When we studied translational motion, a given force applied to a given particle always produced the same result. Because in rotational motion we consider rigid bodies rather than particles, we cannot make such a general statement about the effect of an applied force. For example, if the force is applied to the center of the object, it will not cause the object to rotate. If, however, it is applied to the edge of a rotating object, it can have quite a large effect on the rotation of the object. With this aspect of rotational motion in mind, we define torque to generally describe the effect a force will have on rotational motion.

Consider point P a distance r away from an axis of rotation, and a force F applied to P at an angle of θ to the radial direction, as shown below. If the force is parallel to the radius of the particle (θ = 0), then the force might cause some translational motion of the particle. But, since the force has no component acting in the tangential direction, it causes no change in rotational motion. In addition, if the force is close to the axis of rotation it will cause less change in the rotation of the body than at a farther distance. Thus we define the torque (denoted by τ) accordingly:

The second equation (τ = r×F) expresses the torque in terms of a cross product, an important operation in vector algebra, but not essential for the understanding of torque. With this vector definition, however, we are able to define the direction of the torque. The torque (because it is a cross product) must be perpendicular to both the force applied and the radius of the particle, implying that it points perpendicular to the plane of rotation of the particle.

This definition can be difficult to grasp conceptually, so we will consider some examples to clarify. The best example of a torque is the force applied to opening a door. The easiest way to open the door (in other words, the way to provide maximum torque) is to grab a point the furthest away from the hinges (like the handle), and pull perpendicular to the door itself. In this way, we give a maximum r, and sinθ = 1. The closer to the hinges one pulls, the more force must be exerted to provide the same torque on the door. In addition, the angle with which the torque is applied changes the force necessary for a given torque. The case of pulling perpendicular to the door requires the least force.

Torque plays the same role in rotational motion as force plays in translational motion. In fact, we can restate Newton's First Law to make it apply to rotational motion:

Though this statement helps us to gain a conceptual understanding of exactly how a torque influences rotational motion, we need a rotational analogue to Newton's Second Law, which will serve as a quantitative basis for rotational dynamics.

We know qualitatively how torque effects rotational motion. Our task now is to generate an equation to calculate this effect. We start be examining the torque on a single particle of mass m, a distance r away from the axis of rotation. For simplicity's sake we shall assume the torque acts perpendicular to the radius of the particle. From our definition of torque we know τ = Fr. Newton's Second Law of translational motion states that F = ma and, substituting in our rotational variable, we see that F = mrα. Putting these relations together:

Notice that we have successfully related torque and angular acceleration, as we had hoped to do. However, we need to extend this equation to rigid bodies, as they are the important bodies in rotational dynamics.

Consider a rigid body made up of n particles, each acted upon by a torque. The motion of each particle can be described:

All internal forces between particles in this rigid body cancel out. We can also state that the angular acceleration of each particle is the same (this is one of the properties of the rotation of a rigid body). Thus we may sum over all our particles to generate an equation for the angular acceleration due to a net torque on a rigid body:

This equation looks a lot like Newton's Second Law. We have the axis of rotation and the torque directly related to the angular acceleration, scaled by a proportionality constant that is a property of the rigid body. We shall formally define this constant as the moment of inertia, and denote it by I:

Thus we may simplify our torque equation to give an equation that is mathematically identical to Newton's Second Law:

There we have it! We have generated a simple equation relating a torque with rotational acceleration. The only challenging part of this equation is the quantity I. We may see this quantity as equivalent to mass--it defines the proportion between a physical force or torque and the resulting acceleration. Generally, however, I can only be calculated through calculus. We shall explore how to do so in a calculus-based section at the end of this SparkNote, but in general the moment of inertia of a rigid body will be given in any problem you might be asked to answer.

We have now derived the necessary ingredients for a full study of rotational dynamics. Since the methods are the same as in the linear case, we are able to spend less time going over the concepts of rotational dynamics. Thus we will continue our study by quickly running through work and energy in a rotational system, and looking at the relation between rotational and translational motion.