Torque is the rotational equivalent of linear force in physics and mechanics. Depending on the field of study, it is also known as the moment, moment of force, rotational force, or turning effect. Originally, the concept was based on Archimedes’ studies about levers. Torque can be thought of as a twist to an object around a specific axis, much like a linear force is a push or a pull.

According to another definition, torque is the product of the magnitude of a force and the perpendicular distance of its line of action from the axis of rotation. Typically, torque is symbolized by \(\tau\), the lowercase Greek letter tau. M is commonly used when referring to the moment of force.

Torque is a pseudovector in three dimensions; for point particles, it is given by the cross product of the position vector (distance vector) and the force vector. A rigid body’s torque is influenced by three quantities: the force applied, the lever arm vector connecting the point at which torque is measured to the point of force application, and the angle between the force and lever arm vectors. Symbolically:

$$\vec{\tau}=\vec{r}\times \vec{F}=|{\vec{r}}||\vec{F}|\sin\theta (\hat{n})\\

|\vec{\tau}|=|\vec{r}\times \vec{F}|=|{\vec{r}}||\vec{F}|\sin\theta$$

where

- \(\vec{\tau}\) is the magnitude of the torque,
- \(\vec{r}\) is the position vector (a vector from the point about which the torque is being measured to the point where the force is applied),
- \(\vec{F}\) is the force vector,
- \(‘\times’\) denotes the cross product, which produces a vector that is perpendicular to both \(\vec{r}\) and \(\vec{F}\) following the right-hand rule,
- \(\theta\) is the angle between the force vector and the lever arm vector.

The \(SI\ unit\) for torque is the \(newton-metre\ (N⋅m)\).

**What is Torque Formula?**

Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label \(‘O’\). We will call the force \(‘F’\). The distance from the pivot point to the point where the force acts is called the moment arm and is denoted by \(‘r’\). Note that this distance, \(‘r’\), is also a vector, and points from the axis of rotation to the point where the force acts.

Torque is defined as $$\vec{\tau}=\vec{r}\times \vec{F}=|{\vec{r}}||\vec{F}|\sin\theta (\hat{n})$$

In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, \(‘\theta’\) being the angle between \(r\) and \(F\).

**Direction Of Torque**

**Cross Product**

Using the right-hand rule, we can find the direction of the torque vector. If we put our fingers in the direction of \(r\) and curl them to the direction of \(F\), then the thumb points in the direction of the torque vector.

**Right-Hand Rule**

Imagine pushing a door to open it. The force of your push \(F\) causes the door to rotate about its hinges (the pivot point, O). How hard you need to push depends on the distance you are from the hinges \(r\) (and several other things, but let’s ignore them now). The closer you are to the hinges (i.e. the smaller \(r\) is), the harder it is to push. This is what happens when you try to push open a door on the wrong side. The torque you created on the door is smaller than it would have been had you pushed the correct side (away from its hinges).

Note that the force applied, \(F\), and the moment arm, \(r\), is independent of the object. Furthermore, a force applied at the pivot point will cause no torque since the moment arm would be zero \((r=0)\).

Another way of expressing the above equation is that torque is the product of the magnitude of the force and the perpendicular distance from the force to the axis of rotation (i.e. the pivot point).

Let the force acting on an object be broken up into its tangential \(F_{tan}\) and radial \(F_{rad}\) components. (Note that the tangential component is perpendicular to the moment arm, while the radial component is parallel to the moment arm.)

The radial component of the force has no contribution to the torque because it passes through the pivot point. So, it is only the tangential component of the force that affects torque (since it is perpendicular to the line between the point of action of the force and the pivot point).

There may be more than one force acting on an object, and each of these forces may act at a different point on the object. Then, each force will cause a torque. **The net torque is the sum of the individual torques.**

Rotational Equilibrium is analogous to translational equilibrium, where the sum of the forces is equal to zero. **In rotational equilibrium, the sum of the torques is equal to zero. In other words, there is no net torque on the object.**

$$\sum_{}^{}\tau=0$$

Note that the \(SI\ unit\) of torque is a \(Newton-metre\), which is also a way of expressing a \(Joule\) (the unit for energy). However, **torque is not energy**. So, to avoid confusion, we will use the units \(N.m\), and not \(J\). The distinction arises because energy is a scalar quantity, whereas torque is a vector.

Here is a useful and interesting interactive activity on rotational equilibrium.

**Dimensional Formula Of Torque**

$$\tau = I. \alpha\\

[\tau]=[I]. [\alpha]\\as\ we\ know\ I=Mr^2\ so\ the\ dimensional\ formula\ of\ I\\ [M^1L^2T^0]\\ and,\alpha=\frac{a}{r}\ so\ the\ dimensional\ formula\ of\ \alpha\\ \frac{[M^0L^1T^{-2}]}{[M^0L^1T^0]}=[M^0L^0T^{-2}]\\ Hence\ the\ dimensional\ formula\ of\ \tau=[M^1L^2T^0]\times [M^0L^0T^{-2}]\\ =[M^1L^2T^{-2}]$$

**The Torque Formula Derivation**

Now let’s find the formula or expression.

Rate of change of Angular Momentum in relation to time = \(\frac{\Delta L}{\Delta T}\)

Now, \(\frac{\Delta L}{\Delta T} = \frac{\Delta(I \omega)}{\Delta T} = I.\frac{\Delta \omega}{\Delta T}\)

Here \(I\) is certainly the constant when the mass and shape of the object are unchanged

Now, \(\frac{\Delta \omega}{\Delta T}\) refers to the rate of change of angular velocity with time i.e. angular acceleration \(\alpha\).

So we can write, \(\frac{\Delta L}{\Delta T}=I\alpha\)

\(I\) (moment of inertia) refers to the rotational equivalent of mass(inertia) of linear motion. Similarly, angular acceleration \(\alpha\) (alpha) certainly refers to the rotational motion equivalent of linear acceleration.

So we can get, \(\frac{\Delta L}{\Delta T}=\tau\), this certainly states that the rate of change of angular momentum with time is called Torque.

Torque \(\tau\) refers to the moment of force. \(\vec{\tau}=\vec{r}\times \vec{F}=|{\vec{r}}||\vec{F}|\sin\theta\)

\(\tau = r. F. sin θ = r. ma. sinθ = r. m. \alpha. r. sinθ = mr^2. α. sinθ = I. α. sinθ = I \times α\)

[\(\alpha\) is angular acceleration, \(I\) refers to the moment of inertia and \(\times\) denotes cross product.]\(\tau = I \alpha\)

or, \(\tau = I. \frac{(\omega_2-\omega_1)}{t}\)

Here \(\alpha\) = angular acceleration = time rate of change of the important angular velocity = \(\frac{(\omega_2-\omega_1)}{t}\) where \(\omega_2\) and \(\omega_1\) happen to be the final and initial angular velocities and \(t\) is the time gap.

or, \(\tau. t = I (\omega_2-\omega_1)\)

when, \(\tau = 0\) (i.e., net torque is zero),

\(I. (\omega_2-\omega_1) = 0\)

i.e., \(I.\omega_2=I.\omega_1\)

**Torque and Angular Acceleration**

In this section, we will develop the relationship between torque and angular acceleration. You will need to have a basic understanding of moments of inertia for this section.

**Moment of Inertia**

Imagine a force \(F\) acting on some object at a distance r from its axis of rotation. We can break up the force into tangential \(F_{tan}\), radial \(F_{rad}\). (This is assuming a two-dimensional scenario. For three dimensions, a more realistic, but also more complicated situation, we have three components of force: the tangential component \(F_{tan}\), the radial component \(F_{rad}\), and the z-component \(F_{z}\). All components of force are mutually perpendicular, or normal.)

**Electric Field Between Two Plates: Magnitude, Direction, Examples & More**

From Newton’s Second Law, \(F_{tan}=ma_{tan}\)

However, we know that angular acceleration, \(\alpha\), and the tangential acceleration \(a_{tan}\) are related by:

\(a_{tan}=r\alpha\)

Then,

\(F_{tan}=mr\alpha\)

If we multiply both sides by \(r\) (the moment arm), the equation becomes

\(F_{tan}r=mr^2\alpha\)

**Note that the radial component of the force goes through the axis of rotation, and so has no contribution to torque. The left-hand side of the equation is torque.** For a whole object, there may be many torques. So the sum of the torques is equal to the moment of inertia (of a particle mass, which is the assumption in this derivation), \(I=mr^2\) multiplied by the angular acceleration, \(\alpha\).

\(\sum_{}^{}\tau=I\alpha\)

If we make an analogy between translational and rotational motion, then this relation between torque and angular acceleration is analogous to Newton’s Second Law. Namely, taking torque to be analogous to force, the moment of inertia analogous to mass, and angular acceleration analogous to acceleration, then we have an equation very much like the Second Law.

**Example on Torque Formula**

**A car mechanic applies a force of \(800 N\) to a wrench for the purpose of loosening a bolt. He applies the force which is perpendicular to the arm of the wrench. The distance from the bolt to the mechanic’s hand is \(0.40\ m\). Find out the magnitude of the torque applied?**

The angle between the moment the arm of the wrench and the force is without a doubt \(90°\), and \(\sin 90° = 1\). The torque is:

\(\tau = F \times r \times sinθ\)

Therefore, the magnitude of the torque = \((800N) (0.4m) = 320 N.m\)

Hence, the magnitude of the torque is \(320 N.m.\)

**FAQs**

**How do you calculate torque?**

Mathematically, torque can be written as \(\tau = F \times r \times \sin(\theta)\), and it has units of Newton-meters. When the sum of all torques acting on an object equals zero, it is in rotational equilibrium.

**What is torque write its formula?**

\(\tau = F \times r \times \sin(\theta)\)

\(\tau\) = torque. F = linear force. r = distance measured from the axis of rotation to where the application of linear force takes place. theta = the angle between F and r. In this formula, \(\sin(\theta)\) has no units, r has units of meters (m), and F happens to have units of Newtons (N).

**Why do we calculate torque?**

When studying how objects rotate, it quickly becomes necessary to figure out how a given force results in a change in the rotational motion. The tendency of a force to cause or change rotational motion is called torque, and it’s one of the most important concepts to understand in resolving rotational motion situations.

**What is torque and its SI unit?**

Torque, also called the moment of a force, in physics, is the tendency of a force to rotate the body to which it is applied. Torque is measured in newton-meters in SI units.