Question 10.9: A thin homogeneous rectangular plate in Fig. 10.9 rotates ab...
A thin homogeneous rectangular plate in Fig. 10.9 rotates about a diagonal axis with angular velocity wand angular acceleration \dot{\omega} (i) What is the total torque exerted on the plate about C, referred to the principal body frame \varphi=\left\{C ; \mathbf{i}_k\right\}? (ii) Determine the total torque about C referred to the body frame \varphi^{\prime}=\left\{C ; \mathbf{i}_k^{\prime} \right\}, and identify the restraining torques acting on the plate. (iii) Discuss the role of the static bearing reaction force s, derive an equation relating the drive torque T to the rotation, and determine the dynamic bearing react ion forces. (iv) Suppose that T(t) is specified; find the angular speed.
Solution of (i). The total torque M_C acting on the plate about C is determined by Euler’s equations (10.66). With this in mind, select the principal body frame \varphi=\left\{C ; \mathbf{i}_k\right\} at the center of mass and, for convenience, write
\begin{aligned}&M_1^Q=I_{11}^Q \dot{\omega}_1 + \omega_2 \omega_3\left(I_{33}^Q – I_{22}^Q\right), \\&M_2^Q=I_{22}^Q \dot{\omega}_2 + \omega_3 \omega_1\left(I_{11}^Q – I_{33}^Q\right), \\&M_3^Q=I_{33}^Q \dot{\omega}_3 + \omega_1 \omega_2\left(I_{22}^Q – I_{11}^Q\right).\end{aligned} (10.66)
\sigma \equiv \sin \theta=\frac{w}{\sqrt{\ell^2 + w^2}}, \quad \gamma \equiv \cos \theta=\frac{\ell}{\sqrt{\ell^2 + w^2}}. (10.79a)
Referred to φ, the angular velocity and angular acceleration are given by
\boldsymbol{\omega}=\omega(-\sigma \mathbf{i} + \gamma \mathbf{j}), \quad \dot{\boldsymbol{\omega}}=\dot{\omega}(-\sigma \mathbf{i} + \gamma \mathbf{j}); (10.79b)
and from (9.27), with \left(\mathbf{i}_1^*, \mathbf{i}_2^*, \mathbf{i}_3^*\right)=(\mathbf{j},-\mathbf{i}, \mathbf{k}), the principal moments of inertia about C for the homogeneous thin plate are
\mathbf{I}_C=\frac{m \ell^2}{12} \mathbf{i} \otimes \mathbf{i} + \frac{m w^2}{12} \mathbf{j} \otimes \mathbf{j} + \frac{m}{12}\left(\ell^2 + w^2\right) \mathbf{k} \otimes \mathbf{k}. (10.79c)
The total torque on the plate is now given by Euler’s principal axis equations (10.66) for Q = C:
\mathbf{M}_C=\frac{\dot{\omega} m}{12}\left(-\sigma \ell^2 \mathbf{i}+\gamma w^2 \mathbf{j}\right) + \frac{\omega^2 m \sigma \gamma}{12}\left(\ell^2 – w^2\right) \mathbf{k}. (10.79d)
This result shows that a moment about the k-axis perpendicular to the plane of the plate is required to sustain the motion even when the angular velocity is constant. This torque rotates with the k-axis normal to plate and produces alternating reactions at the support bearings. A square plate (\ell=w), however, can spin at a constant angular speed without application of any torque whatsoever.
Solution of (ii). To relate M_C to the plate frame \varphi^{\prime}=\left\{C ; \mathbf{i}_k^{\prime}\right\} for which \mathbf{j}^{\prime} is the axis of rotation and \mathbf{k}^{\prime}=\mathbf{k} in Fig. 10.9, we use (10.79a) and the vector transformation law (3.107a): M_C^{\prime}=A M_C, where A=\left[A_{j k}\right]=\left[\cos \left\langle\mathbf{i}_j^{\prime}, \mathbf{i}_k\right\rangle\right], to obtain
M_C^{\prime}=\left[\begin{array}{ccc}\gamma & \sigma & 0 \\-\sigma & \gamma & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}M_1 \\M_2 \\M_3\end{array}\right]=\left[\begin{array}{c}\gamma M_1 + \sigma M_2 \\-\sigma M_1 + \gamma M_2 \\M_3\end{array}\right] \text {, } (10.79e)
in which M_k are the components in (10.79d). Thus, referred to \varphi^{\prime}, the total torque exerted on the plate is
\mathbf{M}_C=\frac{m w \ell\left(\ell^2 – w^2\right)}{12\left(\ell^2 + w^2\right)}\left(-\dot{\omega} \mathbf{i}^{\prime} + \omega^2 \mathbf{k}^{\prime}\right) + \frac{\dot{\omega} m w^2 \ell^2}{6\left(\ell^2 + w^2\right)} \mathbf{j}^{\prime} . (10.79f)
The torque about the bearing axle \mathbf{j}^{\prime} is related to the applied drive torque T, and the remaining components, which arise from the asymmetrical distribution of mass about the plate diagonal , are restraining torques supplied by the bearings at A and B.
Solution of (iii). We next explore the role of the static bearing reaction forces . The free body diagram of the plate is shown in Fig. 10.10. When the plate is at rest, each shaft bearing support exerts an equal force \mathbf{A}_S=\mathbf{B}_S=-(\mathbf{W}) / 2 on the plate at A and B, equal to one-half its weight W. These static loads are equipollent to zero, and therefore they contribute nothing to the total force or to the total torque about the center of mass C in the dynamics problem. Henceforward, these statically balanced forces may be ignored .
Now let us relate the drive torque \mathbf{T}=T \mathbf{j}^{\prime} to the rotational motion and determine the resultant dynamic bearing reaction forces \mathbf{A}=A_k \mathbf{i}_k^{\prime} \text { and } \mathbf{B}=B_k \mathbf{i}_k^{\prime}, exerted by the shaft at A and B, respectively, and which we shall suppose, for simplicity, act at the comers of the plate in Fig. 10.10. Euler’s first law (10.26) requires that \mathbf{A} + \mathbf{B}=m \mathbf{a}^*=\mathbf{0}; thus, A = –B, so the bearing reaction force system forms a couple with moment arm 2 \mathbf{x}=\left(\ell^2+w^2\right)^{1 / 2} \mathbf{j^{\prime}}, where x is the position vector of A from C. Therefore, the total applied torque on the plate about C is
\mathbf{F}(\mathscr{B}, t)=m(\mathscr{B}) \mathbf{a}^*(\mathscr{B}, t) (10.26)
\mathbf{M}_C=\mathbf{T} + 2 \mathbf{x} \times \mathbf{A}=T \mathbf{j}^{\prime} + \sqrt{\ell^2 + w^2}\left(A_3 \mathbf{i}^{\prime} – A_1 \mathbf{k}^{\prime}\right) . (10.79g)
Equating the corresponding components in (10.79f) and ( 10.79g), we obtain an equation relating the drive torque to the rotation and two relations for the dynamic bearing reaction force components:
\begin{aligned}T &=\frac{\dot{\omega} m w^2 \ell^2}{6\left(\ell^2 + w^2\right)}, \quad B_1=-A_1=\frac{m \omega^2 w \ell\left(\ell^2 – w^2\right)}{12\left(\ell^2 + w^2\right)^{3 / 2}}, \\ B_3 &=-A_3=\frac{T\left(\ell^2 – w^2\right)}{2 w \ell\left(\ell^2 + w^2\right)^{1 / 2}}.\end{aligned} (10.79h)
The axial components of the forces exerted by the bearing s must satisfy B_2=-A_2; otherwise, this axial force is undetermined by this analysis, and nothing is lost by putting it equal to zero. Usually, however, thrust bearings are used at the shaft ends to secure any axle drift of the drive shaft. Since drag torque s due to friction in the bearings are ignored , when the drive torque is removed, the plate will spin indefinitely with constant angular speed and the only nonzero bearing reaction force components in (10.79h) are A_1=-B_1.
For a square plate (w=\ell) all dynamic bearing reaction forces in (10.79h) vanish. Therefore, when the drive torque is removed, the force systemis equipollent to zero: \mathbf{F}(\mathscr{B}, t)=\mathbf{0} \text { and } \mathbf{M}_C(\mathscr{B}, t)=\mathbf{0}, but the square plate is not in equilibrium; it continues to spin with a constant angular speed.
Solution of (iv). Suppose the applied driving torque T (t) is given and \omega(0)=\omega_0 initially. Then integration of the first equation in (10.79h) yields the angular speed
\omega(t)=\omega_0 + \frac{6\left(\ell^2 + w^2\right)}{m w^2 \ell^2} \int_0^t T(t) d t. (10.79i)
The dynamic bearing reactions are then determined by their equations in (10.79h).
