Question 8.6: A pipeline valve handle consists of three equally spaced han...
A pipeline valve handle consists of three equally spaced handle grips of equal mass m attached to the valve body of mass M by thin rigid torque bars of equal length ℓ and negligible mass. The handle, initially at rest, is turned by forces f_1 and f_2 of equal constant magnitude F applied at two grips in the plane of, and perpendicular to the torque bars, as shown in Fig. 8.7. The handle turns freely without friction. (a) Apply the work–energy principle to find as a function of time the angular speed ω(t) of the handle. (b) Derive the same result by use of the moment of momentum principle.
Solution of (a). We model the handle as a rigid system of three particles of equal mass m attached by massless rigid rods to the valve body, which is a particle of mass M at C, the center of mass of the system. Since the handle turns freely without friction at C, the reaction force of the valve body is equipollent to a single force R at C that does no work in the motion; and, of course, the normal gravitational force also is workless. The work done by the remaining applied external forces, relative to the center of mass, is determined by the first equation in (8.66), and hence, with reference to Fig. 8.7,
\mathscr{W}_{r C}=\int_{\mathscr{C}_1} \mathbf{f}_1 \cdot d \rho_1 + \int_{\mathscr{6}_2} \mathbf{f}_2 \cdot d \rho_2=\int_0^\phi F \mathbf{e}_2 \cdot \ell d \phi \mathbf{e}_2 + \int_0^\phi F \mathbf{n}_2 \cdot \ell d \phi \mathbf{n}_2.
Hence,
\mathscr{W}_{r C}=2 F \ell \phi. (8.67a)
Notice that since C is fixed in \Phi, \mathscr{W}^*=0 in (8.55); hence, by (8.63), \mathscr{W}=\mathscr{W}_{r C} also is the total work done on the rigid system.
\mathscr{W}^* \equiv \int_{\mathscr{C}^*} \mathbf{F}\left(\mathbf{x}^*\right) \cdot d \mathbf{x}^*=\int_{t_0}^t \mathbf{F}\left(\mathbf{x}^*\right) \cdot \mathbf{v}^* d t (8.55)
\mathscr{W}=\mathscr{W}^* + \mathscr{W}_{r C} (8.63)
Since each of the three handle particles has the same speed \left|\dot{\boldsymbol{\rho}}_1\right|=\ell \dot{\phi} relative to C, and because the system is at rest initially, the change in kinetic energy (8.52) relative to the center of mass is
K_{r C}(\beta, t) \equiv \sum_{k=1}^n \frac{1}{2} m_k \dot{\rho}_k \cdot \dot{\rho}_k (8.52)
\Delta K_{r C}=\frac{3}{2} m \ell^2 \dot{\phi}^2. (8.67b)
Use of (8.67a) and (8.67b) in the work–energy equation (8.66) thus yields
\mathscr{W}_{r C}=\sum_{k=1}^n \int_{\mathscr{G}_k} \mathbf{f}_k \cdot d \rho_k=\int_{t_0}^t \sum_{k=1}^n \mathbf{f}_k \cdot \dot{\boldsymbol{\rho}}_k d t=\Delta K_{r C} (8.66)
\omega(\phi)=\dot{\phi}=\sqrt{\frac{4 F \phi}{3 m \ell}}. (8.67c)
To determine ω(t) as a function of time, we integrate this equation with Φ(0) = 0 initially to obtain the angular placement
\phi(t)=\alpha t^2, \quad \text { with } \quad \alpha \equiv \frac{F}{3 m \ell} . (8.67d)
Now (8.67c) yields the desired result:
\omega(t)=2 \alpha t . (8.67e)
Solution of (b). The same result may be obtained from the principle of moment of momentum relative to the center of mass in (8.45). First, we note that the applied torque about C is \mathbf{M}_C=\rho_1 \times \mathbf{f}_1 + \rho_2 \times \mathbf{f}_2=2 F \ell \mathbf{k}. The moment about C of the momenta relative to C is \mathbf{h}_{r C}=\rho_1 \times m_1 \dot{\rho}_1 + \rho_2 \times m_2 \dot{\rho}_2 + \rho_3 \times m_3 \dot{\rho}_3=3 m \ell^2 \phi \mathbf{k}, and d \mathbf{h}_{r C} / d t=3 m \ell^2 \phi \mathbf{k}. Hence, (8.45) yields 2 F \ell \mathbf{k}=3 m \ell^2 \ddot{\phi} \mathbf{k}; that is,
\mathbf{M}_C(\beta, t)=\frac{d \mathbf{h}_C(\beta, t)}{d t}=\frac{d \mathbf{h}_r(\beta, t)}{d t} (8.45)
\ddot{\phi}=\frac{2 F}{3 m \ell}=2 \alpha . (8.67f)
Integration with respect to time, with \dot{\phi}(0)=0 initially, returns (8.67e).