Kinetic energy and work-energy theorem for a rigid-body rotating about a fixed axis: The kinetic energy of a rigid body rotating with angular speed ω is obtained by calculating the energy of small mass element in the body and adding it up. This mass element is rotating in a plane perpendicular to the axis of rotation. This gives (using the notation of figure 2)

The corresponding work-energy theorem for the motion considered here is that the change in kinetic energy is equal to the work done on the body. Let us first calculate the work done on a body, which can only rotate about an axis, when an external force is applied on it. To do this, I would first like you to prove a result (look at figure 2 for reference): when a body rotates by an angle Δθ about an axis in the unit vector direction , the corresponding change in position of a particle in the body at position vector is

The total work done on the body by a net external force composed of forces acting at each point is

By using , we can write the work done as

where τ_Z is the component of the external torque along the axis of rotation. Thus the total work done is

Now the work energy theorem can be expressed as follows:

 

This pretty much concludes what all I have to say about the rotations about a fixed axis. One question that may be asked at this point is: Why is it what describing dynamics in term of angular momentum, torque etcetera rather than momentum and force is more useful in discussing rotational motion. This is because in rotational motion, force, momenta etcetera are distributed and taking their moments by considering the angular momenta and torques automatically takes care of this distribution. We conclude this lecture by drawing a comparison between linear and rotational motion about a fixed axis.

This correspondence will help in understanding and getting relationships to solve most of the problems involving rotations about a fixed axis, particularly if you have solved many problems involving linear momentum.