Question 10.3: A connecting rod QR of a simple machine shown in Fig. 10.6 i...
A connecting rod QR of a simple machine shown in Fig. 10.6 is modeled as a homogeneous thin rod of length \ell, uniform cross section, and mass σ per unit length. The rod is hinged at Q at a distance α from the center O of a flywheel that turns with a constant angular velocity \Omega=\Omega \mathbf{k}, as shown, in the inertial ground frame 0=\{O ; \mathbf{I}, \mathbf{J}, \mathbf{k}\} fixed at O. Determine (i) the moment of momentum of the rod relative to Q and (ii) its moment of momentum about Q.
Solution of (i). The moment of momentum \mathbf{h}_{r Q} of the rod relative to Q is determined by (10.37), in which the rigid body velocity of a rod particle P relative to Q in the ground frame 0=\{O ; \mathbf{I}, \mathbf{J}, \mathbf{k}\} is given by
\mathbf{h}_{r Q}(\mathscr{B}, t) \equiv \int_{\mathscr{B}} \mathbf{r}(P, t) \times \dot{\mathbf{r}}(P, t) d m(P) (10.37)
\dot{\mathbf{r}}(P, t)=\mathbf{v}(P, t) – \mathbf{v}_Q=\boldsymbol{\omega} \times \mathbf{r}(P, t), (10.41a)
where \mathbf{r}(P, t)=r \mathbf{i}, referred to the rod frame 2=\{Q ; \mathbf{i}, \mathbf{j}, \mathbf{k}\}, and the total angular velocity of the rod frame 2 in the ground frame 0 is \boldsymbol{\omega} \equiv \boldsymbol{\omega}_{20} \text {. Let } \boldsymbol{\omega}_{21}=\dot{\beta} \mathbf{k} denote the angular velocity of the rod frame 2 relative to the flywheel frame 1=\{O ; \mathbf{a}, \mathbf{b}, \mathbf{k}\} whose angular velocity is \omega_{10}=\Omega relative to frame 0. Then \omega=\omega_{21} + \omega_{10}, that is,
\boldsymbol{\omega}=(\dot{\beta} + \Omega) \mathbf{k}, (10.41b)
and (10.41a) gives the relative rigid body velocity \dot{\mathbf{r}}(P, t)=r(\dot{\beta} + \Omega) \mathbf{j}. Therefore, with d m=\sigma d r, where d_r is the elemental length of the rod , (10.37) yields
\mathbf{h}_{r Q}(\mathscr{B}, t)=\sigma(\dot{\beta} + \Omega) \mathbf{k} \int_0^{\ell} r^2 d r=\sigma(\dot{\beta} + \Omega) \frac{\ell^3}{3} \mathbf{k}. (10.41c)
The total mass of the rod is m(\mathscr{B})=\sigma \ell, and hence the moment of momentum of the rod relative to Q is
\mathbf{h}_{r Q}(\mathscr{B}, t)=\frac{m \ell^2}{3}(\dot{\beta} + \Omega) \mathbf{k}. (10.41d)
Solution of (ii). Now consider the moment of momentum h_Q of the rod about Q defined by (10.33) in which \mathbf{v}_Q=\boldsymbol{\Omega} \times \mathbf{x}=\Omega \mathbf{k} \times \alpha \mathbf{a} yields
\mathbf{h}_Q(\mathscr{B}, t)=\int_{\mathscr{B}} \mathbf{r}(P, t) \times \mathbf{v}(P, t) d m(P) (10.33)
\mathbf{v}_Q=\alpha \Omega \mathbf{b}, (10.41e)
referred to frame 1. Use of (10.41e) in (10.41a) gives \mathbf{v}(P, t)=\alpha \Omega(\sin \beta \mathbf{i} + \cos \beta \mathbf{j})+r(\dot{\beta} + \Omega) \mathbf{j}, the velocity of P in frame 0 but referred to the rod frame 2; and with \mathbf{r}(P, t)=r \mathbf{i}, (10.33) becomes
\mathbf{h}_Q(\mathscr{B}, t)=\int_0^{\ell} \sigma\left[\alpha \Omega r \cos \beta + r^2(\dot{\beta} + \Omega)\right] d r \mathbf{k}. (10.41f)
An easy integration yields
\mathbf{h}_Q(\mathscr{B}, t)=\left[\frac{m \ell}{2} \alpha \Omega \cos \beta + \frac{m \ell^2}{3}(\dot{\beta} + \Omega)\right] \mathbf{k}. (10.41g)
See Problem 10.4.
