Use the results for Example 10.3, page 424, to determine (i) the total torque about the hinge Q, and (ii) the total torque about the center of mass C, required to sustain the motion of the thin rod.

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Accepted Answer

Solution of (i). The total torque about the moving point Q in Fig. 10.6 can be found from either (10.42) or (10.46) . Let us consider (10.46), note that

[latex]\mathbf{M}_Q(\mathscr{B}, t)=\dot{\mathbf{h}}_Q(\mathscr{B}, t) + \mathbf{v}_Q imes \mathbf{p}^*(\mathscr{B}, t)[/latex] (10.42)

[latex]\mathbf{M}_Q(\mathscr{B}, t)=\dot{\mathbf{h}}_{r Q}(\mathscr{B}, t)+\mathbf{r}^*(\mathscr{B}, t) imes m(\mathscr{B}) \mathbf{a}_Q[/latex] (10.46)

[latex]\mathbf{r}^*(\mathscr{B}, t) imes m \mathbf{a}_Q=-\frac{\ell}{2} \mathbf{i} imes m \alpha \Omega^2 \mathbf{a}=\frac{m \alpha \ell}{2} \Omega^2 \sin \beta \mathbf{k},[/latex] (10.50a)

and recall [latex]\mathbf{h}_{r Q}=\left(m \ell^2 / 3 ight)(\dot{\beta} + \Omega) \mathbf{k}[/latex], which is a vector referred to the moving rod frame 2 whose total angular velocity is [latex]\omega_f=(\Omega + \dot{\beta}) \mathbf{k}[/latex] in (10.41b).

[latex]\boldsymbol{\omega}=(\dot{\beta} + \Omega) \mathbf{k}[/latex] (10.41b)

Because [latex]\omega_f[/latex] is parallel to [latex]\mathbf{h}_{r Q}[/latex], (10.49) simplifies to

[latex]\mathbf{M}_Q(\mathscr{B}, t)=\frac{d \mathbf{h}_{r Q}(\mathscr{B}, t)}{d t}=\frac{\delta \mathbf{h}_{r Q}(\mathscr{B}, t)}{\delta t} + \omega_f imes \mathbf{h}_{r Q}(\mathscr{B}, t)[/latex] (10.49)

[latex]\dot{\mathbf{h}}_{r Q}(\mathscr{B}, t)=\frac{\delta \mathbf{h}_{r Q}(\mathscr{B}, t)}{\delta t}=\frac{m \ell^2}{3} \ddot{\beta} \mathbf{k} .[/latex] (10.50b)

Hence , use of (10.50a) and (10.50b) in (10.46) delivers the total torque about Q required to sustain the motion of the connecting rod:

[latex]\mathbf{M}_Q(\mathscr{B}, t)=\left(\frac{m \ell^2}{3} \ddot{\beta} + \frac{m \alpha \ell}{2} \Omega^2 \sin \beta ight) \mathbf{k} .[/latex] (10.50c)

The reader may show that (10.42) delivers the same result.

Solution of (ii). Now let us consider Euler’s simple rule (10.48) for the moving center of mass and recall the definition (10.39) in which [latex] ho(P, t)= ho \mathbf{i}[/latex] denotes the position vector of a material point P at ρ from the center of mass C. Then , with the aid of (10.41b), the relative rigid body velocity of [latex]\dot{ ho}(P, t)=\boldsymbol{\omega} imes ho= ho(\dot{\beta}+\Omega) \mathbf{j}, d m=\sigma d ho[/latex], and by (10.39), the moment of momentum relative C may be written as

[latex]\mathbf{M}_C(\mathscr{B}, t)=\dot{\mathbf{h}}_{r C}(\mathscr{B}, t)[/latex] (10.48)

[latex]\mathbf{h}_C(\mathscr{B}, t)=\int_{\mathscr{B}} ho(P, t) imes \dot{ ho}(P, t) d m(P)=\mathbf{h}_{r C}(\mathscr{B}, t)[/latex] (10.39)

[latex]\mathbf{h}_{r c}=\sigma(\dot{\beta} + \Omega) \int_{-\ell / 2}^{\ell / 2} ho^2 d ho \mathbf{k}=m(\dot{\beta} + \Omega) \frac{\ell^2}{12} \mathbf{k}[/latex] (10.50d)

This vector is parallel to [latex]\boldsymbol{\omega}_f=\boldsymbol{\omega}[/latex], and hence (10.48) and (10.49) yield the total torque about C needed to sustain the rod’s motion in the machine frame, i.e.

[latex]\mathbf{M}_C(\mathscr{B}, t)=\dot{\mathbf{h}}_{r C}(\mathscr{B}, t)=\frac{m \ell^2}{12} \ddot{\beta}(t) \mathbf{k}[/latex]. (10.50e)

Question 10.4: Use the results for Example 10.3, page 424, to determine (i)...

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