Torque

We have now identified the rotational equivalent of velocity--namely, angular velocity--and the rotational equivalent of mass--namely, moment of inertia. But, what is the rotational equivalent of force?

Consider a bicycle wheel of radius $b$ which is free to rotate around a perpendicular axis passing through its centre. Suppose that we apply a force ${\bf f}$, which is coplanar with the wheel, to a point $P$ lying on its circumference. See Fig. 79. What is the wheel's subsequent motion?

Let us choose the origin $O$ of our coordinate system to coincide with the pivot point of the wheel--i.e., the point of intersection between the wheel and the axis of rotation. Let ${\bf r}$ be the position vector of point $P$, and let $\theta$ be the angle subtended between the directions of ${\bf r}$ and ${\bf f}$. We can resolve ${\bf f}$ into two components--namely, a component $f \cos\theta$ which acts radially, and a component $f \sin\theta$ which acts tangentially. The radial component of ${\bf f}$ is canceled out by a reaction at the pivot, since the wheel is assumed to be mounted in such a manner that it can only rotate, and is prevented from displacing sideways. The tangential component of ${\bf f}$ causes the wheel to accelerate tangentially. Let $v$ be the instantaneous rotation velocity of the wheel's circumference. Newton's second law of motion, applied to the tangential motion of the wheel, yields

$$M \dot{v} = f \sin\theta,$$ (360)

where $M$ is the mass of the wheel (which is assumed to be concentrated in the wheel's rim).

Figure 79: A rotating bicycle wheel.

Let us now convert the above expression into a rotational equation of motion. If $\omega$ is the instantaneous angular velocity of the wheel, then the relation between $\omega$ and $v$ is simply

$$v = b \omega.$$ (361)

Since the wheel is basically a ring of radius $b$, rotating about a perpendicular symmetric axis, its moment of inertia is

$$I = M b^2.$$ (362)

Combining the previous three equations, we obtain

$$I \dot{\omega} = \tau,$$ (363)

where

$$\tau = f b \sin\theta.$$ (364)

Equation (363) is the angular equation of motion of the wheel. It relates the wheel's angular velocity, $\omega$, and moment of inertia, $I$, to a quantity, $\tau$, which is known as the torque. Clearly, if $I$ is analogous to mass, and $\omega$ is analogous to velocity, then torque must be analogous to force. In other words, torque is the rotational equivalent of force.

It is clear, from Eq. (364), that a torque is the product of the magnitude of the applied force, $f$, and some distance $l=b \sin\theta$. The physical interpretation of $l$ is illustrated in Fig. 80. If can be seen that $l$ is the perpendicular distance of the line of action of the force from the axis of rotation. We usually refer to this distance as the length of the lever arm.

In summary, a torque measures the propensity of a given force to cause the object upon which it acts to twist about a certain axis. The torque, $\tau$, is simply the product of the magnitude of the applied force, $f$, and the length of the lever arm, $l$:

$$\tau = f l.$$ (365)

Of course, this definition makes a lot of sense. We all know that it is far easier to turn a rusty bolt using a long, rather than a short, wrench. Assuming that we exert the same force on the end of each wrench, the torque we apply to the bolt is larger in the former case, since the perpendicular distance between the line of action of the force and the bolt (i.e., the length of the wrench) is greater.

Figure 80: Definition of the length of the level arm, $l$.

Since force is a vector quantity, it stands to reason that torque must also be a vector quantity. It follows that Eq. (365) defines the magnitude, $\tau$, of some torque vector, $\mbox{\boldmath$\tau$}$. But, what is the direction of this vector? By convention, if a torque is such as to cause the object upon which it acts to twist about a certain axis, then the direction of that torque runs along the direction of the axis in the sense given by the right-hand grip rule. In other words, if the fingers of the right-hand circulate around the axis of rotation in the sense in which the torque twists the object, then the thumb of the right-hand points along the axis in the direction of the torque. It follows that we can rewrite our rotational equation of motion, Eq. (363), in vector form:

$$I \frac{d\mbox{\boldmath$\omega$}}{dt} = I \mbox{\boldmath$\alpha$}= \mbox{\boldmath$\tau$},$$ (366)

where $\mbox{\boldmath$\alpha$}= d\mbox{\boldmath$\omega$}/dt$ is the vector angular acceleration. Note that the direction of $\mbox{\boldmath$\alpha$}$ indicates the direction of the rotation axis about which the object accelerates (in the sense given by the right-hand grip rule), whereas the direction of $\mbox{\boldmath$\tau$}$ indicates the direction of the rotation axis about which the torque attempts to twist the object (in the sense given by the right-hand grip rule). Of course, these two rotation axes are identical.

Although Eq. (366) was derived for the special case of a torque applied to a ring rotating about a perpendicular symmetric axis, it is, nevertheless, completely general.

It is important to appreciate that the directions we ascribe to angular velocities, angular accelerations, and torques are merely conventions. There is actually no physical motion in the direction of the angular velocity vector--in fact, all of the motion is in the plane perpendicular to this vector. Likewise, there is no physical acceleration in the direction of the angular acceleration vector--again, all of the acceleration is in the plane perpendicular to this vector. Finally, no physical forces act in the direction of the torque vector--in fact, all of the forces act in the plane perpendicular to this vector.

Consider a rigid body which is free to pivot in any direction about some fixed point $O$. Suppose that a force ${\bf f}$ is applied to the body at some point $P$ whose position vector relative to $O$ is ${\bf r}$. See Fig. 81. Let $\theta$ be the angle subtended between the directions of ${\bf r}$ and ${\bf f}$. What is the vector torque $\mbox{\boldmath$\tau$}$ acting on the object about an axis passing through the pivot point? The magnitude of this torque is simply

$$\tau = r f \sin\theta.$$ (367)

In Fig. 81, the conventional direction of the torque is out of the page. Another way of saying this is that the direction of the torque is mutually perpendicular to both ${\bf r}$ and ${\bf f}$, in the sense given by the right-hand grip rule when vector ${\bf r}$ is rotated onto vector ${\bf f}$ (through an angle less than $180^\circ$ degrees). It follows that we can write

$$\mbox{\boldmath$\tau$}= {\bf r}\times {\bf f}.$$ (368)

In other words, the torque exerted by a force acting on a rigid body which pivots about some fixed point is the vector product of the displacement of the point of application of the force from the pivot point with the force itself. Equation (368) specifies both the magnitude of the torque, and the axis of rotation about which the torque twists the body upon which it acts. This axis runs parallel to the direction of $\mbox{\boldmath$\tau$}$, and passes through the pivot point.

Figure 81: Torque about a fixed point.