Question 11.5: (a) Derive the Lagrange equations of motion for a heavy bead...

Solution of (a). Introduce the spherical coordinates  (a, \phi, \theta)  of the bead in its constrained motion referred to a moving, spherical reference frame   \psi=\left\{O ; \mathbf{e}_r, \mathbf{e}_\phi, \mathbf{e}_\theta\right\}  in which  \mathbf{e}_k  are in the  directions of their increasing coordinates, and note that the angle between k and   \mathbf{e}_r,  in the problem diagram is  \pi-\theta \text {. } The workless scleronomic constraint is r = a, constant, and the working  rheonomic constraint is  \dot{\phi}=\omega, \text { a constant },  that is,  \phi=\phi_0  +  \omega t.  Notice, alternatively, that a rheonomic constraint   \dot{\phi}=\omega(t),  a specified function of t , also is holonomic with  \phi=\phi_0  +  \int_{t_0}^t \omega(t) d t.  In either instance,   \phi  is a specified function of time; it is not a generalized coordinate. The radial component of the  nonconservative force   \mathbf{F}=-N \mathbf{e}_r  +  R \mathbf{e}_\phi,  exerted by the tube on the mass m is workless, but due to the moving tube constraint the component R perpendicular to the plane of the tube is not, so the system is nonconservative. Nevertheless, the virtual work done by this force in the virtual displacement   \delta \mathbf{x}=a \delta \theta \mathbf{e}_\theta  +  a \sin \theta \delta \phi \mathbf{e}_\phi,  compatible with the constraint   \delta \phi=\omega \delta t \equiv 0  vanishes,   \delta \mathscr{W}=\mathbf{F} \cdot \delta \mathbf{x}=R a \sin \theta \delta \phi=0,  because the moving constraint in the  virtual displacement is frozen . The bead has one degree of freedom described by  q_1=\theta  the corresponding generalized force  Q_\theta^N=0.  The gravitational potential energy of m is given by

V=m g a(1  –  \cos \theta).         (11.39a)

The absolute velocity of the bead is   \mathbf{v}=a \dot{\phi} \sin \theta \mathbf{e}_\phi  +  a \dot{\theta} \mathbf{e}_\theta,  and  hence its total kinetic energy is

T=\frac{1}{2} m a^2\left(\dot{\phi}^2 \sin ^2 \theta  +  \dot{\theta}^2\right).                    (11.39b)

Now introduce  \dot{\phi}=\omega,  a constant, form the Lagrangian function (11.34),

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 a^2\left(\omega^2 \sin ^2 \theta  +  \dot{\theta}^2\right)  –  m g a(1  –  \cos \theta),                        (11.39c)

and thus derive

\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{\theta}}\right)=m a^2 \ddot{\theta}, \quad \frac{\partial L}{\partial \theta}=m a^2 \omega^2 \sin \theta \cos \theta  –  m g a \sin \theta.              (11 .39d)

Collecting the results in (11.38), we reach the Lagrange equation of  motion for the bead :

\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=Q_k^N                  (11.38)

\ddot{\theta}  +  \sin \theta\left(\frac{g}{a}  –  \omega^2 \cos \theta\right)=0 .                   (11.39e)

Notice that the result is independent of the mass m

Equation (11.39e) written as  \ddot{\theta}=\dot{\theta} d \dot{\theta} / d \theta=f(\theta)  has an easy first  integral given by

\dot{\theta}^2=\dot{\theta}_0^2  +  \omega^2\left(\sin ^2 \theta  –  \sin ^2 \theta_0\right)  +  \frac{2 g}{a}\left(\cos \theta  –  \cos \theta_0\right),                         (11.39f)

in which \dot{\theta}_0=\dot{\theta}(0) and   \theta_0=\theta(0)  denote the initial values. For small angular hoop speeds the  term in  \omega^2  may be neglected to obtain from (11.39e) and (11.39f )  the equation of motion and its first integral for the finite amplitude  oscillations of the bead as an equivalent simple pendulum.

The constraint force R is not determined; it is inconsequential to  the determination of the general motion of the bead by Lagrange’s method. Nevertheless, we can find R easily by writing the moment of momentum equation about the z-axis, The bead is at the instantaneous distance  a \sin \theta  from this axis, and its momentum in the direction perpendicular to the plane of the tube is   \operatorname{ma\omega } \sin \theta,  Therefore, the moment of momentum of the bead about the axis of rotation is  h_z=m a^2 \omega \sin ^2 \theta.  The constraint driving torque relation about the spin axis is given by   \dot{h}_z=R a \sin \theta;  and hence

R=2 m a \omega \dot{\theta} \cos \theta.                (11.39g)

Solution of (b). Now consider a different situation in which we ignore the rheonomic constraint and treat   \phi(t)  as an independent variable . The bead now has two degrees of freedom described by  \left(q_1, q_2\right)=(\theta, \phi)  with the corresponding nonconservative generalized forces   \left(Q_\theta^N, Q_\phi^N\right).  The virtual work done by these forces is given by  \delta \mathscr{W}=\mathbf{F} \cdot \delta \mathbf{x}=R a \sin \theta \delta \phi=Q_\theta^N \delta \theta  +  Q_\phi^N \delta \phi  for all  \delta \theta  and  \delta \phi,  and hence

Q_\theta^N=0, \quad Q_\phi^N=R a \sin \theta.                    (11.39h)

The Lagrangian  L=\frac{1}{2} m a^2\left(\dot{\phi}^2 \sin ^2 \theta  +  \dot{\theta}^2\right)  –  m g a(1  –  \cos \theta)  is the  same as (11.39c) in which   \omega=\dot{\phi}.  In addition to (11.39d), we now have

\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{\phi}}\right)=m a^2 \ddot{\phi} \sin ^2 \theta  +  2 m a^2 \dot{\phi} \dot{\theta} \sin \theta \cos \theta, \quad \frac{\partial L}{\partial \phi}=0 .                         (11.39i)

Assembling the results (11.39d), (11.39h), and (11.39i) in (11.38), we obtain the θ-equation (11.39e) , as before, and the new  \phi \text {-equation: }

m a \ddot{\phi} \sin \theta  +  2 m a \dot{\phi} \dot{\theta} \cos \theta=R \text {. }                  (11.39j)

If  R(\dot{\theta}, \dot{\phi}, \theta, \phi, t)  is some specified function, (11.39j) is a nonlinear differential equation for   \phi(t)  coupled with (11.39e) in which  \omega=\dot{\phi}(t).  On the other hand, for a specified function  \phi(t),  and with   \theta(t)  determined by (11.39e) , (11.39j) is an equation that determines R(t). In particular, for  \dot{\phi}=\omega,  a constant, (11.39j) gives  R=2 m a \omega \dot{\theta} \cos \theta,  which is the same nonconservative, rheonomic  constraint force obtained in (11.39g) . The original Lagrange method eliminates the need to determine the inconsequential working rheonomic constraint force, which may be found by other methods, if needed.