Torque (Moment of force)

We have learnt that the motion of a rigid body, in general, is a combination of rotation and translation. If the body is fixed at a point or along a line, it has only rotational motion. We know that force is needed to change the translational state of a body, i.e. to produce linear acceleration. We may then ask, what is the analogue of force in the case of rotational motion? To look into the question in a concrete situation let us take the example of opening or closing of a door. 

A door is a rigid body which can rotate about a fixed vertical axis passing through the hinges. What makes the door rotate? It is clear that unless a force is applied the door does not rotate. But any force does not do the job. 

A force applied to the hinge line cannot produce any rotation at all, whereas a force of given magnitude applied at right angles to the door at its outer edge is most effective in producing rotation. It is not the force alone, but how and where the force is applied is important in rotational motion.

The rotational analogue of force in linear motion is moment of force. It is also referred to as torque or couple. (We shall use the words moment of force and torque interchangeably.) We shall first define the moment of force for the special case of a single particle. Later on we shall extend the concept to systems of particles including rigid bodies. We shall also relate it to a change in the state of rotational motion, i.e. is angular acceleration of a rigid body.

Fig.1

Fig.1: τ = r × F, τ is perpendicular to the plane containing r and F, and its direction is given by the right handed screw rule.

If a force acts on a single particle at a point P whose position with respect to the origin O is given by the position vector r (Fig. 1), the moment of the force acting on the particle with respect to the origin O is defined as the vector product

τ = r × F

The moment of force (or torque) is a vector quantity. The symbol τ stands for the Greek letter tau. The magnitude of τ is

τ = rF sinθ

where r is the magnitude of the position vector r, i.e. the length OP, F is the magnitude of force F and θ is the angle between r and F as shown.

Moment of force has dimensions $ML^2T^{-2}$. Its dimensions are the same as those of work or energy. It is, however, a very different physical quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of moment of force is newton metre (N.m). The magnitude of the moment of force may be written

τ = (r sin θ)F = r⊥F

or τ = rF sin θ = rF⊥

where r⊥ = r sinθ is the perpendicular distance of the line of action of F from the origin and F⊥( F sin θ) is the component of F in the direction perpendicular to r. Note that τ = 0 if r = 0, F = 0 or θ = $0^0$ or 180$^0$ . Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin.

One may note that since r × F is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same.

Example 1 

Find the torque of a force 7$\mathbf{\hat{i}}$ + 3$\mathbf{\hat{j}}$ – 5$\mathbf{\hat{k}}$ about the origin. The force acts on a particle whose position vector is $\mathbf{\hat{i}}$ – $\mathbf{\hat{j}}$ + $\mathbf{\hat{k}}$.

Answer

Here $\mathbf{r}=\mathbf{\hat{i}}$ – $\mathbf{\hat{j}}$ + $\mathbf{\hat{k}}$

and $\mathbf{F}$ = 7$\mathbf{\hat{i}}$ + 3$\mathbf{\hat{j}}$ – 5$\mathbf{\hat{k}}$

We shall use the determinant rule to find the torque τ = r × F 

$\tau=\begin{vmatrix}\hat{\mathbf{i}} &\hat{\mathbf{j}}  &\hat{\mathbf{k}}\\ 1& -1 & 1\\7& 3 & -5\end{vmatrix}=2\mathbf{\hat{i}}+12\mathbf{\hat{j}}+10\mathbf{\hat{k}}$

• Torque and angular momentum for a system of particles