rotational dynamics - Friction coupling dish

I'm trying to model the friction of a coupling dish. The coupling dish is constructed of two rotating disc's, rotating in opposite direction, with perfect alignment. Does anyone know how to model this friction. An answer which includes moments would be ideal, since they can be used in my simulation model.

The surfaces sliding are two greasy cast iron surfaces, these have an approximate sliding friction coefficient of $\mu_s=0.15$.

Kind regards, Thijs van de Wiel

Answer

You can split the disc into rings as shown below:

You can split the ring into pieces of the shape shown in the figure:

To keep the discs in place, you'd probably have tension applied from the top and bottom of the disc system.

For the sake of simplicity, we can combine both the tensions and consider it to be acting from the top.

Total tension per unit area is given by:

$$T_{per\space unit\space area} = \frac{T_{total}}{\pi r^2}$$

The total force acting on the upper disc which is responsible for causing friction is the sum of the tension and the weight of the upper disc. Let the weight of the upper disc be $W$ and let $\phi$ be the total force acting on the lower disc per unit area.

$$\phi = T_{per\space unit\space area} + \frac{W}{\pi r^2}$$

Now go back to the figure two. The direction of frictional force will be in such a direction which opposes the rotational motion. The direction of frictional force for the differential element $ABCD$ will be perpendicular to the radius vector at that point.

Force on the differential ring element is given by,

$$dF = \mu \phi dA_{ring-element}$$

$$d\tau = \mu \phi r dA_{ring-element}$$

where $\mu$ is the coefficient of friction for the two surfaces.

If you consider a ring as a whole, at each element of the ring, the force will act perpendicularly to the radius line which passes through the element. Since it is always perpendicular to every element in the ring, the net torque on the differential ring is numerically additive.

$$d\tau_{ring} = \mu \phi r dA_{ring}$$

Since all the elements of the ring are equally distant from the axis of revolution (center of the discs), the torque can be obtained by multiplying the differential force by the radius.

$$d\tau = \mu \phi r dA_{ring}$$

The area of the ring is given by,

$$dA = (circumference) \times dr = (2\pi r)dr$$

The final differential equation will look like,

$$d\tau = \mu \phi r (2\pi r)dr = 2\mu \pi \phi r^2 dr$$

Integrating the equation with limits as 0 to R, we get,

$$\tau = \int_{0}^{R}2\mu \pi \phi r^2 dr$$

$$\tau = 2\mu \pi \phi \int_{0}^{R}r^2dr$$

$$\tau = \frac{2}{3}\mu\pi\phi R^3$$

If you want to maintain a constant angular velocity between the discs, you have to apply an external torque equal to the magnitude which was calculated above.

If you need to calculate the angular deceleration of the disc, you can do the following:

$$\tau = I\alpha$$

$$\alpha = \frac{I}{\tau} = \frac{3I}{2\mu\pi\phi R^3}$$