2\pi \sqrt{\frac{2}{3}} \\ {T}_{80}\approx 5.13 \end{gathered}$$

The ratio of period becomes as follows.

$$\begin{gathered} \frac{T}{T_{80}}\approx \frac{5.94}{5.13} \\ \frac{T}{T_{80}}\approx 1.16 \end{gathered}$$

Hence, the solution is mentioned below.

The period is approximately 1.16 times the period for small oscillations.