11.1: Work on a rigid body

If an external force \(\overrightarrow{\boldsymbol{F}}_{B}\) acts on a rigid body on a single point of action, then the work it does is equal to the work it would do on a point mass that is located on the point of action \(\overrightarrow{\boldsymbol{r}}_{B}\). So, we can just use Equation 7.2 to evaluate the work done by such a force.

Concept: Work of a force on a rigid body

\[W_{B, 1 \rightarrow 2}=\int_{\overrightarrow{\boldsymbol{r}}_{1}}^{\overrightarrow{\boldsymbol{r}}_{2}} \overrightarrow{\boldsymbol{F}}_{B, \mathrm{ext}} \cdot \mathrm{d} \overrightarrow{\boldsymbol{r}}_{B} \tag{11.1} \label{11.1}\]

It is important to note that because a rigid body is not deformable, the work done by internal forces on the rigid body is zero. This is obvious when evaluating the work done in a reference frame where the rigid body is at rest, since there all displacements \(\mathrm{d} \overrightarrow{\boldsymbol{r}}\) are zero, and is still valid when the rigid body moves as a consequence of Newton’s third law. Since it is not deformable, a rigid body also cannot store potential energy. Forces from deformable structures, like springs, can perform work, even if they are internal components in a system. So, if you have a system consisting of multiple elements, e.g. two rigid bodies connected by a spring, then you do have to consider the internal forces of that system when determining the work done on the system.

11.1.2 Work of a couple on a rigid body

Concept: Work of a couple

The work done by a pure moment or couple on a rigid body is given by the integral over the dot product between the moment vector and the infinitesimal angular displacement vector \(\mathrm{d} \overrightarrow{\boldsymbol{\phi}}=\mathrm{d} \phi \hat{\boldsymbol{k}}\).

\[W_{\text {couple }, 1 \rightarrow 2}=\int_{\phi_{1}}^{\phi_{2}} \overrightarrow{\boldsymbol{M}}_{\text {couple }} \cdot \mathrm{d} \vec{\phi} \tag{11.2} \label{11.2}\]

Derivation: Work of a couple

A simple couple \(\overrightarrow{\boldsymbol{M}}_{\text {couple }}\) is generated by 2 equal and opposite external forces \(\overrightarrow{\boldsymbol{F}}_{A}=-\overrightarrow{\boldsymbol{F}}_{B}\) that act on points \(A\) and \(B\) on the rigid body. We derive the work done by the couple when the rigid body purely rotates around fixed point \(A\) (e.g. because the point is chosen to be at the CoM: \(A=G\) ) such that the work done by \(\overrightarrow{\boldsymbol{F}}_{A}\) is zero and we only need to consider the work done by \(\overrightarrow{\boldsymbol{F}}_{B}\).

\[\begin{align} W_{\text {couple }, 1 \rightarrow 2} & =\int_{\overrightarrow{\boldsymbol{r}}_{1}}^{\overrightarrow{\boldsymbol{r}}_{2}} \overrightarrow{\boldsymbol{F}}_{B} \cdot \mathrm{d} \overrightarrow{\boldsymbol{r}}_{B} \tag{11.3} \label{11.3}\\[4pt] & =\int_{t_{1}}^{t_{2}} \overrightarrow{\boldsymbol{F}}_{B} \cdot \overrightarrow{\boldsymbol{v}}_{B} \mathrm{~d} t \tag{11.4} \label{11.4}\\[4pt] & =\int_{t_{1}}^{t_{2}} \overrightarrow{\boldsymbol{F}}_{B} \cdot\left(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / A}\right) \mathrm{d} t \tag{11.5} \label{11.5}\\[4pt] & =\int_{t_{1}}^{t_{2}}\left(\overrightarrow{\boldsymbol{r}}_{B / A} \times \overrightarrow{\boldsymbol{F}}_{B}\right) \cdot \overrightarrow{\boldsymbol{\omega}} \mathrm{d} t \tag{11.6} \label{11.6}\\[4pt] & =\int_{\phi_{1}}^{\phi_{2}} \overrightarrow{\boldsymbol{M}}_{\text {couple }} \cdot \mathrm{d} \overrightarrow{\boldsymbol{\phi}} \tag{11.7} \label{11.7}\end{align}\]

Here we used the vector identity \(\overrightarrow{\boldsymbol{a}} \cdot(\overrightarrow{\boldsymbol{b}} \times \overrightarrow{\boldsymbol{c}})=(\overrightarrow{\boldsymbol{c}} \times \overrightarrow{\boldsymbol{a}}) \cdot \overrightarrow{\boldsymbol{b}}\) and used Equation 10.53. Note that the moment of the couple, and the resulting expression for the work of a couple are independent of the reference point for rotations.

11.1.3 Work of a force field

The situation becomes more complicated when the rigid body moves in a force field. Then the total work of the force field can be determined by integrating or summing the work over all point masses in the rigid body. This can sometimes be quite complex if the force field changes with position. But it is easier for
a constant gravitational force field which we already analysed in Equation 7.43, where we found that:

\[W_{g, 1 \rightarrow 2}=-\Delta V_{g}=-m_{\mathrm{tot}} \Delta \overrightarrow{\boldsymbol{r}}_{G} \cdot \hat{\boldsymbol{\jmath}}=-m_{\mathrm{tot}} \Delta h_{G} \tag{11.8} \label{11.8}\]

So the work done on the whole rigid body by the gravitational force is equal to the height reduction of the CoM times the total mass of the rigid body. Note that the moment of the gravitational force on a rigid body is zero when the CoM is taken as reference point, therefore it does no work when the rigid body rotates around its CoM. This can be derived as follows.

Derivation: The moment of gravity on a system or rigid body is zero with respect to its CoM.

\[\begin{align} \overrightarrow{\boldsymbol{M}}_{g / G} & =\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / G} \times \overrightarrow{\boldsymbol{F}}_{i, g}=\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / G} \times\left(-m_{i} g \hat{\boldsymbol{\jmath}}\right) \tag{11.9} \label{11.9}\\[4pt] & =\left(\sum_{i} m_{i}\left(\overrightarrow{\boldsymbol{r}}_{i}-\overrightarrow{\boldsymbol{r}}_{G}\right)\right) \times(-g \hat{\boldsymbol{\jmath}}) \tag{11.10} \label{11.10}\\[4pt] & =\left[m_{\mathrm{tot}}\left(\overrightarrow{\boldsymbol{r}}_{G}-\overrightarrow{\boldsymbol{r}}_{G}\right)\right] \times(-g \hat{\boldsymbol{\jmath}})=\overrightarrow{\mathbf{0}} \tag{11.11} \label{11.11}\end{align}\]

where we used the property of the CoM that \(m_{\mathrm{tot}} \overrightarrow{\boldsymbol{r}}_{G}=\sum_{i} m_{i} \overrightarrow{\boldsymbol{r}}_{i}\).