(a) Derive the Lagrange equations of motion for a heavy bead of mass m that slides freely in a smooth circular tube of radius a, as the tube spins with constant angular speed Φ = ω about its fixed vertical axis, as shown in the diagram for Problem 6.66. Obtain the first integral of the equation

Question

(a) Derive the Lagrange equations of motion for a heavy bead of mass m that slides freely in a smooth circular tube of radius a, as the tube spins with constant angular speed [latex]\dot{\phi}=\omega[/latex] about its fixed vertical axis, as shown in the diagram for Problem 6.66. Obtain the first integral of the equation of motion, and find the tangential constraint force normal to the plane of the tube. (b) Relax the rheonomic constraint, treat [latex]\phi(t)[/latex] as an independent generalized coordinate, and derive the Lagrange equations of motion for the bead.

Accepted Answer

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

[latex]V=m g a(1 - \cos heta)[/latex]. (11.39a)

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

[latex]T=\frac{1}{2} m a^2\left(\dot{\phi}^2 \sin ^2 heta + \dot{ heta}^2 ight)[/latex]. (11.39b)

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

[latex]L\left(\dot{q}_r, q_r, t ight) \equiv T\left(\dot{q}_r, q_r, t ight) - V\left(q_r ight)[/latex] (11.34)

[latex]L=\frac{1}{2} m a^2\left(\omega^2 \sin ^2 heta + \dot{ heta}^2 ight) - m g a(1 - \cos heta)[/latex], (11.39c)

and thus derive

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{ heta}} ight)=m a^2 \ddot{ heta}, \quad \frac{\partial L}{\partial heta}=m a^2 \omega^2 \sin heta \cos heta - m g a \sin heta.[/latex] (11 .39d)

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

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k} ight)-\frac{\partial L}{\partial q_k}=Q_k^N[/latex] (11.38)

[latex]\ddot{ heta} + \sin heta\left(\frac{g}{a} - \omega^2 \cos heta ight)=0 .[/latex] (11.39e)

Notice that the result is independent of the mass m

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

[latex]\dot{ heta}^2=\dot{ heta}_0^2 + \omega^2\left(\sin ^2 heta - \sin ^2 heta_0 ight) + \frac{2 g}{a}\left(\cos heta - \cos heta_0 ight)[/latex], (11.39f)

in which [latex]\dot{ heta}_0=\dot{ heta}(0) [/latex] and [latex] heta_0= heta(0)[/latex] denote the initial values. For small angular hoop speeds the term in [latex]\omega^2[/latex] 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 [latex]a \sin heta[/latex] from this axis, and its momentum in the direction perpendicular to the plane of the tube is [latex]\operatorname{ma\omega } \sin heta[/latex], Therefore, the moment of momentum of the bead about the axis of rotation is [latex]h_z=m a^2 \omega \sin ^2 heta[/latex]. The constraint driving torque relation about the spin axis is given by [latex]\dot{h}_z=R a \sin heta[/latex]; and hence

[latex]R=2 m a \omega \dot{ heta} \cos heta[/latex]. (11.39g)

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

[latex]Q_ heta^N=0, \quad Q_\phi^N=R a \sin heta[/latex]. (11.39h)

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

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{\phi}} ight)=m a^2 \ddot{\phi} \sin ^2 heta + 2 m a^2 \dot{\phi} \dot{ heta} \sin heta \cos heta, \quad \frac{\partial L}{\partial \phi}=0 .[/latex] (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 [latex]\phi ext {-equation: }[/latex]

[latex]m a \ddot{\phi} \sin heta + 2 m a \dot{\phi} \dot{ heta} \cos heta=R ext {. }[/latex] (11.39j)

If [latex]R(\dot{ heta}, \dot{\phi}, heta, \phi, t)[/latex] is some specified function, (11.39j) is a nonlinear differential equation for [latex]\phi(t)[/latex] coupled with (11.39e) in which [latex]\omega=\dot{\phi}(t)[/latex]. On the other hand, for a specified function [latex]\phi(t)[/latex], and with [latex] heta(t)[/latex] determined by (11.39e) , (11.39j) is an equation that determines R(t). In particular, for [latex]\dot{\phi}=\omega[/latex], a constant, (11.39j) gives [latex]R=2 m a \omega \dot{ heta} \cos heta[/latex], 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.

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

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