1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B How to find torque of rotational friction on an axle?

  1. Dec 31, 2016 #1
    Hi, I'm trying to program a simple physics engine from scratch as an exercise, and I'm starting with manipulating a stick with the mouse pointer. As of now, it dangles from one end demonstrating simple pendulum physics. Now, I want to add a friction component to the "axle" it's rotating on to avoid perpetual motion, like when you put a pencil through the end of a ruler and swing it back and forth.

    I can create this illusion by making a line that says "omega = omega * 0.9" (omega being angular velocity) each frame, because right now I'm bypassing torque altogether and skipping straight to acceleration (which I realized this morning was a terrible idea... I know how to fix this though).

    However, even though that appears right when I run the program, it doesn't seem to line up with real-life physics at all. What I need is a torque that goes against the torque of gravity (-m*g*L*sin (theta)); I just don't know how to find what that torque is. I'm pretty sure there's one equation I'm looking for, but I forget how friction works mathematically and I can't find it online or in my high school physics notes for the life of me! Anyone got it?
  2. jcsd
  3. Dec 31, 2016 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    The simplest model just has the retarding torque proportional to how tight the turning section fits the axle.
    The next most complicated has the retarding torque proportional to the angular velocity.

    I'd use the first one for the case of a ruler swinging on a pencil ... the "tightness" is the normal force at the contact point. The second works better for larger systems, and where there are bearings, lube, and so on. Also see:
  4. Jan 3, 2017 #3
    Thanks for the response! The links you provided certainly do have a lot of information, but maybe too much! If I have to go through it all I will, but I feel like this is a lot simpler... I was thinking along the same lines of getting the tightness at the contact point, and originally tried to just think up the analog for rotational friction from translational (which, if I remember right, is Fnet = Fexternal + Ffriction for Fexternal = ma and Ffriction = -μFnormal) which would go to τnet = τexternal + τfriction for τexternal = Iα and τfriction = -μτnormal but I can't figure out what the "normal torque" (?) would be for that joint.

    Come to think of it, I forget how regular normal force works... Because I remember it's just Fnormal = mg sinθ, but if that was the case, let's say Fexternal = 0 then Fnet would just be -μmg sinθ, meaning it would decelerate past 0 and accelerate in the other direction! That can't be right...?
  5. Jan 3, 2017 #4


    User Avatar
    Science Advisor
    Gold Member

    In a dry bearing the pivot pin only touches the bearing over a relatively small area . The normal force on that area and the coefficient of friction give you the tangential friction force . Force times radius of pivot pin gives you the friction torque .

    The mechanics of what happens in a dry bearing can be more complex than this in reality but the simple calculation is what is commonly used and gives reasonably accurate results .

    In a lubricated bearing friction torque is mainly due to viscous drag effects . Can be estimated theoretically but calculations are not easy for real bearing systems .

    Retarding torque on pendulum arm may in any case be due to air resistance as well as pivot friction .
  6. Jan 3, 2017 #5
    Ahh, ok. The tangential friction force times radius makes perfect sense, but now how do I apply that? I can't simply subtract it from the external torque, because unless I'm doing something wrong that would make it continue to accelerate in the opposite direction even after decelerating to a complete stop.
  7. Jan 4, 2017 #6


    User Avatar
    Science Advisor
    Gold Member

    Friction only acts to oppose motion .

    Try drawing a free body diagram for the pendulum arm .
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted