Question 11.3: Derive the equations of motion for a particle of mass m movi...

The holonomic constraint z = 0 is evident, and the velocity vector is   \mathbf{v}=\dot{r} \mathbf{e}_r  +  r \dot{\phi} \mathbf{e}_\phi.  Hence, the kinetic energy is given by

T=\frac{1}{2} m\left(\dot{r}^2  +  r^2 \dot{\phi}^2\right)                           (11.37a)

The central force is conservative with potential energy

V=-\mu m / r,                            (11.37b)

as shown in (7.62) with   \mu \equiv M G.  Thus, the Lagrangian (11.34) for  this conservative scleronomic dynamical system is

V(r)=V_0  –  \frac{G m M}{r}                       (7.62)

L\left(\dot{q}_r, q_r, t\right) \equiv T\left(\dot{q}_r, q_r, t\right)  –  V\left(q_r\right),                              (11.34)

L=\frac{1}{2} m\left(\dot{r}^2  +  r^2 \dot{\phi}^2\right)  +  \frac{\mu m}{r}                     (11.37c)

The generalized coordinates are  \left(q_1, q_2\right)=(r, \phi),  the generalized  velocity components are  \left(\dot{q}_1, \dot{q}_2\right)=(\dot{r}, \dot{\phi}),  and with (11.35) in mind we use (11.37c) to obtain

\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right)  –  \frac{\partial L}{\partial q_k}=0,                                 (11.35)

\frac{\partial L}{\partial \dot{r}}=m \dot{r}, \quad \frac{\partial L}{\partial \dot{\phi}}=m r^2 \dot{\phi}, \quad \frac{\partial L}{\partial r}=m r \dot{\phi}^2  –  \frac{\mu m}{r^2}, \quad \frac{\partial L}{\partial \phi}=0 .                                (11.37d)

Lagrange’s equations (11.35) thus yield the equations of motion for  this conservative system with two degrees of freedom:

\ddot{r}  –  r \dot{\phi}^2  +  \frac{\mu}{r^2}=0, \quad \frac{d}{d t}\left(r^2 \dot{\phi}\right)=0 .                         (11.37e)