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en/3.1.14.md
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| + | ### Statement | ||
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| + | $3.1.14.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Problem | ||
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| + | $3.1.14^*.$ A weightless rod of length $L > l$ with masses $m$ at its ends hangs on threads of length $h$ located at a distance $l$ from each other. The rod is horizontal. Show that when the rod is rotated about its vertical axis of symmetry by a small angle $\varphi$, the torque of the forces acting on the masses is proportional to $\varphi$, and the change in the potential energy of the masses is proportional to $\varphi^2$. Find the maximum angular velocity of the rod if it is released after being initially rotated by an angle $\varphi_0$. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Let's analyze the geometry of the displacement when the rod is rotated by an angle $\varphi$. The point of attachment of the thread moves along an arc. The horizontal displacement of the thread from its vertical position is given by the chord $A'C$. | ||
| + | |||
| + | From the side view, we can express this displacement as: | ||
| + | $$ A'C = h \sin\alpha $$ | ||
| + | where $\alpha$ is the angle of deviation of the thread from the vertical. | ||
| + | |||
| + | Considering the top view, let the center of rotation be $O'$. The distance to the thread attachment point is $O'C = O'A' = \frac{l}{2}$. | ||
| + | |||
| + | Using the Law of Sines in the isosceles triangle $\triangle O'CA'$: | ||
| + | $$ \frac{A'C}{\sin\varphi} = \frac{O'C}{\sin\left(\frac{\pi}{2} - \frac{\varphi}{2}\right)} $$ | ||
| + | |||
| + | Solving for $A'C$: | ||
| + | $$ A'C = O'C \frac{\sin\varphi}{\cos\left(\frac{\varphi}{2}\right)} = \frac{l}{2} \cdot \frac{\sin\varphi}{\cos\left(\frac{\varphi}{2}\right)} $$ | ||
| + | |||
| + | Equating the two expressions for $A'C$, we find the relationship between $\alpha$ and $\varphi$: | ||
| + | $$ \sin\alpha = \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)} $$ | ||
| + | |||
| + | Next, let's find the torque $\vec{M}$ caused by the tension force $\vec{T}$. Using cylindrical coordinates, the position vector of the thread attachment point is $\vec{r} = \frac{l}{2} \vec{e}_r$. The tension force can be decomposed as: | ||
| + | $$ \vec{T} = -T \sin\alpha \vec{e}_\varphi + T \cos\alpha \vec{e}_z $$ | ||
| + | |||
| + | The torque is $\vec{M} = [\vec{r} \times \vec{T}]$. The vertical component $M_z$, which twists the rod back to equilibrium, is: | ||
| + | $$ M_z = -\frac{l}{2} T \sin\alpha $$ | ||
| + | |||
| + | Substituting $\sin\alpha$: | ||
| + | $$ M_z = -\frac{l T}{2} \cdot \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)} = -\frac{l^2 T \sin\varphi}{4h \sqrt{1 - \sin^2\left(\frac{\varphi}{2}\right)}} $$ | ||
| + | |||
| + | For small angles $\varphi \to 0$, we have $\sin\varphi \approx \varphi$ and the denominator approaches $1$. Thus: | ||
| + | $$ M_z \approx -\frac{l^2 T}{4h} \varphi $$ | ||
| + | This proves that the restoring torque is proportional to the angle: $M \sim \varphi$. | ||
| + | |||
| + | Now, let's calculate the change in potential energy. The vertical displacement $z$ of the rod is: | ||
| + | $$ z = h(1 - \cos\alpha) = h(1 - \sqrt{1 - \sin^2\alpha}) $$ | ||
| + | |||
| + | Substituting our expression for $\sin\alpha$: | ||
| + | $$ z = h \left( 1 - \left[ 1 - \left( \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)} \right)^2 \right]^{1/2} \right) $$ | ||
| + | |||
| + | For small angles $\varphi \to 0$, applying the approximation $(1+x)^a \approx 1 + ax$, we obtain: | ||
| + | $$ z \approx h \left( 1 - \left[ 1 - \left( \frac{l\varphi}{2h} \right)^2 \right]^{1/2} \right) \approx h \left( 1 - 1 + \frac{1}{2}\left(\frac{l}{2h}\right)^2 \varphi^2 \right) = \frac{l^2}{8h} \varphi^2 $$ | ||
| + | This proves that the change in potential energy is proportional to the square of the angle: $U \sim \varphi^2$. | ||
| + | |||
| + | Finally, to find the maximum angular velocity $\omega$, we apply the law of conservation of energy. The maximum kinetic energy at the equilibrium position equals the initial potential energy at the maximum deflection angle $\varphi_0$: | ||
| + | $$ \frac{mv^2}{2} = mgz \implies v = \sqrt{2gz} $$ | ||
| + | |||
| + | The linear velocity of the masses at the ends of the rod is related to the angular velocity by $v = \omega \frac{L}{2}$. Therefore: | ||
| + | $$ \omega = \frac{2v}{L} = \frac{2\sqrt{2gz}}{L} $$ | ||
| + | |||
| + | Substituting $z = \frac{l^2}{8h} \varphi_0^2$ into the equation: | ||
| + | $$ \omega = \frac{2\sqrt{2g \cdot \frac{l^2}{8h} \varphi_0^2}}{L} = \frac{l}{L} \sqrt{\frac{g}{h}} \varphi_0 $$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.1.14.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | |||
| $3.1.14^*.$ A weightless rod of length $L > l$ with masses $m$ at its ends hangs on threads of length $h$ located at a distance $l$ from each other. The rod is horizontal. Show that when the rod is rotated about its vertical axis of symmetry by a small angle $\varphi$, the torque of the forces acting on the masses is proportional to $\varphi$, and the change in the potential energy of the masses is proportional to $\varphi^2$. Find the maximum angular velocity of the rod if it is released after being initially rotated by an angle $\varphi_0$. | |||
| ### Solution | |||
| Let's analyze the geometry of the displacement when the rod is rotated by an angle $\varphi$. The point of attachment of the thread moves along an arc. The horizontal displacement of the thread from its vertical position is given by the chord $A'C$. | |||
| From the side view, we can express this displacement as: | |||
| $$ A'C = h \sin\alpha $$ | |||
| where $\alpha$ is the angle of deviation of the thread from the vertical. | |||
| Considering the top view, let the center of rotation be $O'$. The distance to the thread attachment point is $O'C = O'A' = \frac{l}{2}$. | |||
| Using the Law of Sines in the isosceles triangle $\triangle O'CA'$: | |||
| $$ \frac{A'C}{\sin\varphi} = \frac{O'C}{\sin\left(\frac{\pi}{2} - \frac{\varphi}{2}\right)} $$ | |||
| Solving for $A'C$: | |||
| $$ A'C = O'C \frac{\sin\varphi}{\cos\left(\frac{\varphi}{2}\right)} = \frac{l}{2} \cdot \frac{\sin\varphi}{\cos\left(\frac{\varphi}{2}\right)} $$ | |||
| Equating the two expressions for $A'C$, we find the relationship between $\alpha$ and $\varphi$: | |||
| $$ \sin\alpha = \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)} $$ | |||
| Next, let's find the torque $\vec{M}$ caused by the tension force $\vec{T}$. Using cylindrical coordinates, the position vector of the thread attachment point is $\vec{r} = \frac{l}{2} \vec{e}_r$. The tension force can be decomposed as: | |||
| $$ \vec{T} = -T \sin\alpha \vec{e}_\varphi + T \cos\alpha \vec{e}_z $$ | |||
| The torque is $\vec{M} = [\vec{r} \times \vec{T}]$. The vertical component $M_z$, which twists the rod back to equilibrium, is: | |||
| $$ M_z = -\frac{l}{2} T \sin\alpha $$ | |||
| Substituting $\sin\alpha$: | |||
| $$ M_z = -\frac{l T}{2} \cdot \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)} = -\frac{l^2 T \sin\varphi}{4h \sqrt{1 - \sin^2\left(\frac{\varphi}{2}\right)}} $$ | |||
| For small angles $\varphi \to 0$, we have $\sin\varphi \approx \varphi$ and the denominator approaches $1$. Thus: | |||
| $$ M_z \approx -\frac{l^2 T}{4h} \varphi $$ | |||
| This proves that the restoring torque is proportional to the angle: $M \sim \varphi$. | |||
| Now, let's calculate the change in potential energy. The vertical displacement $z$ of the rod is: | |||
| $$ z = h(1 - \cos\alpha) = h(1 - \sqrt{1 - \sin^2\alpha}) $$ | |||
| Substituting our expression for $\sin\alpha$: | |||
| $$ z = h \left( 1 - \left[ 1 - \left( \frac{l \sin\varphi}{2h \cos\left(\frac{\varphi}{2}\right)} \right)^2 \right]^{1/2} \right) $$ | |||
| For small angles $\varphi \to 0$, applying the approximation $(1+x)^a \approx 1 + ax$, we obtain: | |||
| $$ z \approx h \left( 1 - \left[ 1 - \left( \frac{l\varphi}{2h} \right)^2 \right]^{1/2} \right) \approx h \left( 1 - 1 + \frac{1}{2}\left(\frac{l}{2h}\right)^2 \varphi^2 \right) = \frac{l^2}{8h} \varphi^2 $$ | |||
| This proves that the change in potential energy is proportional to the square of the angle: $U \sim \varphi^2$. | |||
| Finally, to find the maximum angular velocity $\omega$, we apply the law of conservation of energy. The maximum kinetic energy at the equilibrium position equals the initial potential energy at the maximum deflection angle $\varphi_0$: | |||
| $$ \frac{mv^2}{2} = mgz \implies v = \sqrt{2gz} $$ | |||
| The linear velocity of the masses at the ends of the rod is related to the angular velocity by $v = \omega \frac{L}{2}$. Therefore: | |||
| $$ \omega = \frac{2v}{L} = \frac{2\sqrt{2gz}}{L} $$ | |||
| Substituting $z = \frac{l^2}{8h} \varphi_0^2$ into the equation: | |||
| $$ \omega = \frac{2\sqrt{2g \cdot \frac{l^2}{8h} \varphi_0^2}}{L} = \frac{l}{L} \sqrt{\frac{g}{h}} \varphi_0 $$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||