Question 11.4: Derive the equation of motion for a particle P that falls fr...
Derive the equation of motion for a particle P that falls from rest in a Stokes medium .
The kinetic energy of P is T=\frac{1}{2} m \dot{y}^2, its gravitational potential energy is V=-m g y, and the nonconservative Stoke s force is \mathbf{F}=-c \mathbf{v}=-c \dot{y} \mathbf{j}, where j is the downward direction of the motion . Let the reader apply the method of virtual work to find Q_1^N. Here we consider (11.20). Accordingly, with q_1=y, the nonconservative generalized force in (11.38) is Q_1^N=\mathbf{F} \cdot \partial \mathbf{v} / \partial \dot{q}_1=-c \dot{y} \mathbf{j} \cdot \mathbf{j}=-c \dot{y}. Now form the Lagrangian L=T-V=\frac{1}{2} m \dot{y}^2+m g y, and apply (11.38) to obtain the equation of motion m \ddot{y}-m g=-c \dot{y} . That is, with v=\dot{y} and \nu=c / m, we recover (6.34a): d v / d t=g – v v
Q_k=\mathbf{F} \cdot \frac{\partial \mathbf{v}}{\partial \dot{q}_k}. (11.20)
\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right) – \frac{\partial L}{\partial q_k}=Q_k^N . (11.38)
\frac{d v}{d t}=g – v v \equiv F(v) \quad \text { with } \quad v \equiv \frac{c}{m} (6.34a)