4.6 Rotational Kinetic Energy

In order to discuss rotational kinetic energy, imagine a rigid body that rotates about an axis. Again, we will discuss the body in terms of n small units of mass. For some element of mass, ds_i, at a radius of r_i from the axis of rotation, after some length of time it has rotated through an angle of dq . There is a force on this element of mass, and the work done by this force (dw_i) is equal to the product of the acting force (F_it) and the distance through which the element travels (ds_i), which is equal to the product of the acting force, the radius to the element (r_i), and the angle through which it has traveled. In 4.3 it was mentioned that F_itr_i is the torque on element i. Thus, the work done is equal to the product of the torque and the angle through which the element travels.

The work done by the entire body is equal to the product of the net external torque and the angle through which the elements of the body travel

The power of the body (P) is equal to dw/dt, which is equal to the product of torque dq /dt, or t w .

The kinetic energy of the body is equal to the sum of the kinetic energy of all n elements. Thus, kinetic energy (KE) is equal to the sum for all elements from i to n of m_iv_i2, where m_i is mass and v_i is translational velocity.

The translational velocity is equal to the product of the roational velocity and the angle through which the body rotates. Thus,

We can remove ½w ² from the summation, as they are the same for all elements. We now have

Recall Equation 4.14 from 4.3: I = (sum from i to N of m_ir_i²). Thus,

The kinetic energy of a rotating body is equal to the product of one-half the square of its angular velocity and its moment of inertia.