1
$\begingroup$

I have been working on longitudinal tire model for a while and constantly encounter problems with the formulation of rolling resistance. Consider one wheel model coming down from a damp. It is a 2DOF system (V and omega). The FBD of the problem is as such:Free Body Diagram

Defining +x along down the damp leads to Newton's equation: m * g * sin(beta)-F_c=m*V_dot.

Defining +omega along down the damp leads to equation: F_c * R=J * omega_dot.

Define F_c: m * g * mu_k * cos(beta).

Above equations create such a situation: The mass starts rolling and always accelerates, which is not physically meaningful. After a while it should go at constant speed.

To do so, I add rolling resistance as Pajeska's suggestion:

F_r = (q_0 + q_1*abs(V/Vc) + q_2(V/Vc)^4) * N

Adding rolling resistance as above, I get this situation: The mass starts rolling with small slip as expected, then it reaches its maximum linear speed; however, it starts spinning! For, I selected application point of rolling resistance as contact point, which is again Pajeska's suggestion.

So my questions are:

  1. what should be the application point of rolling resistance and why?
  2. does angular velocity (omega) affect rolling resistance?
  3. what will be direction of rolling resistance, inverse of V or omega?
$\endgroup$
5
  • $\begingroup$ I think it would be opposite the direction of rolling. The assumption that eventually it would equal gravity isn't necessarily right. Air resistance would tho. $\endgroup$
    – Tiger Guy
    Commented Sep 14, 2023 at 15:03
  • $\begingroup$ What is the direction of the force of friction on your foot when you push off it to step forward? A wheels is just a continuous foot. Remember, a FBD is the forces being applied TO your foot/wheel, not the forces being applied by your foot/wheel to the ground. $\endgroup$
    – DKNguyen
    Commented Sep 14, 2023 at 15:42
  • $\begingroup$ $F_C = m\cdot g \cdot \mu_k \cos \beta$ defines not the instantaneous value of the friction; but the maximum value. Instantaneous value of friction can be any value between $\pm F_c$. $\endgroup$
    – AJN
    Commented Sep 15, 2023 at 12:08
  • $\begingroup$ @TigerGuy What do you mean by opposite direction of rolling? $\endgroup$
    – daltooon
    Commented Sep 15, 2023 at 12:32
  • $\begingroup$ @AJN That is true, thank you. To simplify the problem I used the maximum value rather than slip defined force. But my question was about rolling resistance. $\endgroup$
    – daltooon
    Commented Sep 15, 2023 at 12:35

1 Answer 1

0
$\begingroup$

Rolling resistance has a few contributing sources:

  • Deformation on both the wheel and the road as per figure, which acts together as a couple of vectors with the vertical wight vector creating a torque opposing rotation. It's as if the wheel is constantly using energy to climb out of the hole it has created.

These deformations are nonelastic so part of the energy is nonrecoverable.

  • Air resistance which at high velocities creates a large drag, the reason the airplanes have retractable landing gear.

'

wheel rotation resistance Source physcs.se

$\endgroup$
4
  • $\begingroup$ So your suggestion is to add rolling resistance at contact point? $\endgroup$
    – daltooon
    Commented Sep 15, 2023 at 14:58
  • $\begingroup$ No, either add it as a resistive torque or add it as a force offset from the contact point forward by dx=r/3 and inclined with an angle from the horizon $\theta$ which you need to try different values to simulate closest to your situation. $\endgroup$
    – kamran
    Commented Sep 15, 2023 at 21:28
  • $\begingroup$ I will try adding as resistive torque. This resistive torque will be function of V (linear speed) but not function of omega (angular speed). Is it so? $\endgroup$
    – daltooon
    Commented Sep 22, 2023 at 14:23
  • $\begingroup$ yes, interestingly thogh the torque will be related to e*v meaning it will not increase linearly with speed but for your task linear approximation would do! $\endgroup$
    – kamran
    Commented Sep 22, 2023 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.