# Direction of friction of a rotating ring/disk

I'm trying to understand friction but I'm getting a bit confused...

If I have a ring rotating on the ground around its COM what is the direction of the friction?

If I have the same ring, but this time it's on a rotating plate (the ring is not moving in the frame of the rotating plate but appears spinning at the same $$\omega$$ as the plate for someone outside that frame).

What is the direction of friction this time? Are there "rules" of how should one determine the direction of friction? The rule is- Friction always tries to stop relative motion. So just look at your ring and observe the direction of the velocities of it's various points.Friction will try to slow it down and bring it at rest with respect to the ground

• But is it acting in the tangent direction? radial direction? I've been told that many times friction is the centripetal force so I'm unsure of the case here – snatchysquid Jul 13 at 11:59
• the tension provides the centripetal force – Dylan Rodrigues Jul 13 at 12:02
• if tension is not sufficient,then friction will take up the job for centripetal force – Dylan Rodrigues Jul 13 at 12:03
• tension of the body itself? and does this mean that the friction is in the tangent direction? – snatchysquid Jul 13 at 12:03
• so in the case of insufficent tension the friction will be in both the radial direction and the tangent? – snatchysquid Jul 13 at 12:05

In your rotating system, distributed friction produces a torque. The torque vector would be directed along the axis of rotation (the only unique direction in a rotating system) and would be opposite to the angular velocity vector. In the second question, if the ring and the plate are rotating with the same angular velocity, there is no friction between them.

Each infinitesimal element of your friction area will contribute to friction, which is summed up to zero. However, the friction in each element will also contribute to torque whose direction is opposed to the angular momentum of the plate, for all elements. Thus, the torque will not sum up to zero. I add torque so I can write the equation of motion.

Case I:

The ring is rotating due to the applied torque $$\tau$$ the friction force act opposite to the rotation $$\omega$$

The EOM:

$$\theta_r\,\dot{\omega}=\tau-F_\mu\,R_r$$

Case II:

The plate is rotating due to torque applied to the plate,you have two friction forces one between the ring and the plate, the other one between the plate and the ground.

The friction force between the plate and the ground act opposite to the rotation $$\omega_p$$ The friction force between the ring and the plate act according to Newton third law also on the plate with opposite sign.

The EOM:

$$\theta_r\,\dot{\omega}_r=\pm F_{\mu r}\,R_r$$ $$\theta_p\,\dot{\omega}_p=\tau \mp F_{\mu r}\,R_r-F_{\mu p}\,R_p$$

with: $$F_{\mu r}\mapsto -\text{signum}(\omega_r-\omega_p)\,F_{\mu r}$$ If you want to cover also the case where $$\tau$$ change sign $$\tau\mapsto\pm \tau$$ then

$$F_\mu\mapsto -\text{signum}(\omega) F_\mu$$ and

$$F_{\mu p}\mapsto -\text{signum}(\omega_p) F_{\mu p}$$