- #1

FranzDiCoccio

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Thread moved from the technical forums to the schoolwork forums

**Summary::**I'd like to check my understanding of standard problems where a billiard ball resting on a plane is hit horizontally at some height above its center

So the situation is that a ball of mass ##m## and radius ##r## is at rest on a horizontal surface. There is friction between the ball and the surface. The ball receives a horizontal impulse at some height ##h## above the surface (##r<h<2r##) and starts moving.

One possible question is: "find ##h## such that the ball rolls without sliding". This is discussed e.g. in thread 1 and thread 2.

After looking at these threads and some more sources I'd consider the angular momentum of the ball wrt an axis that lies on the surface and is perpendicular to the impulse (i.e. to the ball velocity).

This choice gets rid of the contribution of the friction to the rotational impulse on the ball. Friction in principle still affects the overall impulse. However, as suggested in thread 2, one assumes that the impulse is sudden, i.e. it lasts for a very short time. Therefore the impulse of the friction is negligible when compared to the impulse of the cue stick.

Hence, taking into account that the initial momentum is ##\vec p_0=0##, we have ##\vec J = \vec p_1=m \vec v_1##.

Since the impulse is horizontal and at an heigth ##h## above the table, and since the initial angular momentum is ##\vec L_0=0##, we have that

##J h = L_1##.

After the impulse, the ball moves with velocity ##v_1## and rotates about its center with angular velocity ##\omega_1##. Its angular momentum wrt the chosen axis is ##L_1=I \omega_1 + m v r_1##.

If we want the ball to roll without sliding, it should be ##\omega_1 r = v_1##, so that ##L=(1+\kappa) m v_1 r## where ##\kappa =2/5## for a homogeneous ball. Therefore we get ##(1+\kappa) m v_1 r=m v_1 h##, which gives ##h=(1+\kappa) r## (or ##\kappa r## above the center).

In thread 1 the result depends both on the force from the cue stick and the friction. However, I guess that we can say that the former must be extremely large, since it produces an effect on such a short time scale. On the other hand, as we mention, the friction is much smaller.

As far as I understand, the OP of thread 2 obtained the solution like in thread 1, but ignored the friction from the start (and he was not very sure why). I found two discussions on the web (here and here) which also seem to ignore the friction from the table (but perhaps the authors make a comment about it being negligible... I do not understand the language).

A variant of the problem is "what happens if the ball is hit at a different height above its center?". In this case the ball will initially slide on the table, and end up rolling without sliding at a different velocity.

If I'm not missing something the strategy should be similar. The force responsible for the change in velocity is the kinetic friction between the ball and the table. Since the corrisponding torque about the chosen axis vanishes, after the ball starts moving ## \Delta \vec L=0## (##L_2=L_1##).

But on the one hand ##L_1-L_0=L_1= J h = m v_1 h## and on the other hand ##L_2 = m v_2 r (1+\kappa)##. Therefore ##v_2 = v_1 \dfrac{h}{r (1+\kappa)}##.

This makes sense: if the ball is hit above the "critical height" it will end up rolling faster. If the cue stick hits below that height, the ball will end up rolling slower.

It should be possible to hit the ball so that it stops, or even ends up going backwards. I'm not seeing this, but it's very late here and I'm a bit tired. Perhaps this is not possible with an horizontal impulse.

Does all this makes sense?