Bowling: What's the effect of rotation of the ball has on the pin action?

[JonsDuu]

Does more rotation of the bowling ball result in faster pin?

Here's my understanding, please check my physics.

The ball has 2 forms of energy, kinetic due to its momentum and rotational due to its angular momentum. When it hit the pin, there're transfer of energy. The pin will pick up kinetic energy and rotational energy. How much is pick up is based on conservation of momentum (translational & angular) and conservation of energy and the relative angle between the ball original motion to that of the line of force between the ball & pin center of mass.

If the ball & pin has no friction, would I be correct in say the rotation of the ball has no bearing on the pin kinetic & rotational energy?

How would you calculate the pin kinetic & rotational energy as a function of the ball rotation?

 

[Rap]

Does more rotation of the bowling ball result in faster pin?

Here's my understanding, please check my physics.

The ball has 2 forms of energy, kinetic due to its momentum and rotational due to its angular momentum. When it hit the pin, there're transfer of energy. The pin will pick up kinetic energy and rotational energy. How much is pick up is based on conservation of momentum (translational & angular) and conservation of energy and the relative angle between the ball original motion to that of the line of force between the ball & pin center of mass.

If the ball & pin has no friction, would I be correct in say the rotation of the ball has no bearing on the pin kinetic & rotational energy?

How would you calculate the pin kinetic & rotational energy as a function of the ball rotation?

You have it almost right, but there are three objects to consider, not two. There is the ball, which has linear and angular velocity (vb and wb), the pin which has linear and angular velocity (vp and wp), and the Earth (the bowling lane), which has linear and angular velocity (ve and we). Remember that the w's and v's are vectors, they have a magnitude and a direction. The linear velocity of the ball gives it linear momentum (pb=mb*vb) and linear kinetic energy (Elb=mb*vb^2/2). The angular velocity of the ball gives it angular momentum (Lb=Ib*wb) and angular energy (Erb=wb.(Ib*wb)/2) where Ib is the moment of inertia of the ball, and it is a tensor. Likewise there are linear and angular momenta and energies for the pin and the Earth.

The Earth is special - we can assume it has infinite mass. This will make the math simpler, but we cannot ignore it. There are three conservation laws - the conservation of linear momentum, angular momentum, and total (linear plus rotational) energy. We cannot apply these laws unless we take the Earth into account, because the pin and the ball are giving or receiving them from the Earth. When it comes to conservation of energy, the frictional forces convert some kinetic and rotational energy into heat, so that gets quite complicated. I think when you solve this problem, you will not need to take the Earth into account as an object which absorbs or transmits energy or momenta, but that means you cannot use the conservation laws in your solution.

You will probably solve it by using the forces. There is always the force of gravity, downward on the pin and ball, but at the moment the ball strikes the pin, it will not be a factor except to tell you what the friction is between the Earth and the ball and the Earth and the pin. When the ball hits the pin, there is a sudden, momentary force by the ball on the pin which pushes it away from the ball, and and the equal but opposite force is applied to the ball. At the same time, there will be a frictional force on the ball and pin due to the Earth, applied at the point of contact.

I once solved the equations for the collision of two billiard balls on a billiard table, and it was kind of complicated. Solving things for the collision of a bowling ball with a pin will be a much more complicated, unless the center of mass of the pin happens to be the same height off the ground as the center of mass of the ball. Do you know if that is true?

Anyway, I don't remember how I solved it, but I remember it was kind of tricky, not being able to apply conservation laws, and having a force of friction and impact applied over a very short time (the duration of the impact), but its doable. I will try to remember how I did it, if you really want to do it.

To answer your specific questions, the rotation of the ball imparts linear and angular rotation to the pin and vice versa, generally speaking. If there is no friction between ball and pin, then yes, the rotation of the ball will have no bearing on the pin's kinetic and rotational energy or momentum as a result of the collision between the ball and the pin. The pin will nevertheless acquire rotational velocity, however, because of the frictional force of the Earth on the pin.

 

[rcgldr]

If the ball & pin has no friction, would I be correct in say the rotation of the ball has no bearing on the pin kinetic & rotational energy?

There is also the issue of the friction between the ball and the lane. If that friction was significant, then some of the balls angular momentum would get translated into linear momentum during the collisions (conservation of angular momentum would need to include the Earth that the lane is attached to explain this). Based on the few times I've seen bowling on tv, it seems there isn't much friction between the ball and the lane, but I also recall hearing that the lanes are not waxed evenly on those tv shows.

In the case of billiard balls, spin on the cue ball has some effect on the target balls direction from a collision, and a significant effect on the cue ball's direction after collision, because of friction between billard balls and the felt of the table.

 

[Rap]

Based on the few times I've seen bowling on tv, it seems there isn't much friction between the ball and the lane, but I also recall hearing that the lanes are not waxed evenly on those tv shows.

In the case of billiard balls, spin on the cue ball has some effect on the target balls direction from a collision, and a significant effect on the cue ball's direction after collision, because of friction between billard balls and the felt of the table.

There is enough friction between ball and lane to impart a curve to the ball's trajectory. A reasonably strong bowler can strike the center pin by releasing the ball at the edge of the lane and parallel to that edge. Also, the effect of a spinning cue ball on the target ball is significant enough to "throw" the target ball a few millimeters over the distance of a meter or two. Especially if the cue ball is touching the target ball before being struck. Striking the cue ball off center left or right will significantly "throw" the target ball away from the line between their center of masses.

 

[JonsDuu]

Solving things for the collision of a bowling ball with a pin will be a much more complicated, unless the center of mass of the pin happens to be the same height off the ground as the center of mass of the ball. Do you know if that is true?

The CoM of the pin is higher than the CoM of the ball. How much higher, I don't know. It depends on the manufacturer (they're allowed to have hollow space within the pin, the location, shape, volume is up to them).

Friction between lane and ball you can almost take out of the equation. In a nutshell, when the ball hit the pin, the rotation is along the translation line; eg not likely to have rotation normal to translation. I said almost because it really depends on how the bowler tilt the ball at the release point. Most good bowler tilt it in a manner such that the ball will roll toward the end.

It'll be really hard if you take the entire trajectory of the ball into account. At the release point, the bowler impart kinetic energy, rotation energy, and axis tilt to the ball. The friction between the ball is highly variable (amount of oil along the path, the dynamic friction surface of the ball, the ball moment of inertia). These can be generalize that all of these forces act to change the direction of the ball such that when it hit the pin, the ball is rotate along the line of motion.

Friction between the ball and pin will cause the ball to rotate slower (eg transfer some of its rotational energy to the pin). The pin will rotate in the opposite direction.

The thing I can't figure out is how to determine the how significant this is.

 

[JonsDuu]

Based on the few times I've seen bowling on tv, it seems there isn't much friction between the ball and the lane, but I also recall hearing that the lanes are not waxed evenly on those tv shows.

Oiling pattern is up to the event organizer. Oil pattern is never uniform. The lane is 40 boards wide, each board 1 in wide, and 60 feet from foul line to the 1st pin. Oil is laid down from 4 inch after the foul line down to between 35 to 42 feet. Depending on the house, they may strip the backend (eg remove the oil to make it highly frictional) or they can be lazy and don't clean it, in which case there'll be residual oil making it less frictional. There are many different type of ball surface, from the plastic very low co-efficient of friction up to the modern ball, whose co-efficient of friction is very low when in oil and then very high when the ball slightly heat up as it translate the lane. This make the modern ball slide down to about 35-45 feet before the friction take effect, then the ball change direction due to the friction. This is why the shape of the trajectory is not a parabola, but more like a hockey stick.

 

[Rap]

The CoM of the pin is higher than the CoM of the ball. How much higher, I don't know. It depends on the manufacturer (they're allowed to have hollow space within the pin, the location, shape, volume is up to them).

Thinking about it, this may actually make things easier. The ball will kick the pin upwards, (as long as the frictional force is weak enough) and there will be no friction force on the pin due to the ground.

Friction between lane and ball you can almost take out of the equation. In a nutshell, when the ball hit the pin, the rotation is along the translation line; eg not likely to have rotation normal to translation. I said almost because it really depends on how the bowler tilt the ball at the release point. Most good bowler tilt it in a manner such that the ball will roll toward the end.

Ok, that simplifies things too. Just to be clear, let's talk about axis of rotation. A rolling ball has its axis of rotation perpendicular to the velocity and parallel to the surface of the lane. When the ball strikes the pin, the velocity due to rotation at the point of contact will be downward.

It'll be really hard if you take the entire trajectory of the ball into account. At the release point, the bowler impart kinetic energy, rotation energy, and axis tilt to the ball. The friction between the ball is highly variable (amount of oil along the path, the dynamic friction surface of the ball, the ball moment of inertia). These can be generalize that all of these forces act to change the direction of the ball such that when it hit the pin, the ball is rotate along the line of motion.

No, we don't want to take the entire trajectory into account, only what happens upon impact, right?

Friction between the ball and pin will cause the ball to rotate slower (eg transfer some of its rotational energy to the pin). The pin will rotate in the opposite direction.

The thing I can't figure out is how to determine the how significant this is.

The best way is to do some experiments. You have to control things in an experiment, however, so it might get kind of complicated. You could hang a ball and a pin by ropes, pull the ball back, let it hit the pin. You can use a pendulum equation to calculate the velocity of the ball when it hits the pin. Adjust the ropes so it hits the same place it would hit if the pin and ball were on the lane. Then you could rotate the ball a number of times, and measure how much it spins as the rope unwinds a function of time, then hit the pin again with the spinning ball, measure how much it throws the pin away from the direction that the ball hit.

The pin is made out of plastic or wood? Get a ball of the same plastic or wood, same mass as the pin, it will be easier to measure how the plastic or wood ball behaves, rather than trying to follow the pin after it gets hit.

The other way is to find some literature on the subject. Maybe someone has done something like this before. It might be easy to estimate the coefficient of friction, but you will have to know the duration of the collision, and that involves the elastic properties of the ball and pin. That could get complicated.