11.2: Equation of Motion for a Rigid Body in Pure Plane Rotation

An important special case of general plane motion is pure rotation about a fixed pivot or hinge, as depicted in Figure \(\PageIndex{1}\). To analyze this case, let us suppose that reaction point \(A\) of Figure 11.1.1 is the hinge point, and let us position the \(xyz\) origin \(O\) at point \(A\). To emphasize the character of this hinge point, we re-label it as \(H\); in other words, both points \(A\) and \(O\) on Figure 11.1.1 now become hinge point \(H\) on Figure \(\PageIndex{1}\). We can use the previous equations simply by setting position vector \(\mathbf{r}_{\mathrm{OA}}\) on Figure 11.1.1 equal to zero: \(\mathbf{r}_{\mathrm{OA}} =\mathbf{0}=\mathbf{r}_{\mathrm{oc}}+\mathbf{r}_{\mathrm{CA}} \Rightarrow \mathbf{r}_{\mathrm{CA}}=-\mathbf{r}_{\mathrm{OC}} \equiv-\mathbf{r}_{\mathrm{HC}}\). It is helpful also to express \(\mathbf{r}_{\mathrm{CB}}\) on Figure 11.1.1 in terms of \(\mathbf{r}_{\mathrm{o} \mathbf{C}}\): \(\mathbf{r}_{\mathrm{CB}}=\mathbf{r}_{\mathrm{OB}}-\mathbf{r}_{\mathrm{OC}}\). Using these substitutions in Equation 11.1.4 gives

\[J_{C} \ddot{\theta} \mathbf{1}_{\mathbf{z}}=-\mathbf{r}_{\mathbf{O C}} \times \mathbf{F}_{\mathbf{A}}+\left(\mathbf{r}_{\mathrm{OB}}-\mathbf{r}_{\mathbf{O C}}\right) \times \mathbf{F}_{\mathbf{B}}+\left(M_{A}+M_{D}\right) \mathbf{1}_{\mathbf{z}}\label{eqn:11.4a} \]

After we replace subscripts \(O\) and \(A\) with \(H\) and re-arrange terms, Equation \(\ref{eqn:11.4a}\) becomes

\[J_{C} \ddot{\theta} \mathbf{1}_{\mathbf{z}}+\mathbf{r}_{\mathbf{H C}} \times\left(\mathbf{F}_{\mathbf{H}}+\mathbf{F}_{\mathbf{B}}\right)=\mathbf{r}_{\mathbf{H B}} \times \mathbf{F}_{\mathbf{B}}+\left(M_{H}+M_{D}\right) \mathbf{1}_{\mathbf{z}}\label{eqn:11.4b} \]

Next, substituting Equation 11.1.1 into Equation \(\ref{eqn:11.4b}\) gives

\[J_{C} \ddot{\theta} \mathbf{1}_{\mathbf{z}}+\mathbf{r}_{\mathbf{H C}} \times m \mathbf{a}_{\mathbf{C}}=\mathbf{r}_{\mathbf{H B}} \times \mathbf{F}_{\mathbf{B}}+\left(M_{H}+M_{D}\right) \mathbf{1}_{\mathbf{z}}\label{eqn:11.5a} \]

Since hinge point \(H\) is fixed, let us simplify notation by writing \(\mathbf{r}_{\mathbf{H C}} \equiv \mathbf{r}_{\mathbf{C}}\) and \(\mathbf{r}_{\mathrm{HB}} \equiv \mathbf{r}_{\mathrm{B}}\). Also, we can express the acceleration as \(\mathbf{a}_{\mathrm{C}} \equiv \ddot{\mathbf{r}}_{\mathrm{C}}\), so Equation \(\ref{eqn:11.5a}\) becomes

\[J_{C} \ddot{\theta} \mathbf{1}_{\mathrm{z}}+\mathbf{r}_{\mathrm{C}} \times m \ddot{\mathbf{r}}_{\mathrm{C}}=\mathbf{r}_{\mathrm{B}} \times \mathbf{F}_{\mathrm{B}}+\left(M_{H}+M_{D}\right) \mathbf{1}_{\mathrm{z}}\label{eqn:11.5b} \]

We can simplify the \(\ddot{\mathbf{r}}_{\mathrm{C}}\) term on the left-hand side of Equation \(\ref{eqn:11.5b}\) by using the kinematics of rigid-body, plane, pure rotation about point \(H\). For this task, it is convenient to work in polar rather than Cartesian coordinates, so we define on Figure \(\PageIndex{1}\) the orthogonal rotating unit vectors \(\mathbf{1}_{\mathrm{r}}\) and \(\mathbf{1}_{\theta}\); thus, we can express the position vector to mass center \(C\) as \(\mathbf{r}_{\mathrm{C}}=r_{C} \mathbf{1}_{\mathrm{r}}\), where \(r_C\) is the constant radius from hinge point \(H\) to mass center \(C\). We express the rotational velocity vector as \(\boldsymbol{\omega}=\dot{\theta} \mathbf{1}_{\mathbf{z}}\). (In this context, the vector symbol \(\mathbf{\omega}\) denotes rotational velocity, which is standard in dynamics textbooks, whereas everywhere else in this book the scalar symbol \(\omega\) denotes frequency of vibration.) Therefore, the first derivative of the mass-center position vector is

\[\dot{\mathbf{r}}_{\mathrm{C}}=\boldsymbol{\omega} \times \mathbf{r}_{\mathrm{C}}=\dot{\theta} \mathbf{1}_{\mathrm{z}} \times r_{C} \mathbf{1}_{\mathrm{r}}=r_{C} \dot{\theta} \mathbf{1}_{\mathbf{\theta}}\label{eqn:11.6} \]

To obtain the second derivative, we differentiate product Equation \(\ref{eqn:11.6}\), recognizing that \(r_C\) is constant, but that \(\mathbf{1}_{\boldsymbol{\theta}}\) is rotating:

\[\ddot{\mathbf{r}}_{\mathrm{C}}=r_{C}\left[\ddot{\theta} \mathbf{1}_{\mathrm{e}}+\dot{\theta} \mathbf{i}_{\mathrm{\theta}}\right]=r_{C}\left[\ddot{\theta} \mathbf{1}_{\mathrm{\theta}}+\dot{\theta}\left(\dot{\theta} \mathbf{1}_{\mathrm{z}} \times \mathbf{1}_{\mathrm{\theta}}\right)\right]=r_{C}\left[\ddot{\theta} \mathbf{1}_{\mathrm{e}}-\dot{\theta}^{2} \mathbf{1}_{\mathrm{r}}\right]\label{eqn:11.7} \]

With Equation \(\ref{eqn:11.7}\), the vector cross product on the left-hand side of Equation \(\ref{eqn:11.5b}\) becomes

\[\mathbf{r}_{\mathrm{C}} \times \ddot{\mathbf{r}}_{\mathrm{C}}=r_{C} \mathbf{1}_{\mathrm{r}} \times r_{C}\left[\ddot{\theta} \mathbf{1}_{\mathbf{\theta}}-\dot{\theta}^{2} \mathbf{1}_{\mathrm{r}}\right]=r_{C}^{2} \ddot{\theta} \mathbf{1}_{\mathbf{z}}\label{eqn:11.8} \]

With Equation \(\ref{eqn:11.8}\), the left-hand side of Equation \(\ref{eqn:11.5b}\) becomes

\[J_{C} \ddot{\theta} \mathbf{1}_{\mathbf{z}}+\mathbf{r}_{\mathbf{C}} \times m \ddot{\mathbf{r}}_{\mathrm{C}}=\left(J_{C}+m r_{C}^{2}\right) \ddot{\theta} \mathbf{1}_{\mathbf{z}} \equiv J_{H} \ddot{\theta} \mathbf{1}_{\mathbf{z}}\label{eqn:11.9} \]

In Equation \(\ref{eqn:11.9}\), we use the parallel-axis theorem to define the rotational inertia of the rigid body about center of rotation \(H\), \(J_{H}=J_{C}+m r_{C}^{2}\). Thus, Equation \(\ref{eqn:11.5b}\) becomes

\[J_{H} \ddot{\theta} \mathbf{1}_{\mathbf{z}}=\mathbf{r}_{\mathbf{B}} \times \mathbf{F}_{\mathbf{B}}+\left(M_{H}+M_{D}\right) \mathbf{1}_{\mathbf{z}} \equiv \Sigma(\text { all active and reactivemoments about } H) \mathbf{1}_{\mathbf{z}}\label{eqn:11.10} \]

To summarize, for a rigid-body in pure plane rotation about point \(H\), the only degree of freedom is rotation \(\theta(t)\), and the single, relatively simple ODE of motion is

\[J_{H} \ddot{\theta}=\Sigma(\text { allactive and reactivemoments about } H)\label{eqn:11.11} \]

in which \(J_H\) is the rotational inertia of the rigid body about center of rotation \(H\).

Note that the reaction force at hinge \(H\) does not appear in ODE Equation \(\ref{eqn:11.11}\). In fact, to find this reaction force, you would have to solve Equation \(\ref{eqn:11.11}\) for \(\theta(t)\), then find the acceleration of \(C\), then substitute the results back into Equations 11.1.1 or 11.1.2 and 11.1.3.