10.8 Work and power for rotational motion  (Page 2/7)

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.

Rotational work: a pulley

A string wrapped around the pulley in [link] 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.

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

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

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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, [link] 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,