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Pool ball hits wall

by Anony-mouse
Tags: ball, hits, pool, wall
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Anony-mouse
#1
May16-08, 02:52 PM
P: 61
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?
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Lojzek
#2
May16-08, 04:57 PM
P: 249
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
#3
May19-08, 08:58 AM
P: 61
Oh, I'm sorry. Yes, I know M, R, V0 and W0, and I need to know V1 and W1. (I donīt 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.

DaveC426913
#4
May19-08, 09:29 AM
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Pool ball hits wall

This is why there are guidelines about the format of homework questions. You should read em and follow em.
Anony-mouse
#5
May19-08, 09:37 AM
P: 61
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.

2. Relevant 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?

3. The attempt at a solution
I can already calculate alfa0 using V1 and N, but that's it.
Hootenanny
#6
May19-08, 09:51 AM
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Quote Quote by Anony-mouse View Post
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.

2. Relevant 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?

3. 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
#7
May19-08, 09:56 AM
P: 61
Quote Quote by Hootenanny View Post
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
#8
May19-08, 09:58 AM
P: 61
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
#9
May20-08, 12:47 PM
P: 61
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?
[tex]
\ m v_0^2 + I {\omega_0}^2 = m \underline{v_1}^2 + I {\omega_1}^2
[/tex]
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?
Hootenanny
#10
May21-08, 02:42 AM
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Quote Quote by Anony-mouse View Post
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?
[tex]
\ m v_0^2 + I {\omega_0}^2 = m \underline{v_1}^2 + I {\omega_1}^2
[/tex]
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
#11
May21-08, 12:17 PM
P: 61
Quote Quote by Hootenanny View Post
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 [tex]v_1[/tex] in two. So then I have one equation with 2 (or 3) variables to solve. [tex]v_1_y[/tex] and [tex]{\omega_1}^2[/tex]
So I only need to know what [tex]{\omega_1}[/tex] is and I can calculate [tex]v_1_y[/tex] from that.
Hootenanny
#12
May21-08, 12:37 PM
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Quote Quote by Anony-mouse View Post
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 [tex]v_1[/tex] in two. So then I have one equation with 2 (or 3) variables to solve. [tex]v_1_y[/tex] and [tex]{\omega_1}^2[/tex]
So I only need to know what [tex]{\omega_1}[/tex] is and I can calculate [tex]v_1_y[/tex] 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
#13
May21-08, 02:53 PM
P: 61
Quote Quote by Hootenanny View Post
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 [tex] v_1_y = -v_0_y [/tex], now I have
[tex]
\ m v_0^2 + I {\omega_0}^2 = m(\underline{v_1_x^2} - v_0_y^2) + I \underline{{\omega_1}^2}
[/tex]
I have underlined the ones I don't know.
Hootenanny
#14
May22-08, 02:25 AM
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Quote Quote by Anony-mouse View Post
Ok, so since [tex] v_1_y = -v_0_y [/tex], now I have
[tex]
\ m v_0^2 + I {\omega_0}^2 = m(\underline{v_1_x^2} - v_0_y^2) + I \underline{{\omega_1}^2}
[/tex]
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
#15
May22-08, 10:51 AM
P: 61
Quote Quote by Hootenanny View Post
Good. So when the ball loses some angular velocity, something else must increase to conserve energy.
Yes, [tex]v_1_x[/tex] should increase.
So to calculate it, I can use this equation:
[tex]
\sqrt{(m v_0^2 + I {\omega_0}^2 - I \underline{{\omega_1}^2})/m + v_0_y^2} = \underline{|v_1_x|}
[/tex]

Now, to calculate [tex]\omega_1[/tex], I should take the perpendicular velocity and multiply it by the ball mass and some friction coefficient to get how much is [tex]\Delta\omega[/tex] right?
Hootenanny
#16
May22-08, 02:15 PM
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Quote Quote by Anony-mouse View Post
Yes, [tex]v_1_x[/tex] should increase.
So to calculate it, I can use this equation:
[tex]
\sqrt{(m v_0^2 + I {\omega_0}^2 - I \underline{{\omega_1}^2})/m + v_0_y^2} = \underline{|v_1_x|}
[/tex]
Looking good

The equation would be much easier to solve if you could relate the final angular velocity to the final linear velocity ...
Lojzek
#17
May22-08, 03:31 PM
P: 249
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
#18
May23-08, 09:18 AM
P: 61
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 [tex]r=(0, \pm R, 0)[/tex], where R is the radius of the ball)
So the velocity at that point is [tex]v_p = r \times {\omega_0}[/tex]. To this I also have to add the parallel velocity of the ball. Is this correct?
Having that, I think the frictional force is:
[tex] F_f = \mu \frac{m\Delta{v}}{\Delta{t}} \frac{v_p + v_x}{|v_p + v_x|} [/tex]
Here, [tex]\mu[/tex] is the friction coefficient, and [tex]\Delta{t}[/tex] is how long it acts.
[tex]\Delta{v}[/tex] should be the change in momentum of the ball, but I'm not sure what it is exactly...


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