I Modeling Coulomb Friction for a Moving Point Mass in a Circular Bowl

[joey_ra]

TL;DR Summary

model friction on a curved surface

Consider a 2D circular bowl with a point mass moving inside, as shown in the picture. Assume that the normal force N acting on the mass is already known and keeps the mass constrained to the curved surface of the bowl.

I want to model Coulomb friction in this system so that, for example, if I release the mass on one side of the bowl, it loses energy over time and does not reach the opposite edge.

My initial idea is to use a friction model like this:
F = -mu * N * sgn(vel_rel)
where mu is the friction coefficient and v_rel is the relative velocity between the mass and the bowl.

However, since the bowl has a curved surface, should I instead project the relative velocity onto the local tangent of the bowl's surface where the ball is in contact before calculating the friction?

What is the correct way to include Coulomb friction in this scenario?
View attachment 361241

 

[berkeman]

Your attachment did not seem to work. Maybe try again? What file type is the attachment?

 

[A.T.]

should I instead project the relative velocity onto the local tangent of the bowl's surface

Isn't the relative velocity allready parallel to that local tangent? I can't see your attachement.

 

[joey_ra]

Isn't the relative velocity allready parallel to that local tangent? I can't see your attachement.

Yes, I agree. When the ball is in contact with the bowl they share the same normal velocity. Therefore the relative velocity should be parallel. I set up the dynamics equations of the ball in cartesian coordinates in x and y direction. And coulomb friction is not dependent on the relative velocity just: friction_coefficient * normalforce. But let's assume for example the ball is in the equilibrium position of the bowl and has some velocity in y direction. then the friction in y direction is zero and non-zero in x direction. If I just compute the friction with: friction_coefficient * normalforce then there is an error or am I wrong?

Source: https://www.physicsforums.com/threa...moving-point-mass-in-a-circular-bowl.1080341/

 

[joey_ra]

Here I need to model the Frictional force in cartesian coordinates. I already set up the lagrangian and derived the expression for the normalforce. But the Friction in the lagrangian appears on the right hand side of the equation in the non conservative force vector. So I have to set up this seperatly

 

[A.T.]

When the ball is in contact with the bowl they share the same normal velocity.

The bowl has a velocity too?

And coulomb friction is not dependent on the relative velocity just: friction_coefficient * normalforce.

This is kinetic friction. Is the ball not rolling at all, just sliding?

But let's assume for example the ball is in the equilibrium position of the bowl and has some velocity in y direction. then the friction in y direction is zero and non-zero in x direction.

I don't understand what you mean here.

 

[joey_ra]

The bowl has a velocity too?


This is kinetic friction. Is the ball not rolling at all, just sliding?


I don't understand what you mean here.

yes, the bowl accelerates and has different velocities.

the ball is modeled as a pointmass and is therefore only sliding.

Okay I'll try to rephrase this. In my model the ball is described in cartesian coordinates. So it has a acceleration in x and y. Therefore I have two differential equations in the form
ball_mass * x_ddot = ...
ball_mass * y_ddot = ...

So my non-conservative force vector Q also has 2 components, one in x and one in y direction. My initial idea was to just set the Q vector to friction_coefficient * normalforce in both components. But now imagine the ball is in the lowest point in the bowl and has some velocity. at this point the frictional force should only act in x direction not in y. therefore I think I need to project the friction to the local tangent but am not sure how to do that

 

[erobz]

E

Post in thread 'Model a ball in a moving circular bowl using Cartesian coordinates'

Nov 26, 2024

I think you want the cup to have accelerations $\ddot x , \ddot y$ in the inertial (fixed) frame regardless of what the ball is doing in the cup. Forces $F_x,F_y$ in this case are nonsensical( you are ignoring the normal force from the ball, and the cup has no mass). I think you want to investigate 3 coordinates, and their derivatives in the Lagrangian (as has been stated). The two above and the angle $\theta$ between one of those coordinates and a radial line through the ball since the ball is constrained to move along the arc (effectively a bead on a frictionless...

• erobz

• 

A continuation of this thread?

 

[A.T.]

So my non-conservative force vector Q also has 2 components, one in x and one in y direction. My initial idea was to just set the Q vector to friction_coefficient * normalforce in both components.

The friction vector is perpendicular to the normal force vector. But you are just scaling the normal force vector with the friction_coefficient?

The direction of friction on the ball is opposite to velocity of ball relative to bowl. This is already parallel to the local tangential plane.

 

[joey_ra]

The friction vector is perpendicular to the normal force vector. But you are just scaling the normal force vector with the friction_coefficient?

The direction of friction on the ball is opposite to velocity of ball relative to bowl. This is already parallel to the local tangential plane.

okay is this then correct
The friction forces in the tangential direction are modeled as:
$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \hat{T}_x \cdot \text{sgn}(\dot{x}_{\text{rel}})
$$
$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \hat{T}_y \cdot \text{sgn}(\dot{y}_{\text{rel}})
$$
Here:
- $\mu$: coefficient of friction
- $\lambda$: normal contact force
- $\hat{T}_x$, $\hat{T}_y$: components of the unit tangent vector $\hat{\mathbf{T}}$
- $\dot{x}_{\text{rel}}$, $\dot{y}_{\text{rel}}$: components of the relative velocity between the two bodies in contact

To compute the tangent direction, I start with the normal vector at the contact point:
$$
\hat{\mathbf{n}} = \frac{\mathbf{p}_b - \mathbf{p}_c}{\|\mathbf{p}_b - \mathbf{p}_c\|}
$$
where:
$\mathbf{p}_b = (x_b, y_b)$: Position of the Ball

$\mathbf{p}_c = (x_c, y_c)$: Position of the Bowl Center

The tangent vector $\hat{\mathbf{T}}$ is then defined as the counterclockwise rotation of the normal vector by 90° in 2D:
$$
\hat{\mathbf{T}} =
\begin{pmatrix}
\hat{T}_x \\
\hat{T}_y
\end{pmatrix}
=
\begin{pmatrix}
-\hat{n}_y \\
\hat{n}_x
\end{pmatrix}
$$

Can someone confirm that this is correct?

 

[A.T.]

okay is this then correct
The friction forces in the tangential direction are modeled as:

$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \hat{T}_x \cdot \text{sgn}(\dot{x}_{\text{rel}})
$$

$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \hat{T}_y \cdot \text{sgn}(\dot{y}_{\text{rel}})
$$

Here:
- \( \mu \): coefficient of friction
- \( \lambda \): normal contact force
- \( \hat{T}_x, \hat{T}_y \): components of the unit tangent vector \( \hat{\mathbf{T}} \)
- \( \dot{x}_{\text{rel}}, \dot{y}_{\text{rel}} \): components of the relative velocity between the two bodies in contact

To compute the tangent direction,...

As already said, if you have the relative velocity vector, then you don't need to compute the tangent vector. The relative velocity vector is already tangent to the bowl surface, so you can just normalize it, then negate and multiply by friction coefficient and normal force magnitude to get the friction vector.

 

[joey_ra]

As already said, if you have the relative velocity vector, then you don't need to compute the tangent vector. The relative velocity vector is already tangent to the bowl surface, so you can just normalize it, then negate and multiply by friction coefficient and normal force magnitude to get the friction vector.

okay I see, but let's say I leave away the tangent projection and just use:
$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{x}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$
$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{y}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$
then this is not coulomb friction or am I wrong? How I see this in this formulation the friction is dependant on the velocity.

 

[A.T.]

okay I see, but let's say I leave away the tangent projection and just use:
$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{x}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$
$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{y}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$
then this is not coulomb friction or am I wrong? How I see this in this formulation the friction is dependant on the velocity.

It's dependent on the direction of relative velocity, because kinetic friction always opposes relative motion at the contact. It's not dependent on the magnitude of relative velocity.

 

[joey_ra]

It's dependent on the direction of relative velocity, because kinetic friction always opposes relative motion at the contact. It's not dependent on the magnitude of relative velocity.

That doesn't make sense. what you describe is only valid if the contact pair has no curvature! What you describe are the following equations:
$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \text{sgn}(\dot{x}_{\text{rel}})
$$
$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \text{sgn}(\dot{y}_{\text{rel}})
$$
These can't be realistic. As I tried to explain before. Concider the ball being near the equilibrium position, then the friction in x direction should be much larger in x direction then in y direction. In these equations this is not given as the friction would act with equal magnitude in both directions.

 

[A.T.]

That doesn't make sense. what you describe is only valid if the contact pair has no curvature!

What do you mean by "contact pair has no curvature"?

Concider the ball being near the equilibrium position,

What do you mean by "ball being near the equilibrium position"?

 

[joey_ra]

What do you mean by "contact pair has no curvature"?


What do you mean by "ball being near the equilibrium position"?

in the upper example there is no curvature. in the bowl system the contact surface has a curvature. this means in the 2 drawn examples that the frictional force acts in different directions. so the magnitude in y direction for example is in the left example bigger then in the right example. when the ball is in equilibrium position the friction in y direction becomes zero. Do you see my point now?

 

[A.T.]

this means in the 2 drawn examples that the frictional force acts in different directions.

Sure, because relative velocity is in different directions, and kinetic friction is in the opposite direction of relative velocity.

Do you see my point now?

No.

 

[joey_ra]

Sure, because relative velocity is in different directions, and kinetic friction is in the opposite direction of relative velocity.

No.

okay so you think these equations are correct?
$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{x}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$

$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{y}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$

 

[A.T.]

okay so you think these equations are correct?
$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{x}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$

$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \frac{\dot{y}_{\text{rel}}}{\sqrt{\dot{x}_{\text{rel}}^2 + \dot{y}_{\text{rel}}^2}}
$$

Yes, for the simplest model of kinetic friction. It doesn't include static friction and the transition between the two types.

What you describe are the following equations:
$$
F_x^{\text{friction}} = -\mu \cdot \lambda \cdot \text{sgn}(\dot{x}_{\text{rel}})
$$
$$
F_y^{\text{friction}} = -\mu \cdot \lambda \cdot \text{sgn}(\dot{y}_{\text{rel}})
$$

No, this is not equivalent to the above.

 

[joey_ra]

Yes, for the simplest model of kinetic friction. It doesn't include static friction and the transition between the two types.

No, this is not equivalent to the above.

Okay, got it. Thank you! But maybe a follow-up question: that doesn't mean my other computation with the tangent projection is wrong ?

Is my other formulation with the tangent projection also correct, but a little more complicated?

 

[A.T.]

Is my other formulation with the tangent projection also correct, but a little more complicated?

Check if you get the right sign for y-friction in the situation you have drawn in post #16 (ball on the left side going down).