Question 11.10: A rigid body shown in Fig. 11.4 is driven by a torque μ(t) a...
A rigid body shown in Fig. 11.4 is driven by a torque μ(t) about a fixed, principal body axis k in a smooth bearing at H . (i) Apply (11.73) to derive the equationof motionfor the body.(ii) Repeatthe derivation from (11.38). Show that the result has the familiar form of the equation of motion of a driven pendulum. (iii) Apply Euler’s law to obtain the equation of motion.
\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{q}_r}\right) – \frac{\partial T}{\partial q_r}=Q_r, \quad r=1,2, \ldots, n (11.73)
\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right) – \frac{\partial L}{\partial q_k}=Q_k^N (11.38)
Solution of (i). The system is holonomic with one degree of freedom described by q_1=\psi; hence, (11.73) yields
\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{\psi}}\right) – \frac{\partial T}{\partial \psi}=Q_\psi (11.74a)
With the total kinetic energy of the body T=\frac{1}{2} I \dot{\psi}^2, where I is the principal moment of inertia about the body axis at H, (11.74a) becomes
I \ddot{\psi}=Q_\psi. (11.74b)
We next determine the generalized force Q_\psi . The bearing reaction force R is workless, and the total external torque \mathbf{M}_H about H is the sum of the gravitational torque -W \ell \sin \psi \mathbf{k} and the applied driving torque \boldsymbol{\mu}=\mu \mathbf{k}. The virtual work \delta \mathscr{W} done by the total torque in the virtual displacement \delta \psi \equiv \delta \psi \mathbf{k} is thus given by
\delta \mathscr{W}=\mathbf{M}_H \cdot \delta \psi=(-W \ell \sin \psi + \mu) \delta \psi \equiv Q_\psi \delta \psi. (11.74c)
Hence, Q_\psi=-W \ell \sin \psi + \mu, and (11.74b) yields the equation of motion:
I \ddot{\psi} + m g \ell \sin \psi=\mu(t). (11.74d)
Solution of (ii). Application of the Lagrange equations (11.38) yields
\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{\psi}}\right) – \frac{\partial L}{\partial \psi}=Q_\psi^N . (11.74e)
Again, the workless constraint force R need not be considered, and the gravitational force is conservative with total potential energy V=m g \ell(1 – \cos \psi). Therefore, the Lagrangian is
L=T-V=\frac{1}{2} I \dot{\psi}^2 – m g \ell(1 – \cos \psi) . (11.74f)
The virtual work done by the nonconservative generalized force is \delta \mathscr{W}_N=\boldsymbol{\mu}·\delta \psi=\mu \delta \psi=Q_\psi^N \delta \psi. Hence, Q_\psi^N=\mu, and (11.74e) leads to (11.74d) .
With I=m R^2 in terms of the radius of gyration R, (l1.74d) may be written in the form of the equation of motion of a driven pendulum for which p^2 \equiv g \ell / R^2 and \hat{\mu}(t) \equiv \mu(t) / I; namely,
\ddot{\psi}+p^2 \sin \psi=\hat{\mu}(t). (11.74g)
Solution of (iii). Euler’s law for the rotation about a fixed principal axis at H requires \mathbf{M}_H=I_H \dot{\omega}, wherein \mathbf{M}_H=(\mu – W \ell \sin \psi) \mathbf{k} and I_H \dot{\omega}=I \ddot{\psi} \mathbf{k}. This yields the equation of motion I \psi=\mu – W \ell \sin \psi, which is the same as (11.74d) .