Edit to “Solution”

Tete edited
revision #20066 parent #20013 ← older
@@ -20,16 +20,16 @@Solution
\[\rho=\rho_w\cdot\frac{(l-\Delta)[2(R-H)+l-\Delta]}{l[2(R-H-\Delta)+l]}.\]
−b. Suppose mass $m_1$ with density $\rho_1<\rho_w$ and mass $m_2$ with density $\rho_2>\rho_w$ are linked by a thread of length $2d$. Since $m_1$ is pulled inward and $m_2$ is pushed outward, they will align in a radial direction. Let $D$ be the distance between the midpoint of the thread and the axis of the space ship when the system is in equilibrium (if possible). Then, the outward push on $m_2$ and the inward pull on $m_1$ have the same magnitude, i.e.
+b. Suppose mass $m_1$ with density $\rho_1<\rho_w$ and mass $m_2$ with density $\rho_2>\rho_w$ are linked by a thread of length $2d$. Since $m_1$ is pulled inward and $m_2$ is pushed outward, they will align in a radial direction. Let $D$ be the distance between the midpoint of the thread and the axis of the space ship when the system is in equilibrium. Then, the outward push on $m_2$ and the inward pull on $m_1$ have the same magnitude, i.e.
\[m_2\left(1-\frac{\rho_w}{\rho_2}\right)\omega^2(D+d)=m_1\left(\frac{\rho_w}{\rho_1}-1\right)\omega^2(D-d),\]
−which can be solved to obtain an expression for $D$. Under the assumption that
+which is possible when
−\[m_2\left(1-\frac{\rho_w}{\rho_2}\right)>m_1\left(\frac{\rho_w}{\rho_1}-1\right),\]
+\[m_2\left(1-\frac{\rho_w}{\rho_2}\right)<m_1\left(\frac{\rho_w}{\rho_1}-1\right).\]
−if the midpoint of the thread is at a distance $r>D$, then the outward push on $m_2$ is stronger than the inward pull on $m_1$, and the system will sink toward the wall of the spacecraft. On the other hand, if the midpoint of the thread is at a distance $r<D$, then the outward push on $m_2$ is weaker than the inward pull on $m_1$, and the system will float toward the free surface. If the inequality in the assumption is reversed, then the equilibrium where the midpoint of the thread is at the distance $r=D$ is stable.
+If the midpoint of the thread is at a distance $r>D$, then the outward push on $m_2$ is weaker than the inward pull on $m_1$, and the system will be pulled back to equilibruim. On the other hand, if the midpoint of the thread is at a distance $r<D$, then the outward push on $m_2$ is stronger than the inward pull on $m_1$, and the system will be pushed back to equilibrium. Therefore, the equilibrium at $r=D$ is stable.
#### Answer
$\rho=\rho_w\cdot\frac{(l-\Delta)[2(R-H)+l-\Delta]}{l[2(R-H-\Delta)+l]}$