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en/5.4.9.md
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| + | ### Statement | ||
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| + | $5.4.9.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | We can get an estimate of the angular speed $\omega'$ of the outer cylinder as follows. In steady state, an average gas molecule must have tangential velocities of $\omega r_1$ and $\omega' r_2$ at the inner and the outer cylinders, respectively. We consider the motion of one such molecule whose velocity (shown in red) has magnitude $v$ and whose tangential velocity at each point is shown in pink. If the velocity vector makes an angle $\alpha$ (shown in blue) with the tangent to the inner cylinder, then $v=\omega r_1\sec\alpha$. Similarly, if the velocity vector makes an angle $\beta$ (shown in green) with the tangent to the outer cylinder, then $v=\omega'r_2\sec\beta$. Equating these two expressions for $v$, we have | ||
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| + | \[\omega r_1\sec\alpha=\omega'r_2\sec\beta\qquad\Rightarrow\qquad\omega'=\omega\cdot\frac{r_1}{r_2}\cdot\frac{\cos\beta}{\cos\alpha}.\] | ||
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| + | Using the law of sines in $\triangle OAB$, we have | ||
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| + | \[\frac{r_1}{\sin(90^\circ-\beta)}=\frac{r_2}{\sin(90^\circ+\alpha)}\quad\Rightarrow\quad\frac{\cos\beta}{\cos\alpha}=\frac{r_1}{r_2},\] | ||
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| + | and thus $\omega'=\omega(r_1/r_2)^2$. | ||
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| + | Interestingly, if the gas is \underline{not} rarefied and this problem is treated as a Taylor-Coulette flow, we still obtain the same answer. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $5.4.9.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| We can get an estimate of the angular speed $\omega'$ of the outer cylinder as follows. In steady state, an average gas molecule must have tangential velocities of $\omega r_1$ and $\omega' r_2$ at the inner and the outer cylinders, respectively. We consider the motion of one such molecule whose velocity (shown in red) has magnitude $v$ and whose tangential velocity at each point is shown in pink. If the velocity vector makes an angle $\alpha$ (shown in blue) with the tangent to the inner cylinder, then $v=\omega r_1\sec\alpha$. Similarly, if the velocity vector makes an angle $\beta$ (shown in green) with the tangent to the outer cylinder, then $v=\omega'r_2\sec\beta$. Equating these two expressions for $v$, we have | |||
| \[\omega r_1\sec\alpha=\omega'r_2\sec\beta\qquad\Rightarrow\qquad\omega'=\omega\cdot\frac{r_1}{r_2}\cdot\frac{\cos\beta}{\cos\alpha}.\] | |||
| Using the law of sines in $\triangle OAB$, we have | |||
| \[\frac{r_1}{\sin(90^\circ-\beta)}=\frac{r_2}{\sin(90^\circ+\alpha)}\quad\Rightarrow\quad\frac{\cos\beta}{\cos\alpha}=\frac{r_1}{r_2},\] | |||
| and thus $\omega'=\omega(r_1/r_2)^2$. | |||
| Interestingly, if the gas is \underline{not} rarefied and this problem is treated as a Taylor-Coulette flow, we still obtain the same answer. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||