The physics of billiards. (Describing the motion)

[JuanC97]

A. The problem: (If you want the short version just read the underlined statements)
I need to describe the motion of a cue ball over the cloth of a billiard table.
For rule, I can't use equations for rotational motion or 'Work and Energy theorems'. I have some ideas but I'm not sure about the concepts and procedures. I'd appreciate it if you take a look to my reasoning and give me some pieces of advices (I'm just trying to describe traslation)

B. Relevant equations:
Friction force due to bearing (Rolling resistance): F_rr = C_rr* N
C_rr is the Rotational resistance coefficient ; N is the Normal force.
Dynamic (Kinetic / Sliding) friction: Fr_d = μ_d * N

C. The attempt at a solution: (Reasoning)
See the first Image: The white ball is at rest. There are frictional forces.
The cue hits the ball in its center of mass and the force it's aplied parallel to the plane.
Because of the motion it will be a frictional force acting opposite the movement but...
Here's the problem: the ball will roll or it will slide ? (What kind of friction goes into the equations?)

I think these are the most probable situations:
1. The cue hits the ball and its motion is Pure sliding. A few meters later, the motion is pure rolling.
2. The cue hits the ball and its motion combines Sliding and rolling. Then, is just pure rolling.

D. The questions:
* In both cases I have to calculate the distance traveled before starting to roll purely. How?
* How can I find an expression of the minimum force needed to make each of those possiblities real?

It's my first question in PF.
I will appreciate every hint and piece of advice that you could give me. :)

 

[UVCatastrophe]

Draw the free body diagram. Supposing the initial (center of mass) motion is to the right, you need to answer the following questions:

1. Which forces are acting on the ball, and in which direction?
2. Where on the ball are these forces acting?

You should find that the rolling resistance and dynamic friction oppose each other. Which one is larger? It depends on whether ## C_r > \mu_d ## or vice versa; the two possibilities correspond to the two scenarios you described above.

To calculate the distance, use the fact that net torques cause angular acceleration (## \tau = I \alpha ##, and ##I = \frac{2}{5} M R^2## for a sphere) and the kinematic equations for rotational motion.

 

[haruspex]

1. The cue hits the ball and its motion is Pure sliding. A few meters later, the motion is pure rolling.

Pure sliding at first, yes. What is the nature of the motion between then and pure rolling?

In both cases I have to calculate the distance traveled before starting to roll purely.

you need to answer the following questions:

1. Which forces are acting on the ball, and in which direction?
2. Where on the ball are these forces acting?

Agreed.

You should find that the rolling resistance and dynamic friction oppose each other.

Eh? The two will act in the same direction, and will be effectively indistinguishable until pure rolling is achieved, so might as well consider it all to be friction.

 

[JuanC97]

Ok, I'm working on the equations for rolling motion and traslation. (They're almost done but, I'll share them another day). Now I have this question:

I found on Internet that the time that the ball slips (before starting to roll at 100%) is: [tex] \Delta t = \frac{2}{7} \frac{V_o}{\mu g} [/tex]
But... When I try to deduce it I get another thing. Here's what I did:
[tex] \left[Eq. 1\right] \; \; \; I\Delta \omega = \tau \Delta t \; \longrightarrow \; I(\omega_f - \omega_o) = FR\Delta t \\
\left[Eq. 2\right] \; \; \; m\Delta V = F\Delta t \; \longrightarrow \; m(V_f - V_o) = F\Delta t [/tex]
Since V=WR when the ball starts the pure rolling motion and W(0)=0, we can rewrite these equations.
[tex] \left[Eq. 1\right] \; \; \; I\omega_f = FR\Delta t \; \longrightarrow \: \omega_f=FR\Delta t / I
\\
\left[Eq. 2\right] \; \; \; m(\omega_f R - V_o) = F\Delta t \; \longrightarrow \; \frac{5\, m\, R^2 \, F \, \Delta t}{2\, m\, R^2} - mV_o = F\Delta t
\\
\left(\frac{5}{2} - 1\right)F\Delta t = mV_o \; \longrightarrow \; \frac{3}{2}F\Delta t = mV_o \; \longrightarrow \; \Delta t = \frac{2mV_o}{3F} = \frac{2mV_o}{3mg\mu} = \frac{2}{3}\frac{V_o}{g\mu} [/tex]
And this should be true if the ball moves along a plane with friction before starting the phase of pure rolling.
So... which of those " Δt " is the right time that takes the ball to start the pure rolling motion?

 

[haruspex]

You're not being consistent with your signs. I can't point to a specific error unless you define which is the positive sense for forces, velocities, rotations... At the end you should have ##(\frac 52+1)##.
Btw, you can simplify it a little by considering conservation of angular momentum about a fixed point at ground level. The frictional force has no moment about such a point.

 

[JuanC97]

I use this convection:

Forces that goes in the same direction of the linear acceleration are positive.
Torques that goes in the clockwise sense are positive (Because these torques goes in the same direction of the angular acceleration)

 

[haruspex]

Forces that goes in the same direction of the linear acceleration are positive.

Then that's opposite to the direction of travel, right? So what sign are you using for velocities?

Torques that goes in the clockwise sense are positive

That doesn't help me because there's no diagram. Do you mean the direction of angular acceleration is positive?

 

[JuanC97]

what sign are you using for velocities?

I take both velocities as positive (The same for the accelerations). Then, these would be my equations:
[tex] -F_r = ma \\ FrR=I\alpha [/tex]

 

[haruspex]

I take both velocities as positive (The same for the accelerations). Then, these would be my equations:
[tex] -F_r = ma \\ FrR=I\alpha [/tex]

If all forces, accelerations, displacements and linear velocities are positive in the original direction of travel then F = ma. F takes a negative value, resulting in a negative acceleration. Conversely, this negative F will produce a positive angular acceleration, therefore FR = - Iα.

 

[JuanC97]

Hmmm, that would explain a lot of things...
I'll remake some calcs, maybe that was the problem.
Thanks.