The direction of friction does not flip when the ball stops rolling forwards and reverses direction.

The figure below shows a progression with time from right to left. At each stage, it is important to consider where the center of rotation is (blue circle). Above the rotation center, the velocity vectors point to the right, and below the velocity vectors point to the left.

0. Initially the ball moves to the left with a large amount of backspin (clockwise rotation). The slip velocity at the contact contains the effects of the forward motion as well as the rotation. The velocity on the top is small at this stage.

1. As friction slows down the velocity of the ball significantly, the rotation center moves downwards reducing the slip velocity and increasing the velocity on top.

2. At this point the ball has reversed the direction of travel, but the rotation is still in the same sense (clockwise). The rotation center is now below the center of the ball and the contact is still sliding to the left, while most of the ball is moving to the right.

3. At this point the sliding motion ceases and the ball is under pure rolling to the right, with rotation still clockwise. The rotation center is at the contact point (no-slip condition).

For cases 0 and 1, friction force opposes the forward motion of the ball. But for case 2, friction is acting in the same direction as the motion of the ball. This is where you think there is a problem. But in reality, this is not an issue, as friction still opposes the relative motion at the contact.

Doing a quick simulation I found the initial conditions for this to happen. Below the dot is at coordinates $(v,\omega R)$ and as the simulation progresses it moves down and to the left with a slope equaling $I/m R^2$ where $I$ is the mass moment of inertia, $m$ is the mass and $R$ is the radius.

If the graph crosses the x-axis and the ball still has some spin, it will start rolling backward (stages 2 and 3 above). But if the initial spin is not high enough, then the graph is going to cross the x-axis with still some forward speed.

The requirement is thus $\omega > \frac{m R\,I}{v} $ or for a solid ball with $I=\frac{2}{5} m R^2$ $$ \omega > \frac{5}{2} \frac{v}{R}$$

The direction of friction does not flip when the ball stops rolling forwards and reverses direction.

The figure below shows a progression with time from right to left. At each stage, it is important to consider where the center of rotation is (blue circle). Above the rotation center, the velocity vectors point to the right, and below the velocity vectors point to the left.

0. Initially the ball moves to the left with a large amount of backspin (clockwise rotation). The slip velocity at the contact contains the effects of the forward motion as well as the rotation. The velocity on the top is small at this stage.

1. As friction slows down the velocity of the ball significantly, the rotation center moves downwards reducing the slip velocity and increasing the velocity on top.

2. At this point the ball has reversed the direction of travel, but the rotation is still in the same sense (clockwise). The rotation center is now below the center of the ball and the contact is still sliding to the left, while most of the ball is moving to the right.

3. At this point the sliding motion ceases and the ball is under pure rolling to the right, with rotation still clockwise. The rotation center is at the contact point (no-slip condition).

For cases 0 and 1, friction force opposes the forward motion of the ball. But for case 2, friction is acting in the same direction as the motion of the ball. This is where you think there is a problem. But in reality, this is not an issue, as friction still opposes the relative motion at the contact.

Doing a quick simulation I found the initial conditions for this to happen. Below the dot is at coordinates $(v,\omega R)$ and as the simulation progresses it moves down and to the left with a slope equaling $I/m R^2$ where $I$ is the mass moment of inertia, $m$ is the mass and $R$ is the radius.

If the graph crosses the x-axis and the ball still has some spin, it will start rolling backward (stages 2 and 3 above). But if the initial spin is not high enough, then the graph is going to cross the x-axis with still some forward speed.

The requirement is thus $\omega > \frac{m\,v R}{I} $ or for a solid ball with $I=\frac{2}{5} m R^2$ $$ \omega > \frac{5}{2} \frac{v}{R}$$

The direction of friction does not flip when the ball stops rolling forwards and reverses direction.

The figure below shows a progression with time from right to left. At each stage, it is important to consider where the center of rotation is (blue circle). Above the rotation center, the velocity vectors point to the right, and below the velocity vectors point to the left.

0. Initially the ball moves to the left with a large amount of backspin (clockwise rotation). The slip velocity at the contact contains the effects of the forward motion as well as the rotation. The velocity on the top is small at this stage.

1. As friction slows down the velocity of the ball significantly, the rotation center moves downwards reducing the slip velocity and increasing the velocity on top.

2. At this point the ball has reversed the direction of travel, but the rotation is still in the same sense (clockwise). The rotation center is now below the center of the ball and the contact is still sliding to the left, while most of the ball is moving to the right.

3. At this point the sliding motion ceases and the ball is under pure rolling to the right, with rotation still clockwise. The rotation center is at the contact point (no-slip condition).

For cases 0 and 1, friction force opposes the forward motion of the ball. But for case 2, friction is acting in the same direction as the motion of the ball. This is where you think there is a problem. But in reality, this is not an issue, as friction still opposes the relative motion at the contact.

The direction of friction does not flip when the ball stops rolling forwards and reverses direction.

The figure below shows a progression with time from right to left. At each stage, it is important to consider where the center of rotation is (blue circle). Above the rotation center, the velocity vectors point to the right, and below the velocity vectors point to the left.

0. Initially the ball moves to the left with a large amount of backspin (clockwise rotation). The slip velocity at the contact contains the effects of the forward motion as well as the rotation. The velocity on the top is small at this stage.

1. As friction slows down the velocity of the ball significantly, the rotation center moves downwards reducing the slip velocity and increasing the velocity on top.

2. At this point the ball has reversed the direction of travel, but the rotation is still in the same sense (clockwise). The rotation center is now below the center of the ball and the contact is still sliding to the left, while most of the ball is moving to the right.

3. At this point the sliding motion ceases and the ball is under pure rolling to the right, with rotation still clockwise. The rotation center is at the contact point (no-slip condition).

For cases 0 and 1, friction force opposes the forward motion of the ball. But for case 2, friction is acting in the same direction as the motion of the ball. This is where you think there is a problem. But in reality, this is not an issue, as friction still opposes the relative motion at the contact.

Doing a quick simulation I found the initial conditions for this to happen. Below the dot is at coordinates $(v,\omega R)$ and as the simulation progresses it moves down and to the left with a slope equaling $I/m R^2$ where $I$ is the mass moment of inertia, $m$ is the mass and $R$ is the radius.

If the graph crosses the x-axis and the ball still has some spin, it will start rolling backward (stages 2 and 3 above). But if the initial spin is not high enough, then the graph is going to cross the x-axis with still some forward speed.

The requirement is thus $\omega > \frac{m R\,I}{v} $ or for a solid ball with $I=\frac{2}{5} m R^2$ $$ \omega > \frac{5}{2} \frac{v}{R}$$