10.8 Work and Power for Rotational Motion

Work for Rotational Motion

Now that we have determined how to calculate kinetic energy for rotating rigid bodies, we can proceed with a discussion of the work done on a rigid body rotating about a fixed axis. Figure 10.39 shows a rigid body that has rotated through an angle $d\theta$ from A to B while under the influence of a force $\overrightarrow{F}$. The external force $\overrightarrow{F}$ is applied to point P, whose position is $\overrightarrow{r}$, and the rigid body is constrained to rotate about a fixed axis that is perpendicular to the page and passes through O. The rotational axis is fixed, so the vector $\overrightarrow{r}$ moves in a circle of radius r, and the vector $d\overrightarrow{s}$ is perpendicular to $\overrightarrow{r}.$

Figure 10.39 A rigid body rotates through an angle $d\theta$ from A to B by the action of an external force $\overrightarrow{F}$ applied to point P.

From Equation 10.2, we have

Thus,

Note that $d\overrightarrow{r}$ is zero because $\overrightarrow{r}$ is fixed on the rigid body from the origin O to point P. Using the definition of work, we obtain

where we used the identity $\overrightarrow{a}·(\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{b}·(\overrightarrow{c}\times \overrightarrow{a})$. Noting that $(\overrightarrow{r}\times {\displaystyle \sum \overrightarrow{F}})={\displaystyle \sum \overrightarrow{\tau }}$, we arrive at the expression for the rotational work done on a rigid body:

The total work done on a rigid body is the sum of the torques integrated over the angle through which the body rotates. The incremental work is

where we have taken the dot product in Equation 10.27, leaving only torques along the axis of rotation. In a rigid body, all particles rotate through the same angle; thus the work of every external force is equal to the torque times the common incremental angle $d\theta$. The quantity $\left({\displaystyle \sum _i{\tau }_i}\right)$ is the net torque on the body due to external forces.

Similarly, we found the kinetic energy of a rigid body rotating around a fixed axis by summing the kinetic energy of each particle that makes up the rigid body. Since the work-energy theorem $W_i=\text{Δ}{K}_i$ is valid for each particle, it is valid for the sum of the particles and the entire body.

We give a strategy for using this equation when analyzing rotational motion.

Let’s look at two examples and use the work-energy theorem to analyze rotational motion.

Example 10.18

Rotational Work: A Pulley

A string wrapped around the pulley in Figure 10.40 is pulled with a constant downward force $\overrightarrow{F}$ of magnitude 50 N. The radius R and moment of inertia I of the pulley are 0.10 m and $2.5\times {10}^{\mathrm{-3}}{\text{kg-m}}^2$, respectively. If the string does not slip, what is the angular velocity of the pulley after 1.0 m of string has unwound? Assume the pulley starts from rest.

Figure 10.40 (a) A string is wrapped around a pulley of radius R. (b) The free-body diagram.

Strategy

Looking at the free-body diagram, we see that neither $\overrightarrow{B}$, the force on the bearings of the pulley, nor$M\overrightarrow{g}$, the weight of the pulley, exerts a torque around the rotational axis, and therefore does no work on the pulley. As the pulley rotates through an angle $\theta ,$ $\overrightarrow{F}$ acts through a distance d such that $d=R\theta .$

Solution

Since the torque due to $\overrightarrow{F}$ has magnitude $\tau =RF$, we have

$$W=\tau \theta =(FR)\theta =Fd.$$

If the force on the string acts through a distance of 1.0 m, we have, from the work-energy theorem,

$$\begin{array}{ccc}\hfill W_{AB}& =\hfill & K_B-K_A\hfill \\ \hfill Fd& =\hfill & \frac{1}{2}I{\omega }^2-0\hfill \\ \hfill (50.0\text{N})(1.0\text{m})& =\hfill & \frac{1}{2}(2.5\times {10}^{\mathrm{-3}}{\text{kg-m}}^2){\omega }^2.\hfill \end{array}$$

Solving for $\omega$, we obtain

$$\omega =200.0\text{rad}\text{/}\text{s}.$$

Power for Rotational Motion

Power always comes up in the discussion of applications in engineering and physics. Power for rotational motion is equally as important as power in linear motion and can be derived in a similar way as in linear motion when the force is a constant. The linear power when the force is a constant is $P=\overrightarrow{F}·\overrightarrow{v}$. If the net torque is constant over the angular displacement, Equation 10.25 simplifies and the net torque can be taken out of the integral. In the following discussion, we assume the net torque is constant. We can apply the definition of power derived in Power to rotational motion. From Work and Kinetic Energy, the instantaneous power (or just power) is defined as the rate of doing work,

If we have a constant net torque, Equation 10.25 becomes $W=\tau \theta$ and the power is

or

Rotational and Translational Relationships Summarized

The rotational quantities and their linear analog are summarized in three tables. Table 10.5 summarizes the rotational variables for circular motion about a fixed axis with their linear analogs and the connecting equation, except for the centripetal acceleration, which stands by itself. Table 10.6 summarizes the rotational and translational kinematic equations. Table 10.7 summarizes the rotational dynamics equations with their linear analogs.