# Pool ball hits wall

Anony-mouse
Hello, I have a question.
I have a pool ball mass M and radius R, that hits the pool table wall, at speed V0, angular speed W0 at an angle alfa0, and then goes away with an angle alfa1, speed V1 and angular speed W1.

If the ball wasn't spinning then I think it would be easy to solve with linear algebra, but since it can be spinning in weird ways (W0 is a 3d vector), I think the equations involve the friction with the wall, and that is beyond my expertise. Can anyone help me?

Lojzek
What is known and what must be calculated? I would suppose that the parameters after the
recoil are unknown, but you listed them among the known data?

Anony-mouse
Oh, I'm sorry. Yes, I know M, R, V0 and W0, and I need to know V1 and W1. (I don´t know alfa0, but it should be easily calculated using V0 and the normal to the wall right?)

edit: Oh, I would show what I've done so far, but I have no idea where to start. Any pointers would be appreciated.

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Gold Member
This is why there are guidelines about the format of homework questions. You should read em and follow em.

Anony-mouse
I'm sorry. Here is the reformulated question:

I have a pool ball mass M and radius R, that hits the pool table wall, at speed V0, angular speed W0 at an angle alfa0, and then goes away with an angle alfa1, speed V1 and angular speed W1.

Things I know: M, R, V0 and W0. I also know the normal to the wall (N). I'm guessing since it's a pool ball, that the collision is elastic.
Things I want to know: V1 and W1.

## Homework Equations

If the ball wasn't spinning then I think it would be easy to solve with linear algebra, but since it can be spinning in weird ways (W0 is a 3d vector), I think the equations involve the Friction with the wall, and that is beyond my expertise. Can anyone help me with the equations I need to use?

## The Attempt at a Solution

I can already calculate alfa0 using V1 and N, but that's it.

Staff Emeritus
Gold Member
I'm sorry. Here is the reformulated question:

I have a pool ball mass M and radius R, that hits the pool table wall, at speed V0, angular speed W0 at an angle alfa0, and then goes away with an angle alfa1, speed V1 and angular speed W1.

Things I know: M, R, V0 and W0. I also know the normal to the wall (N). I'm guessing since it's a pool ball, that the collision is elastic.
Things I want to know: V1 and W1.

## Homework Equations

If the ball wasn't spinning then I think it would be easy to solve with linear algebra, but since it can be spinning in weird ways (W0 is a 3d vector), I think the equations involve the Friction with the wall, and that is beyond my expertise. Can anyone help me with the equations I need to use?

## The Attempt at a Solution

I can already calculate alfa0 using V1 and N, but that's it.
If the collision is elastic (i.e. the wall does not deform and the ball does not slip) then what do you know about the total energy of the collision?

Anony-mouse
If the collision is elastic (i.e. the wall does not deform and the ball does not slip) then what do you know about the total energy of the collision?

The total energy should stay the same. Some energy of the rotation should be lost because of friction, but I don't know where it would go (to the velocity of the ball?).

Anony-mouse
Oh, I just found out that I didn't see something. The wall is axis aligned, so it should really simplify the equations.

Anony-mouse
Ok, so this is what I got. I need to have the same energy before and after the collision. But some of the angular velocity will be lost because of friction with the wall. So first I need to see what the angular velocity is after the collision, and then I can use something like this right?
$$\ m v_0^2 + I {\omega_0}^2 = m \underline{v_1}^2 + I {\omega_1}^2$$
Where I is the moment of inertia of a sphere. From this I can calculate V1. Is this right?
What I still need to figure out is how to calculate the angular velocity lost because of friction. Any pointers?

Staff Emeritus
Gold Member
Ok, so this is what I got. I need to have the same energy before and after the collision. But some of the angular velocity will be lost because of friction with the wall. So first I need to see what the angular velocity is after the collision, and then I can use something like this right?
$$\ m v_0^2 + I {\omega_0}^2 = m \underline{v_1}^2 + I {\omega_1}^2$$
Where I is the moment of inertia of a sphere. From this I can calculate V1. Is this right?
What I still need to figure out is how to calculate the angular velocity lost because of friction. Any pointers?
You're looking good so far. What can you say about the [perpendicular/parallel] components of the linear velocity before and after the collision? Do they both change?

Anony-mouse
You're looking good so far. What can you say about the [perpendicular/parallel] components of the linear velocity before and after the collision? Do they both change?
The perpendicular component would have to change, because otherwise the ball would go right through the wall, so I'm guessing the parallel component stays the same?
So, let's say the wall is horizontal. The y component would change, while the x component would stay the same.
Then, I can take the equation from my last post and split $$v_1$$ in two. So then I have one equation with 2 (or 3) variables to solve. $$v_1_y$$ and $${\omega_1}^2$$
So I only need to know what $${\omega_1}$$ is and I can calculate $$v_1_y$$ from that.

Staff Emeritus
Gold Member
The perpendicular component would have to change, because otherwise the ball would go right through the wall, so I'm guessing the parallel component stays the same?
So, let's say the wall is horizontal. The y component would change, while the x component would stay the same.
Then, I can take the equation from my last post and split $$v_1$$ in two. So then I have one equation with 2 (or 3) variables to solve. $$v_1_y$$ and $${\omega_1}^2$$
So I only need to know what $${\omega_1}$$ is and I can calculate $$v_1_y$$ from that.
Perhaps my previous hint was a little misleading. Both components of the velocity do change, however, since the only force acting in the perpendicular direction is the normal reaction force, the magnitude of the perpendicular component of the velocity doesn't change, but obviously it's direction is reversed. The magnitude of the parallel component, however, does change.

Anony-mouse
Perhaps my previous hint was a little misleading. Both components of the velocity do change, however, since the only force acting in the perpendicular direction is the normal reaction force, the magnitude of the perpendicular component of the velocity doesn't change, but obviously it's direction is reversed. The magnitude of the parallel component, however, does change.
Ok, so since $$v_1_y = -v_0_y$$, now I have
$$\ m v_0^2 + I {\omega_0}^2 = m(\underline{v_1_x^2} - v_0_y^2) + I \underline{{\omega_1}^2}$$
I have underlined the ones I don't know.

Staff Emeritus
Gold Member
Ok, so since $$v_1_y = -v_0_y$$, now I have
$$\ m v_0^2 + I {\omega_0}^2 = m(\underline{v_1_x^2} - v_0_y^2) + I \underline{{\omega_1}^2}$$
I have underlined the ones I don't know.
Good. So when the ball loses some angular velocity, something else must increase to conserve energy.

Anony-mouse
Good. So when the ball loses some angular velocity, something else must increase to conserve energy.

Yes, $$v_1_x$$ should increase.
So to calculate it, I can use this equation:
$$\sqrt{(m v_0^2 + I {\omega_0}^2 - I \underline{{\omega_1}^2})/m + v_0_y^2} = \underline{|v_1_x|}$$

Now, to calculate $$\omega_1$$, I should take the perpendicular velocity and multiply it by the ball mass and some friction coefficient to get how much is $$\Delta\omega$$ right?

Staff Emeritus
Gold Member
Yes, $$v_1_x$$ should increase.
So to calculate it, I can use this equation:
$$\sqrt{(m v_0^2 + I {\omega_0}^2 - I \underline{{\omega_1}^2})/m + v_0_y^2} = \underline{|v_1_x|}$$
Looking good

The equation would be much easier to solve if you could relate the final angular velocity to the final linear velocity ...

Lojzek
It is possible to obtain another simple equation connecting the change of parallel velocity and the change of angular velocity during the collision.
Hint: although you don't know the friction force and how long it acts, you do know that it
causes both acceleration and angular acceleration, because it does not act on the center of mass, but a distance R away (so it creates a torque).

Anony-mouse
Ok, so I need the friction force right?
For this, I need the velocity of the ball at the point of the collision.
To do this, I take a vector extending from the center of the ball to the point of the collision (if the wall is horizontal, it should be $$r=(0, \pm R, 0)$$, where R is the radius of the ball)
So the velocity at that point is $$v_p = r \times {\omega_0}$$. To this I also have to add the parallel velocity of the ball. Is this correct?
Having that, I think the frictional force is:
$$F_f = \mu \frac{m\Delta{v}}{\Delta{t}} \frac{v_p + v_x}{|v_p + v_x|}$$
Here, $$\mu$$ is the friction coefficient, and $$\Delta{t}$$ is how long it acts.
$$\Delta{v}$$ should be the change in momentum of the ball, but I'm not sure what it is exactly...

Staff Emeritus
Gold Member
See my previous hint: Can you relate $|\omega_1|$ to $|v_1|$?

Lojzek
See my previous hint: Can you relate $|\omega_1|$ to $|v_1|$?
I hope you don't mean v=omega*R? This equation is incompatible with the conservation of energy! If the collision is totaly elastic and the ball does not slip, then it's initial rotation (if different from initial parallel velocity/R) would cause deformation (including the change of non-diagonal components of strain tenzor), which will then release it's elastic energy and rotate the ball with angular velocity that does not equal v/R.

Staff Emeritus
Gold Member
See my previous hint: Can you relate $|\omega_1|$ to $|v_1|$?
I hope you don't mean v=omega*R? This equation is incompatible with the conservation of energy! If the collision is totaly elastic and the ball does not slip, then it's initial rotation (if different from initial parallel velocity/R) would cause deformation (including the change of non-diagonal components of strain tenzor), which will then release it's elastic energy and rotate the ball with angular velocity that does not equal v/R.
Sorry, that was a typo! Those should be time derivatives, I missed off the dots...

Edit: And of course I meant $v_\parallel$ rather than $v_1$. Furthermore, there should be no subscript on the angular velocity. I should type slower in future...

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Lojzek
To anony-mouse:

No, this is wrong: if you assume that the energy is conserved (no slipping), then you can't assume that the friction force equals $$\mu$$ *perpendicular force: all you know is that it is smaller (and directed in the opposite of slipping velocity). But as I sad, you don't really need the friction force. You just define unknown variable p as the time integral of friction force and then you express it from:

1. the change of parallel velocity (depends on p)
2. the change of angular momentum

This gives you a relation between these two quantities.
Just remember how the change of momentum is related to force and how the change of angular momentum is related to torque.

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Anony-mouse
See my previous hint: Can you relate $|\omega_1|$ to $|v_1|$?
I thought I was relating them... using $$F_f$$ I can calculate $$\Delta{\omega}$$, and $$F_f$$ depends on $$v_\parallel$$.
Unless there is another relation I'm missing?

Phizyk
I think, what we required to consider two situations, in the first, the ball slips during the entire impact time, and in the second the slipping stops before the end of the impact time...

Lojzek
I think, what we required to consider two situations, in the first, the ball slips during the entire impact time, and in the second the slipping stops before the end of the impact time...
This is a possible approach. However in those two cases we need to drop the conserved energy equation (energy is lost due friction), while we get one new equation instead (the last Anony-mouse's equation in fact corresponds to the situation where the ball slipps during the entire impact time).
The other approach is constant energy approximation: if the friction is high enough and the ball is soft enough, then there is almost no slipping and energy is conserved (however part of the energy could be transformed into vibration energy).
But no matter which way we chose, we still need to connect time derivative of momentum with time derivative of angular momentum.

Phizyk
We need not use the conservation of energy... Torque's friction is $$F(t)R=I\frac{d\omega}{dt}$$
but
$$F(t)=-\frac{dp_{||}}{dt}$$
and
$$R(t)=\frac{dp_{_}}{dt}$$
$$T(t)=\mu{R(t)}$$
we can integrate these equations and get rid of time...

Phizyk
$$p$$ is of course component of momentum, perpendicular to the wall...

Phizyk
My solution:
$$v_{X0}=v_{0}cos\alpha_{0}$$
$$v_{Y0}=v_{0}sin\alpha_{0}$$
$$N(t)$$- force normal to the wall
we have:
$$\frac{dp_{X}}{dt}=-\mu{N(t)}$$
$$I\frac{d\omega}{dt}=\mu{N(t)R}$$
$$\frac{dp_{Y}}{dt}=N(t)$$
so
$$\frac{dp_{X}}{dt}=-\mu{\frac{dp_{Y}}{dt}}$$
and
$$I\frac{d\omega}{dt}=\mu{R\frac{dp_{Y}}{dt}}$$

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Phizyk
First case - ball slips all the time
so
$$v_{X1}=v_{X0}-2\mu{v_{Y0}}$$
and
$$\omega_{1}=\omega_{0}+2v_{Y0}\mu{\frac{mR}{I}}$$
for
$$v_{X0}<\omega_{0}R+2v_{Y0}\mu{(\frac{mR^{2}}{I}+1)}$$

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Phizyk
Second case - slipping stops before end collision:
so $$\omega_{1}=\frac{v_{X1}}{R}$$
angular momentum is constant
$$R\frac{dp_{X}}{dt}+I\frac{d\omega}{dt}=0$$
we have
$$v_{X1}=\frac{v_{X0}+\frac{I}{mR}\omega_{0}}{\frac{I}{mR^{2}}+1}$$
for $$v_{X0}>\omega_{0}R+2v_{Y0}\mu{(1+\frac{mR^{2}}{I})}$$

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Phizyk
You have to use Pythagoras
$$v_{1}=\sqrt{v_{X1}^{2}+v_{Y1}^{2}}$$
where
$$v_{Y1}=v_{Y0}=v_{0}sin\alpha_{0}$$ in both cases and $$v_{X1}$$, which I counted.

Anony-mouse
Thanks everyone for your help, and thanks Phizyk for the equations. There is something I didn't understand.
$$\omega$$ is a vector right?
I don't see how to do this comparision $$v_{X0}<\omega_{0}R+2v_{Y0}\mu{(\frac{mR^{2}}{I}+1) }$$

Also, It seems weird to me that in the equation $$v_{X1}=v_{X0}-2\mu{v_{Y0}}$$, the angular velocity isn't involved in calculation $$v_{X1}$$. Is there something I'm missing?