# Torque

The force which causes rotation of a body is called torque. The motion along a straight line (translational motion) is caused by the force given by Newton's second law but the rotation about an axis is caused by the torque which is denoted by $\tau$ (tau).

For example in figures Figure 1 and Figure 2 an old nut tight in a bolt is being loosen by a wrench. It is easier to loosen the nut if you apply a force on the handle of the wrench at a larger distance from the centre of the nut.

When you rotate the nut, the axis of rotation passes through the centre of the nut and in this case it is towards you perpendicular to the plane of the nut. We have chosen the centre of the nut as our origin.

In Figure 1 a force $\vec F$ acts at a point $p$ on the handle of the wrench and makes an angle $\theta$ with the direction of position vector $\vec r$ of the point. The torque $\vec \tau$ due to the force $\vec F$ at the point $p$ is,

\[\vec \tau = \vec r \times \vec F = rF\sin \theta \hat {{\kern 1pt} n} \tag{1} \label{1}\]

The Eq. \eqref{1} shows that the torque is the cross product of two vectors; one is the force and other is the position vector of the point where the force acts. The direction of the vector product is given by the right hand rule which is perpendicular to the plane of the two vectors. The unit vector $\hat {{\kern 1pt} n}$ gives the direction of the vector product. The magnitude of the torque is

\[\tau = rF\sin \theta \tag{2} \label{2}\]

In Eq. \eqref{2}, $F\sin \theta $ is the component of $\vec F$ perpendicular to $\vec r$, that is ${F_ \bot } = F\sin \theta $. Or similarly $r\sin \theta $ is the component of $\vec r$ perpendicular to $\vec F$. The parallel component of $\vec F$ which is ${F_\parallel }$ produces no torque ( the angle $\theta$ is zero for the parallel component).

As you can see in Figure 2, forces $\vec F_1$ and $\vec F_2$ perpendicular to the handle of the wrench are applied separately at different distances from the axis of rotation (centre of the nut). Both forces $\vec F_1$ and $\vec F_2$ produce torques in clockwise direction. And let $\vec r_1$ be the position vector of point $p_1$ where the force $\vec F_1$ is applied. Similarly let $\vec r_2$ be the position vector of point $p_2$ where the force $\vec F_2$ is applied.

Suppose the magnitude of both vectors is the same and both apply torque in the same direction separately. The magnitude of torque produced by the force $\vec F_1$ is ${\tau _1} = {r_1}{F_1}\sin \theta = {r_1}{F_1}$ and that produced by the force $\vec F_2$ is ${\tau _2} = {r_2}{F_2}\sin \theta = {r_2}{F_2}$.

Note that the forces are perpendicular to the handle of the wrench and the angle $\theta$ is ${90^ \circ }$. Since ${r_2} > {r_1}$, ${\tau _2} > {\tau _1}$ and it's easier to loosen the nut by the force $\vec F_2$ than by the force $\vec F_1$. Note that both forces have the same magnitude and therefore it's easier to loosen the nut by applying the force at a larger distance from the rotation axis (larger distance means greater torque).

## Rotational Analog of Newton's Second Law

You can express Newton's second law for a body of mass $m$ moving with acceleration $\vec a$ as $\vec F = m\vec a $. Here we find the relationship of torque on a rigid body in terms of moment of inertia (the measure of mass of a rotating rigid body) and angular acceleration.

In Figure 3 a rigid body rotates about y-axis in anticlockwise direction due to a torque applied on it. Note that we are saying *a torque applied* not *a force applied*. Remember that torque is for rotational motion.

If you consider any particle of mass $m_i$ on the body, the particle moves in its own circle of radius $r_i$ centred on the axis. The net force $\vec F_i$ on the particle has two components; one is perpendicular to $\vec r_i$ and another is parallel to $\vec r_i$.

The parallel component ${F_{i - \parallel }}$ does not produce any torque on the particle as the angle between $\vec r_i$ and ${F_{i - \parallel }}$ is zero. The torque is produced by the perpendicular component ${F_{i - \bot }}$ only. Obviously from our definition of torque, the torque $\tau_i$ on the particle is

\[{\tau _i} = {r_i}{F_i}\sin \theta = {r_i}{F_{i - \bot }} \]

In the above expression for $\tau_i$, ${F_{i - \bot }}$ is the force tangential to the circle in which the particle moves which can be written as ${F_{i - \bot }} = {m_i}{a_{i - \tan }}$ where ${a_{i - \tan }}$ is the tangential acceleration of the particle. Also from this article, we have ${a_{i - \tan }} = {r_i}\alpha $. And therefore,

\[{\tau _i} = {r_i}{m_i}{a_{i - \tan }} = m{r_i}^2\alpha \]

We know that $m{r_i}^2$ is the moment of inertia ${I_i}$ of the particle. We can write the above expression as, \[{\tau _i} = {I_i}\alpha \]

The total torque on the rigid body is then the sum of the torques on all particles. You should not forget that the angular acceleration for all particles is the same. The total torque of n number of particles is,

\[\tau = \sum\limits_{i = 1}^n {{I_i}\alpha } = I\alpha \tag{3} \label{3}\]

In Eq. \eqref{3}, the torque on the rigid body about a particular axis is the product of the angular acceleration of the body and the moment of inertia of the body about the corresponding axis.

Note that we have considered a rigid body because its moment of inertia about a particular axis is constant. The moment of inertia of a non-rigid body is not constant even if the axis of rotaion is the same. The Eq. \eqref{3} is analogous to the Newton's second law that is, $F = ma$.