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en/2.3.22.md
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| + | ### Statement | ||
| + | |||
| + | $2.3.22.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $2.3.22.$ A small ball is suspended on a thread. The thread is brought to a horizontal position and then the ball is released. At what point of the trajectory is its acceleration directed vertically upwards? vertically downwards? horizontally? | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | <div align="center"> | ||
| + | <img src="https://savchenkosolutions.com/2/2.3.22/drawing.png" alt="For problem 2.3.22"> | ||
| + | </div> | ||
| + | |||
| + | Let $L$ be the length of the thread, and $\alpha$ be the angle of the thread with the vertical. | ||
| + | The total acceleration of the ball $\vec{a}$ consists of two components: | ||
| + | 1. <b>Tangential acceleration</b> ($a_\tau$), arising due to the change in the magnitude of velocity. It is directed tangentially to the trajectory (perpendicular to the thread) and is caused by the gravity projection: $a_\tau = g \sin\alpha$. | ||
| + | 2. <b>Centripetal acceleration</b> ($a_n$), arising due to the curvature of the trajectory. It is directed along the thread towards the suspension point: $a_n = \frac{v^2}{L}$. | ||
| + | |||
| + | Let's consider the three cases of interest. | ||
| + | |||
| + | <b>1. Acceleration is directed vertically downwards</b> | ||
| + | At the initial moment, the thread is horizontal ($\alpha = 90^\circ$). The velocity of the ball is zero ($v = 0$), hence the centripetal acceleration is $a_n = 0$. | ||
| + | Only the tangential acceleration remains, $a_\tau = g \sin 90^\circ = g$, which is directed strictly vertically downwards. | ||
| + | <b>Answer:</b> at the initial (highest) point. | ||
| + | |||
| + | <b>2. Acceleration is directed vertically upwards</b> | ||
| + | At the lowest point of the trajectory, the thread is vertical ($\alpha = 0^\circ$). The tangential acceleration is $a_\tau = g \sin 0^\circ = 0$ (velocity is maximum, its derivative is zero). | ||
| + | Only the centripetal acceleration remains, $a_n = \frac{v^2}{L}$, which is directed along the thread towards the suspension point, i.e., strictly vertically upwards. | ||
| + | <b>Answer:</b> at the lowest point of the trajectory. | ||
| + | |||
| + | <b>3. Acceleration is directed horizontally</b> | ||
| + | For the total acceleration to be directed horizontally, its vertical projection must be zero. | ||
| + | Let's project $a_n$ and $a_\tau$ onto the vertical axis. The centripetal acceleration gives an upward projection ($a_n \cos\alpha$), and the tangential gives a downward projection ($a_\tau \sin\alpha$). Equating them: | ||
| + | $$a_n \cos\alpha = a_\tau \sin\alpha \quad (1)$$ | ||
| + | |||
| + | We find the velocity at an arbitrary point from the law of conservation of energy. If we take the lowest point as the zero level of potential energy: | ||
| + | $$mgL = \frac{mv^2}{2} + mgL(1 - \cos\alpha)$$ | ||
| + | $$v^2 = 2gL\cos\alpha$$ | ||
| + | Then the centripetal acceleration is: | ||
| + | $$a_n = \frac{v^2}{L} = 2g\cos\alpha \quad (2)$$ | ||
| + | |||
| + | Substitute $a_n$ from $(2)$ and $a_\tau = g\sin\alpha$ into equation $(1)$: | ||
| + | $$2g\cos\alpha \cdot \cos\alpha = g\sin\alpha \cdot \sin\alpha$$ | ||
| + | $$2\cos^2\alpha = \sin^2\alpha$$ | ||
| + | Replace $\sin^2\alpha$ with $1 - \cos^2\alpha$: | ||
| + | $$2\cos^2\alpha = 1 - \cos^2\alpha$$ | ||
| + | $$3\cos^2\alpha = 1 \implies \cos\alpha = \frac{1}{\sqrt{3}}$$ | ||
| + | Hence, the deviation angle from the vertical is $\alpha = \arccos\left(\frac{1}{\sqrt{3}}\right)$. | ||
| + | (This can also be expressed via tangent: $\tan^2\alpha = 2 \implies \tan\alpha = \sqrt{2}$). | ||
| + | |||
| + | <i>Note: There is an error in the original edition of Savchenko's problem book — the answer states an angle of $45^\circ$. At $45^\circ$, only the horizontal projections of $a_\tau$ and $a_n$ are equal, but the total acceleration itself is directed at an angle to the horizon. The correct answer is $\arccos(1/\sqrt{3})$.</i> | ||
| + | |||
| + | #### Answer | ||
| + | Vertically upwards: at the lowest point of the trajectory. | ||
| + | Vertically downwards: at the initial (highest) point. | ||
| + | Horizontally: at the point where the thread makes an angle $\arccos(1/\sqrt{3})$ (or $\arctan\sqrt{2}$) with the vertical. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.3.22.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $2.3.22.$ A small ball is suspended on a thread. The thread is brought to a horizontal position and then the ball is released. At what point of the trajectory is its acceleration directed vertically upwards? vertically downwards? horizontally? | |||
| ### Solution | |||
| <div align="center"> | |||
| <img src="https://savchenkosolutions.com/2/2.3.22/drawing.png" alt="For problem 2.3.22"> | |||
| </div> | |||
| Let $L$ be the length of the thread, and $\alpha$ be the angle of the thread with the vertical. | |||
| The total acceleration of the ball $\vec{a}$ consists of two components: | |||
| 1. <b>Tangential acceleration</b> ($a_\tau$), arising due to the change in the magnitude of velocity. It is directed tangentially to the trajectory (perpendicular to the thread) and is caused by the gravity projection: $a_\tau = g \sin\alpha$. | |||
| 2. <b>Centripetal acceleration</b> ($a_n$), arising due to the curvature of the trajectory. It is directed along the thread towards the suspension point: $a_n = \frac{v^2}{L}$. | |||
| Let's consider the three cases of interest. | |||
| <b>1. Acceleration is directed vertically downwards</b> | |||
| At the initial moment, the thread is horizontal ($\alpha = 90^\circ$). The velocity of the ball is zero ($v = 0$), hence the centripetal acceleration is $a_n = 0$. | |||
| Only the tangential acceleration remains, $a_\tau = g \sin 90^\circ = g$, which is directed strictly vertically downwards. | |||
| <b>Answer:</b> at the initial (highest) point. | |||
| <b>2. Acceleration is directed vertically upwards</b> | |||
| At the lowest point of the trajectory, the thread is vertical ($\alpha = 0^\circ$). The tangential acceleration is $a_\tau = g \sin 0^\circ = 0$ (velocity is maximum, its derivative is zero). | |||
| Only the centripetal acceleration remains, $a_n = \frac{v^2}{L}$, which is directed along the thread towards the suspension point, i.e., strictly vertically upwards. | |||
| <b>Answer:</b> at the lowest point of the trajectory. | |||
| <b>3. Acceleration is directed horizontally</b> | |||
| For the total acceleration to be directed horizontally, its vertical projection must be zero. | |||
| Let's project $a_n$ and $a_\tau$ onto the vertical axis. The centripetal acceleration gives an upward projection ($a_n \cos\alpha$), and the tangential gives a downward projection ($a_\tau \sin\alpha$). Equating them: | |||
| $$a_n \cos\alpha = a_\tau \sin\alpha \quad (1)$$ | |||
| We find the velocity at an arbitrary point from the law of conservation of energy. If we take the lowest point as the zero level of potential energy: | |||
| $$mgL = \frac{mv^2}{2} + mgL(1 - \cos\alpha)$$ | |||
| $$v^2 = 2gL\cos\alpha$$ | |||
| Then the centripetal acceleration is: | |||
| $$a_n = \frac{v^2}{L} = 2g\cos\alpha \quad (2)$$ | |||
| Substitute $a_n$ from $(2)$ and $a_\tau = g\sin\alpha$ into equation $(1)$: | |||
| $$2g\cos\alpha \cdot \cos\alpha = g\sin\alpha \cdot \sin\alpha$$ | |||
| $$2\cos^2\alpha = \sin^2\alpha$$ | |||
| Replace $\sin^2\alpha$ with $1 - \cos^2\alpha$: | |||
| $$2\cos^2\alpha = 1 - \cos^2\alpha$$ | |||
| $$3\cos^2\alpha = 1 \implies \cos\alpha = \frac{1}{\sqrt{3}}$$ | |||
| Hence, the deviation angle from the vertical is $\alpha = \arccos\left(\frac{1}{\sqrt{3}}\right)$. | |||
| (This can also be expressed via tangent: $\tan^2\alpha = 2 \implies \tan\alpha = \sqrt{2}$). | |||
| <i>Note: There is an error in the original edition of Savchenko's problem book — the answer states an angle of $45^\circ$. At $45^\circ$, only the horizontal projections of $a_\tau$ and $a_n$ are equal, but the total acceleration itself is directed at an angle to the horizon. The correct answer is $\arccos(1/\sqrt{3})$.</i> | |||
| #### Answer | |||
| Vertically upwards: at the lowest point of the trajectory. | |||
| Vertically downwards: at the initial (highest) point. | |||
| Horizontally: at the point where the thread makes an angle $\arccos(1/\sqrt{3})$ (or $\arctan\sqrt{2}$) with the vertical. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||