# Torque to rotate a object on the ground about its own axis

• gaz097

#### gaz097

TL;DR Summary
Does a equation exist for calculating torque for rotating a object about its cog axis where the only reaction forces are its own weight at a COF.
Hi,

I have an object sitting on the ground, with a coefficient of friction (COF) of 0.3.
Lets say it is a square block, and will rotate on its central axis.

How much torque is required to rotate this block? I am ignoring inertia weights as will be rotating very slowly.

I can solve my problem by breaking the assumed footprint up into sections (Quadrants) and applying the force to slide the quadrant at its center point at a radius to the original shape, which works in theory but continues to increase in value the more sections it is broken up into and i imagine converging to a value. This sounds like a integral that i have forgotten many years ago.

Is this a constant for different COF values, shapes?
I assume an equation could exist for different equal shapes of torque, or at least for a circle.

Welcome to PF.
The block down force will be; F = m*g.
The force needed to translate the block will be; F= m*g*CoF .
The torque needed to rotate the block will be less, since there is less area, and an advantage near the neutral axis.

For a disc of radius r, and mass m.
Torque = 2/3 * CoF * m * g * r .
https://engineeringstatics.org/Chapter_09-disc-friction.html

Even though you are not concerned with inertia, the moment of inertia of a rotating square will be useful in determining the radial area distribution of the footprint.

Last edited:
• berkeman
Summary: Does a equation exist for calculating torque for rotating a object about its cog axis where the only reaction forces are its own weight at a COF.

I can solve my problem by breaking the assumed footprint up into sections (Quadrants) and applying the force to slide the quadrant at its center point at a radius to the original shape, which works in theory but continues to increase in value the more sections it is broken up into and i imagine converging to a value. This sounds like a integral that i have forgotten many years ago.
Not sure what you mean there -- applying the force to slide the quadrant at its center point at a radius to the original shape.

There are two ways to do this, actually a few more, but for a circle, the easier ones are,
Break the circle into thin circumferential r + dr's from 0 to R, find the torque on each, and add them all up.
Break the circle into thin wafer sized wedges, rdθ, find the torque on each, and add them all up.

It doesn't matter how far the circle is rotated, as the torque does not depend upon that, the same as for a translation, the force required to move the object any distance depends upon the friction f = μW.

so for the circle, and the wedge, if rotated dθ, the friction df on the wedge will be μdW.
Adding up the df's of all the wedges the friction will add up to μW ---- f = μW.

@gaz097 Can you find the torque required?