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

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. Tangential acceleration ($a_\tau$), 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. Centripetal acceleration ($a_n$), 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.

1. Acceleration is directed vertically downwards
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.
Answer: at the initial (highest) point.

2. Acceleration is directed vertically upwards
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.
Answer: at the lowest point of the trajectory.

3. Acceleration is directed horizontally
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: $\tan^2\alpha = 2 \implies \tan\alpha = \sqrt{2}$).

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 $\arccos(1/\sqrt{3})$.