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| ### Statement |
| ### Statement |
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| $2.3.22.$ [Insert the problem statement] |
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| ### Solution |
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| ### Statement |
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| $2.3.22.$ 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? |
| $2.3.22.$ 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? |
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| ### Solution |
| ### Solution |
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| <div align="center"> |
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| <img src="https://savchenkosolutions.com/2/2.3.22/drawing.png" alt="For problem 2.3.22"> |
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| </div> |
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| Let $L$ be the length of the thread, and $\alpha$ be the angle of the thread with the vertical. |
| 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: |
| 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: $a_\tau = g \sin\alpha$. |
| 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: $a_n = \frac{v^2}{L}$. |
| 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: $a_n = \frac{v^2}{L}$. |
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| Let's consider the three cases of interest. |
| Let's consider the three cases of interest. |
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| <b>1. Acceleration is directed vertically downwards</b> |
| <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 $a_n = 0$. |
| 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. |
| Only the tangential acceleration remains, $a_\tau = g \sin 90^\circ = g$, which is directed strictly vertically downwards. |
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| <b>Answer:</b> at the initial (highest) point. |
| <b>Answer:</b> at the initial (highest) point. |
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| <b>2. Acceleration is directed vertically upwards</b> |
| <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). |
| 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. |
| 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. |
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| <b>Answer:</b> at the lowest point of the trajectory. |
| <b>Answer:</b> at the lowest point of the trajectory. |
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| <b>3. Acceleration is directed horizontally</b> |
| <b>3. Acceleration is directed horizontally</b>\ |
| For the total acceleration to be directed horizontally, its vertical projection must be zero. |
| 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: |
| 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: |
| $$a_n \cos\alpha = a_\tau \sin\alpha \quad (1)$$ |
| $$a_n \cos\alpha = a_\tau \sin\alpha \quad (1)$$ |
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| 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: | | 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: |
| $$mgL = \frac{mv^2}{2} + mgL(1 - \cos\alpha)$$ | | $$mgL = \frac{mv^2}{2} + mgL(1 - \cos\alpha)$$ |
| $$v^2 = 2gL\cos\alpha$$ | | $$v^2 = 2gL\cos\alpha$$ |
| Then the centripetal acceleration is: | | Then the centripetal acceleration is: |
| $$a_n = \frac{v^2}{L} = 2g\cos\alpha \quad (2)$$ | | $$a_n = \frac{v^2}{L} = 2g\cos\alpha \quad (2)$$ |
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| Substitute $a_n$ from $(2)$ and $a_\tau = g\sin\alpha$ into equation $(1)$: | | Substitute $a_n$ from $(2)$ and $a_\tau = g\sin\alpha$ into equation $(1)$: |
| $$2g\cos\alpha \cdot \cos\alpha = g\sin\alpha \cdot \sin\alpha$$ | | $$2g\cos\alpha \cdot \cos\alpha = g\sin\alpha \cdot \sin\alpha$$ |
| $$2\cos^2\alpha = \sin^2\alpha$$ | | $$2\cos^2\alpha = \sin^2\alpha$$ |
| Replace $\sin^2\alpha$ with $1 - \cos^2\alpha$: | | Replace $\sin^2\alpha$ with $1 - \cos^2\alpha$: |
| $$2\cos^2\alpha = 1 - \cos^2\alpha$$ | | $$2\cos^2\alpha = 1 - \cos^2\alpha$$ |
| $$3\cos^2\alpha = 1 \implies \cos\alpha = \frac{1}{\sqrt{3}}$$ | | $$3\cos^2\alpha = 1 \implies \cos\alpha = \frac{1}{\sqrt{3}}$$ |
| Hence, the deviation angle from the vertical is $\alpha = \arccos\left(\frac{1}{\sqrt{3}}\right)$. | | Hence, the deviation angle from the vertical is $\alpha = \arccos\left(\frac{1}{\sqrt{3}}\right)$. |
| (This can also be expressed via tangent: $\tan^2\alpha = 2 \implies \tan\alpha = \sqrt{2}$). | | (This can also be expressed via tangent: $\tan^2\alpha = 2 \implies \tan\alpha = \sqrt{2}$). |
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| <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> | | <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> |
