# Angular friction: torque needed to let a shaft in floor contact spin around its axis

1. Feb 19, 2010

### Sumtwit

Hi everyone.

This is not a homework problem. I am very familiar with the case of linear(direction) friction.
$$F_{friction}=\mu F_{normal}$$ ...(1)
However my question is closer to something like this:
$$\tau_{friction}=\beta F_{normal}$$ ...(2)
Where $$\tau_{friction}$$ is a torque

Imagine if a shaft were to be making contact with the floor at its stub. Assuming the normal force that the shaft exerts on the ground is known. Then how much torque would need to be applied around the shafts axis to make it overcome friction and start turning.

Maybe another analogy would be a drill with a flat tip, how much torque would the flat tip need to overcome friction ?

Is equation 2 of the right form ? what name does this effect go by because so far I could only find references to linear friction.

2. Feb 20, 2010

### nvn

Frictional torque on flat tip of round solid rod, T = mu*(d/3)*N, where d = rod diameter, and N = normal force rod exerts on floor. This assumes the rod is normal to the floor surface.

3. Feb 21, 2010

### Sumtwit

Re: angular friction: torque needed to let a shaft in floor contact spin around its a

Hi nvn, thanks that helps a lot. I see "frictional torque" doesn't appear to be a very well covered topic. I was surprised to see that it is a function of the dimensions of the contacting area.
Could you please guide me to a derivation of frictional torque ? I would like to find an expression for other shapes as well.

4. Feb 21, 2010

### nvn

Pressure p = N/A; rod diameter D = 2*r2.

$$\begin{equation*}\begin{split}T &= \int\int r{\cdot}(\mu\,p)\,dA\\[0.5mm] &=\ \mu\,p\int\int r\,dr\,(r\,d\theta)\\[2mm] &=\ \mu\,p\int_{0}^{2\,\pi}\int_{0}^{r_2}r^2\,dr\:d\theta\\[3mm] &=\ (D/3)\,\mu\,N\end{split}\end{equation*}$$

5. Feb 22, 2010

### Sumtwit

Re: angular friction: torque needed to let a shaft in floor contact spin around its a

OK Thanks! that helps a lot :)