Question

As shown in figure, a non-stretchable string of length $2l$ has one end fixed to a nail at point A, and a small ball is tied to the other end. The small ball can move in the vertical plane. Initially, the ball is on the same horizontal line with A, and is launched with an initial velocity $v_0$ straight downward. To the left of point A, at a distance $l$ from A, there is another nail at point B. When the string passes over point B, the ball will move upward. The minimum value of the initial speed $v_0$ required for the ball to eventually collide with the nail at point A is $\sqrt{\alpha gl}$. The value of $\alpha$ is:

Answer 1

3

Answer 2

The problem describes a ball attached to a string of length $2l$, fixed at point A. Initially, the ball is at a horizontal distance $2l$ from A, and is launched downward with speed $v_0$. There is a nail at point B, at a distance $l$ to the left of A. We assume A is at the origin (0,0) and B is at (-l, 0). The initial position of the ball is P at (2l, 0). The string length is $2l$. The ball is launched with $v_0$ downward.

Let's assume the transition to swinging around B occurs at the point (-2l, 0). The initial position is (2l, 0). The vertical drop is 0. By energy conservation, speed at (-2l, 0) is $v_0$. At this point, the ball is at a distance $l$ from B. So it starts swinging around B with radius $l$ from this point. For the ball to collide with A, it must reach A. The point A is at a distance $l$ from B. So A is on the circle of radius $l$ around B.

The ball swings around B from (-2l, 0) to A(0,0). Both points are on the horizontal line through B. The lowest point on the path is (-l, -l).

For the ball to reach A, it must pass through the lowest point (-l, -l). The minimum speed at the lowest point to reach the same height is 0. However, to reach A, it must go up from the lowest point. For the ball to reach A, it must have sufficient energy to reach A.

Let's consider the condition for completing the circle around B. The speed at the top (-l, l) must be at least $\sqrt{gl}$. By conservation of energy from (-2l, 0) to (-l, l) while swinging around B:

$\frac{1}{2}mv_0^2 + mg(0) = \frac{1}{2}m(\sqrt{gl})^2 + mg(l)$

$\frac{1}{2}mv_0^2 = \frac{1}{2}mgl + mgl = \frac{3}{2}mgl$

$v_0^2 = 3gl$.

So $v_0 = \sqrt{3gl}$.

This assumes the transition happens at (-2l, 0). Is it possible to reach (-2l, 0) by swinging downwards from (2l, 0)? Yes, by swinging through an angle of $\pi$.

Let's verify if the string can be constrained by B at (-2l, 0). At (-2l, 0), the string is from A(0,0) to (-2l, 0). This is a horizontal line segment of length $2l$. B is at (-l, 0). B is on this line segment. So the string passes over B at this point.

Therefore, $v_0 = \sqrt{3gl}$. Comparing with $\sqrt{\alpha gl}$, we have $\alpha = 3$.