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

Question

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.

[latex]\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[/latex] (11.73)

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right) - \frac{\partial L}{\partial q_k}=Q_k^N[/latex] (11.38)

Accepted Answer

Solution of (i). The system is holonomic with one degree of freedom described by [latex]q_1=\psi[/latex]; hence, (11.73) yields

[latex]\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{\psi}}\right) - \frac{\partial T}{\partial \psi}=Q_\psi[/latex] (11.74a)

With the total kinetic energy of the body [latex]T=\frac{1}{2} I \dot{\psi}^2,[/latex] where I is the principal moment of inertia about the body axis at H, (11.74a) becomes

[latex]I \ddot{\psi}=Q_\psi[/latex]. (11.74b)

We next determine the generalized force [latex]Q_\psi .[/latex] The bearing reaction force R is workless, and the total external torque [latex]\mathbf{M}_H[/latex] about H is the sum of the gravitational torque [latex]-W \ell \sin \psi \mathbf{k}[/latex] and the applied driving torque [latex]\boldsymbol{\mu}=\mu \mathbf{k}[/latex]. The virtual work [latex]\delta \mathscr{W}[/latex] done by the total torque in the virtual displacement [latex]\delta \psi \equiv \delta \psi \mathbf{k}[/latex] is thus given by

[latex]\delta \mathscr{W}=\mathbf{M}_H \cdot \delta \psi=(-W \ell \sin \psi + \mu) \delta \psi \equiv Q_\psi \delta \psi[/latex]. (11.74c)

Hence, [latex]Q_\psi=-W \ell \sin \psi + \mu[/latex], and (11.74b) yields the equation of motion:

[latex]I \ddot{\psi} + m g \ell \sin \psi=\mu(t)[/latex]. (11.74d)

Solution of (ii). Application of the Lagrange equations (11.38) yields

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{\psi}}\right) - \frac{\partial L}{\partial \psi}=Q_\psi^N .[/latex] (11.74e)

Again, the workless constraint force R need not be considered, and the gravitational force is conservative with total potential energy [latex]V=m g \ell(1 - \cos \psi)[/latex]. Therefore, the Lagrangian is

[latex]L=T-V=\frac{1}{2} I \dot{\psi}^2 - m g \ell(1 - \cos \psi) .[/latex] (11.74f)

The virtual work done by the nonconservative generalized force is [latex]\delta \mathscr{W}_N=\boldsymbol{\mu}·\delta \psi=\mu \delta \psi=Q_\psi^N \delta \psi[/latex]. Hence, [latex]Q_\psi^N=\mu[/latex], and (11.74e) leads to (11.74d) .

With [latex]I=m R^2[/latex] 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 [latex]p^2 \equiv g \ell / R^2[/latex] and [latex]\hat{\mu}(t) \equiv \mu(t) / I[/latex]; namely,

[latex]\ddot{\psi}+p^2 \sin \psi=\hat{\mu}(t)[/latex]. (11.74g)

Solution of (iii). Euler’s law for the rotation about a fixed principal axis at H requires [latex]\mathbf{M}_H=I_H \dot{\omega}[/latex], wherein [latex]\mathbf{M}_H=(\mu - W \ell \sin \psi) \mathbf{k}[/latex] and [latex]I_H \dot{\omega}=I \ddot{\psi} \mathbf{k}[/latex]. This yields the equation of motion [latex]I \psi=\mu - W \ell \sin \psi[/latex], which is the same as (11.74d) .

Question 11.10: A rigid body shown in Fig. 11.4 is driven by a torque μ(t) a...

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