18.3 Rotation  (Page 4/5)

Torque has the unit of Newton - meter.

Rotational torque .vs. torque about a point

We have seen that rotational torque is calculated with the component of force which lies in the plane of rotation. This is required as the rigid body is free to move only about the fixed axis perpendicular to the plane of rotation. Accordingly, we evaluate vector expression of torque by measuring position vector from center of circle and using component of force in the plane of rotation. However, if we evaluate the vector expression of torque r X F with respect to the origin of reference lying on the axis of rotation as is the general case of torque about a point, then we get torque which needs not be along the axial direction.

Clearly, rotational torque is a subset of torque defined for a point on the axis of rotation. This fact gives an alternate technique to determine rotational torque. We can approach the problem of determining rotational torque as described in the module, wherein we first resolve the given force in the plane of rotation containing the point of application. Then, we find the moment arm or perpendicular force component and determine the rotational torque as already explained. Alternatively, we can determine the torque about a point on the axis using vector expression of torque and then consider only the component of torque in the direction of axis of rotation.

In order to fully understand the two techniques, we shall work out an example here to illustrate the two approaches discussed here.

Let us consider a force F = (2 i + 2 j – 3 k ) Newton which acts on a rigid body at a point r = ( i + j – k ) meters as measured from the origin of reference. In order to determine the rotational torque say about x-axis as the axis of rotation, we first proceed by considering only the force in the plane of rotation. The figure below shows the components of force and their perpendicular distances from three axes.

We find the position of application of force by first identifying A (1,1) in xy-plane, then we find the position of the particle, B(1,1,-1) by moving “-1” in negative z-direction as in the figure below.

Torque on the particle

Since the rigid body rotates about x-axis, the plane of rotation is yz plane. We, therefore, do not consider the component of force in x-direction. The components of force relevant here are (i) the component of force in y-direction Fy = 2 N at a perpendicular distance z = 1 m from axis of rotation and (ii) the component of force in z-direction Fz = 3 N at a perpendicular distance y = 1 m. For determining torque about x-axis, we multiply perpendicular distance with force. We apply appropriate sign, depending on the sense of rotation about the axis. The torque about x-axis due to component of force in y - direction is anticlockwise and hence is positive :

$$\begin{array}{l}{\tau }_1=z{F}_y=1x2=2Nm\end{array}$$

The torque about x-axis due to the component of force in z - direction is clockwise and hence is negative :

$$\begin{array}{l}{\tau }_2=-y{F}_z=-1x3=-3Nm\end{array}$$

Since both torques are acting along same axis i.e. x-axis, we can obtain net torque about x-axis by algebraic sum :