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

• gaz097
In summary, to rotate a square block with a coefficient of friction of 0.3, the required torque will be less than if the block was not rotating.
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
gaz097 said:
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?

## 1. What is torque?

Torque is a measure of the force that causes an object to rotate about an axis. It is typically measured in units of Newton-meters (N*m) or foot-pounds (ft-lb).

## 2. How is torque calculated?

Torque is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied. The formula for torque is T = F * d, where T is torque, F is force, and d is distance.

## 3. How does torque affect an object's rotation?

The greater the torque applied to an object, the faster it will rotate. This is because torque is directly proportional to an object's angular acceleration, which is a measure of how quickly the object's rotational speed changes.

## 4. What factors can affect torque?

The two main factors that affect torque are the magnitude of the force applied and the distance from the axis of rotation to the point where the force is applied. Other factors that can affect torque include the angle at which the force is applied and the mass and shape of the object.

## 5. How is torque used in everyday life?

Torque is used in many everyday objects and activities, such as opening a door, riding a bike, and using a wrench. It is also important in the design and operation of machines and vehicles, such as cars and airplanes.

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