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Section 9.14 Rotational Work

Consider rotating a rigid wheel about its axle by applying a force \(\vec F\) at the rim as shown in Figure 9.14.1. In time \(\Delta t\text{,}\) the force acts parallel to the displacement \(\Delta s\) at the rim. Therefore, the work done by the force will be

\begin{equation*} W = F \Delta s. \end{equation*}

Figure 9.14.1.

This work can be written in terms of torque and angle \(\Delta \theta\) of rotation. With \(R\) as radius of the wheel, arc distance, \(\Delta s = R\Delta\theta\text{.}\)

\begin{equation*} W = F\, \Delta s = F\, R\, \Delta\theta. \end{equation*}

The torque by the force applied at the rim is \(\tau = FR\text{.}\) Therefore, the work by the force in rotating the wheel by and angle \(\Delta\theta\) is

\begin{equation*} W = \tau \Delta \theta. \end{equation*}

This work is called rotational work.

A \(12\text{-inch}\)-long wrench is used to tighten a bolt by applying a steady force of \(5\text{ N}\) at the end of the handle which is approximately \(10\text{ inches}\) from the center of the bolt.

Figure 9.14.3.

(a) How much work is done for each quarter turn?

(b) Suppose, you use a \(6\)-in wrench, and apply the same force, except that the force now would act at a distance of about \(5\text{ inches}\) from the center of the bolt. How much work would now be done for each quarter turn?


(a) Use \(W = \tau\Delta \theta\) after expressing quantities in SI units. (b) Similarly as (a).


(a) \(2.0\text{ N.m} \text{,}\) (b) \(1.0\text{ N.m}\text{.}\)

Solution 1 (a)

(a) The lever arm of the \(5\text{ N}\) force is

\begin{equation*} r_\perp = 10\text{ inches} \times 0.0254\text{ m/inch} = 0.254\text{ m}. \end{equation*}

Therefore, the torque is

\begin{equation*} \tau = 5\text{ N} \times 0.254\text{ m} = 1.28\text{ N.m}. \end{equation*}

The angle to be rotated at this torque level is

\begin{equation*} \Delta \theta = 0.25\text{ turn} \times 2\pi\text{ rad/turn} = \dfrac{\pi}{2}\text{ rad}. \end{equation*}

Therefore, the rotational work is

\begin{equation*} W_\text{rot} = \tau\Delta \theta = 2.0\text{ N.m}. \end{equation*}
Solution 2 (b)

(b) The lever arm of the \(5\text{ N}\) force is now half as much. This factor will continue into the work.

\begin{equation*} W_\text{rot} = 1.0\text{ N.m}. \end{equation*}