PHY.7:5 - Archetypal Grounding

PHY.7:5.1 - Represent a constraint, then move its support

A point mass m moves in a vertical plane on an ideal rigid, massless rod of length l. Its frictionless pivot is initially fixed. Gravity is uniform with acceleration g. Let theta be the angle from the downward vertical; relative to the pivot,

x=l*sin(theta); y=-l*cos(theta).

The rod constraint is built into this configuration. For its ideal reaction, virtual motion along the circle has no radial displacement and the reaction does no virtual work. Kinetic energy is T=m*l^2*theta_dot^2/2, and potential energy is V=-m*g*l*cos(theta). Thus

L=m*l^2*theta_dot^2/2+m*g*l*cos(theta).

Varying theta with fixed temporal endpoints gives

m*l^2*theta_ddot+m*g*l*sin(theta)=0.

For l=1 m, g=10 m/s^2 and initial theta=pi/6, the angular acceleration is -5 rad/s^2. The radial reaction need not be solved to obtain that consequence. It can be recovered from the physical acceleration if a later question concerns the rod load.

Changed support. Prescribe a horizontal pivot position X(t). The physical position becomes x=X(t)+l*sin(theta) while y is unchanged. Differentiating the complete position gives

T=m*(X_dot^2+2*X_dot*l*cos(theta)*theta_dot+l^2*theta_dot^2)/2.

Keeping the same gravitational potential and varying theta gives

m*l^2*theta_ddot+m*l*X_ddot*cos(theta)+m*g*l*sin(theta)=0.

With pivot acceleration X_ddot=2 m/s^2 at the same angle, the angular acceleration is about -6.732 rad/s^2. Omitting the pivot term from the velocity would preserve the old answer while losing a real forcing. The pivot’s prescribed motion may do work, so the fixed-pivot mechanical-energy conservation claim does not automatically transfer.

Changed interaction. For the fixed pivot, add an established damping torque Q=-b*theta_dot, with b>=0. The virtual-work equation gives m*l^2*theta_ddot+b*theta_dot+m*g*l*sin(theta)=0. The mechanical energy has derivative -b*theta_dot^2. Appending this dissipative torque to a conservative scalar potential would require a different physical account. PHY.6 supplies the corresponding balance and receiving energy form; a more elaborate dissipative action is useful only when the intended work needs it.

The general move is to construct the allowed configuration and its velocities, derive the consequence, and rebuild only the affected contribution when the support or interaction changes.

PHY.7:5.2 - Recover an interaction that energy alone would miss

A nonrelativistic particle with mass m and charge e moves in the xy plane in a prescribed uniform magnetic field B perpendicular to it. Electric fields, radiation reaction and the particle’s alteration of the source field are neglected. The electromagnetic coupling is the physical premise; the field does no mechanical work but changes the direction of motion.

Choose a vector potential A=(-B*y/2,B*x/2,0), whose curl is the specified magnetic field. With zero electric scalar potential, the physical Lagrangian is

L=m*(x_dot^2+y_dot^2)/2 + e*B*(x*y_dot-y*x_dot)/2.

Its derivatives give

d/dt(L_xdot)=m*x_ddot-e*B*y_dot/2; L_x=e*B*y_dot/2,

and the corresponding y expressions. The Euler-Lagrange equations are therefore

m*x_ddot=e*B*y_dot; m*y_ddot=-e*B*x_dot.

For e*B/m=2 per second, initial x_dot=3 m/s and y_dot=0, the initial acceleration is (0,-6) m/s^2. Substitution into the kinetic-energy derivative gives zero. Conservation of kinetic energy alone would also allow straight uniform motion; it does not determine the magnetic turning. Using only T with zero scalar potential would miss the interaction.

Changed representation. Let chi=B*x*y/2 and use A_new=A+grad(chi)=(0,B*x,0). The new Lagrangian is L_new=m*(x_dot^2+y_dot^2)/2+e*B*x*y_dot. Its difference from L is e*d(chi)/dt, so the fixed-endpoint equations are unchanged. The canonical momenta L_xdot and L_ydot do change; the physical velocity and magnetic field do not. Comparing those canonical expressions as though they were two observed mechanical momenta would invent a physical discrepancy.

The action’s stationary paths depend on the physical interaction; equivalent gauge descriptions preserve them under the stated endpoint rule.

PHY.7:5.3 - Derive an interior law and change the endpoint condition

A taut string has uniform tension T and mass per unit length mu. Its transverse displacement u(x,t) is small enough that slopes can be treated to leading order; changes of tension and longitudinal motion are neglected. The kinetic energy per length is mu*u_t^2/2. Expanding the extra length to second order in slope gives the stored elastic contribution T*u_x^2/2. For length l, use

S[u]=integral over time and 0<=x<=l of (mu*u_t^2-T*u_x^2)/2 dx dt.

Take variations eta that vanish at the two temporal endpoints. Integration by parts gives the interior coefficient -mu*u_tt+T*u_xx and the spatial boundary contribution

integral over time of [-T*u_x*eta] at x=0,l dt.

Thus the interior equation is mu*u_tt=T*u_xx. At a fixed endpoint, eta is zero. At an unloaded endpoint free to move transversely in this model, eta is arbitrary and the corresponding slope must be zero.

For mu=0.01 kg/m, T=100 N and l=1 m, wave speed is sqrt(T/mu)=100 m/s. Two fixed endpoints admit the lowest nonzero spatial mode sin(pi*x/l), giving frequency 50 Hz. Keep the left endpoint fixed and free the right endpoint transversely while maintaining its axial tension; the lowest mode becomes sin(pi*x/(2*l)) and its frequency is 25 Hz. The interior equation did not change. Reusing the fixed-end spectrum would miss the changed boundary.

Loaded endpoint. Attach a massless transverse spring of stiffness kappa at x=l. Add -integral kappa*u(l,t)^2/2 dt to the action. With the right endpoint variation free, its coefficient gives T*u_x(l,t)+kappa*u(l,t)=0. The spring changes the boundary condition through its stored energy. If an endpoint mass is consequential, its kinetic term must be included too; the massless condition would no longer supply that boundary’s dynamics.

Retain the boundary contribution until the physical freedom or load determines its use.