Question

Маленький шарик, заполненный водой, подвешен на нити длиной L. В нижней точке шарика имеется маленькое отверстие, из которого вытекают капельки воды. Под шариком на расстоянии h от него расположена горизонтальная плоскость. Нить с шариком отклоняют от вертикали на угол φ0 (φ0 << 1) и отпускают. При каком значении h капелька, оторвавшаяся от шарика в его нижнем положении, попадёт в ту же точку плоскости, что и капелька, оторвавшаяся в момент максимального отклонения шарика от вертикали? Сопротивлением воздуха пренебречь. Примечание: при φ << 1 sin φ ≈ φ, cos φ ≈ (1 – φ 2/2), где угол φ выражен в радианах.

Answer 1

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Answer 2

Problem Analysis

We have a small ball filled with water, hanging from a string of length L. The ball has a small hole at its bottom, from which water droplets fall. There is a horizontal plane located at a distance h below the ball. The ball is initially displaced from the vertical by an angle φ0 (where φ0 << 1) and then released. We need to find the value of h at which a droplet that detaches from the ball at its lowest position will hit the same point on the plane as a droplet that detaches from the ball at the moment of maximum deviation from the vertical.

Solution

To solve this problem, we need to consider the motion of the ball and the droplets. Let's break down the solution into steps:

1. Find the time it takes for a droplet to fall from the ball to the plane. 2. Find the time it takes for the ball to reach its maximum deviation from the vertical. 3. Find the horizontal distance traveled by the ball during this time. 4. Set up an equation equating the horizontal distance traveled by the ball to the horizontal distance traveled by the droplet. 5. Solve the equation to find the value of h.

Step 1: Time for a Droplet to Fall

The time it takes for a droplet to fall from the ball to the plane can be found using the equation for free fall:

h = (1/2) * g * t^2,

where h is the distance fallen, g is the acceleration due to gravity, and t is the time taken. Rearranging the equation, we get:

t = sqrt((2 * h) / g).

Step 2: Time for the Ball to Reach Maximum Deviation

The time it takes for the ball to reach its maximum deviation from the vertical can be found using the equation for simple harmonic motion:

t = (2 * π) / ω,

where ω is the angular frequency of the ball's motion. The angular frequency can be approximated as:

ω ≈ sqrt(g / L).

Step 3: Horizontal Distance Traveled by the Ball

The horizontal distance traveled by the ball during the time it takes to reach maximum deviation can be found using the equation for simple harmonic motion:

d = L * sin(φ0).

Step 4: Setting up the Equation

We can set up an equation equating the horizontal distance traveled by the ball to the horizontal distance traveled by the droplet:

d = v * t,

where v is the horizontal velocity of the droplet and t is the time it takes for the droplet to fall. The horizontal velocity of the droplet can be approximated as:

v ≈ L * φ0.

Substituting the expressions for d, v, and t, we get:

L * sin(φ0) = L * φ0 * sqrt((2 * h) / g).

Step 5: Solving the Equation

Simplifying the equation, we get:

sin(φ0) = φ0 * sqrt((2 * h) / g).

Using the approximation sin(φ) ≈ φ for small angles, we can rewrite the equation as:

φ0 ≈ φ0 * sqrt((2 * h) / g).

Simplifying further, we get:

1 ≈ sqrt((2 * h) / g).

Squaring both sides of the equation, we get:

1 ≈ (2 * h) / g.

Rearranging the equation, we find:

h ≈ g / 2.

Therefore, the value of h at which a droplet that detaches from the ball at its lowest position will hit the same point on the plane as a droplet that detaches from the ball at the moment of maximum deviation from the vertical is approximately h ≈ g / 2.

Conclusion

The value of h at which a droplet that detaches from the ball at its lowest position will hit the same point on the plane as a droplet that detaches from the ball at the moment of maximum deviation from the vertical is approximately h ≈ g / 2.

Please note that this solution assumes that the ball is released from rest and that the motion is idealized without considering factors such as air resistance.