Newton's Second Law for Rotational Motion

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:

τ = Fr = (mrα)r = (mr2)α    

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.

Second Law of Rotational Motion for Rigid Bodies

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:

τ = (mr2)α    

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:

I = mr2    

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

τ =    

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.