18.10 Work - kinetic energy theorem for rotational motion  (Page 2/3)

This understanding of correspondence of expressions helps us to remember expressions of rotational motion, but it does not provide the insight into the angular quantities used to describe pure rotation. For example, we can not understand how torque accomplishes work on a rigid body, which does not have any linear displacement ! For this reason, we shall develop expressions of work by considering rotation of a particle or particle like object in the next section.

We need to emphasize here another important underlying concept. The description of motion in pure translation for a single particle and a rigid body differ in one very important aspect. Recall that the motion of a rigid body, in translation, is described by assigning motional quantities to the "center of mass(COM)". If we look at the expression of center of mass in x-direction, then we realize that the concept of COM is actually designed to incorporate distribution of mass in the body.

We must note here that such distinction arising due to distribution of mass between a particle and rigid body does not exist for pure rotation. It is so because the effect of the distribution of mass is incorporated in the definition of moment of inertia itself :

For this reason, there is no corresponding "center of rotational inertia" or "center of moment of inertia" for rigid body in rotation. This is one aspect in which there is no correspondence between two motion types. It follows, then, that the expressions of various quantities of a particle or a rigid body in rotation about a fixed axis should be same. Same expressions would, therefore, determine work and kinetic energy of a particle or a rigid body. The difference in two cases will solely arise from the differences in the values of moments of inertia.

Work done in rotation

We consider a particle like object, attached to a "mass-less" rod, to simulate rotation of a particle. The particle like object rotates in a plane perpendicular to the plane of screen or paper as shown in the figure. Here, we shall evaluate work by a force, "F", in xy-plane, which is perpendicular to the axis of rotation.

We know that it is only the tangential component of force that accelerates the particle and does the work. The component of force perpendicular to displacement does not perform work. For small linear displacement "ds", we have :

Rotation

$$\begin{array}{l}đW=F_xđx=F_tđs\end{array}$$

By geometry,

$$\begin{array}{l}=rđ\theta \end{array}$$

Combining two equations, we have :

$$\begin{array}{l}đW=F_{t}rđ\theta =\tau đ\theta \end{array}$$

Work done by a torque in rotating the particle like body by an angular displacement, we have :

$$\begin{array}{l}\Rightarrow W=\int \tau đ\theta \end{array}$$

If work is done by a constant torque, then we can take torque out of the integral sign and :

$$\begin{array}{l}\Rightarrow W=\tau \int đ\theta =\tau \Delta \theta \end{array}$$

As discussed earlier, this same expressions is valid for work done on a rigid body in pure rotation.

Power in rotation

Power measures the rate at which external torque does work on a particle like object or a rigid body. It is equal to the first time differential of work done,

We must note the correspondence of this expression with its linear counterpart, where :