Applied forces and dissipation

1. Forces that cannot be put into a position potential

A pendulum slows in air, and a driven oscillator can gain energy from a motor. Their constraints are just as useful as before, but a position potential cannot represent all their forces. Friction depends on the direction of motion and usually converts mechanical energy to internal energy. An external drive can supply work on every cycle.

The Lagrangian still describes the kinetic energy and the conservative part of the interaction. The remaining forces enter through $Q_i$. There is no universal rule “subtract the work lost to friction from $L$”: dissipated work depends on the actual history, whereas an ordinary position potential is a function of configuration.

2. Generalized forces by virtual work

For $\mathbf r_a(q,t)$, a virtual displacement at fixed time is $\delta\mathbf r_a=\sum_i(\partial\mathbf r_a/\partial q_i)\delta q_i$. Inserting it into the remaining-force work gives

The coefficient of $\delta q_i$ is $Q_i$. This definition works for velocity-dependent, time-dependent, and dissipative forces. The distinction “generalized” concerns the coordinates, not whether a force is conservative.

For a pendulum angle, a tangential force $F_t$ acts through an arc displacement $a\delta\theta$, so $Q_\theta=aF_t$. A horizontal force $F_x$ on the same bob instead gives $Q_\theta=F_x a\cos\theta$. It is the work projection, not simply the force magnitude, that belongs on the right-hand side.

For a stationary coordinate transformation, actual velocity is $\mathbf v_a=\sum_i(\partial\mathbf r_a/\partial q_i)\dot q_i$, so the instantaneous power of the remaining forces is $\sum_iQ_i\dot q_i$. If the transformation depends explicitly on time, add the transport power $\sum_a\mathbf F_a^{\rm rem}\cdot\partial_t\mathbf r_a$. The energy function developed in the next chapter must then be distinguished from laboratory mechanical energy.

3. Viscous damping and Rayleigh's function

Suppose a particle on a line experiences $F_d=-c\dot x$ with $c>0$. Define

$\mathcal R$ is the Rayleigh dissipation function. It has dimensions of power, not energy. For linear damping in several coordinates one may write

Differentiating the two velocity factors gives $\partial\mathcal R/\partial\dot q_i=\sum_jc_{ij}\dot q_j$. If the matrix $c_{ij}$ is positive semidefinite, the dissipative power is

This proves the sign of mechanical-energy removal for this model. With an additional drive $Q_i^{(d\!r)}$, the equations are

The familiar quadratic $\mathcal R$ applies to linear viscous damping. Dry friction must be modeled separately. During sliding, its direction opposes velocity; at rest, static friction takes the value needed for equilibrium up to its bound. A single fixed-sign kinetic friction formula is not valid through a reversal or sticking interval.

4. Applications

Example 1 · A damped spring

A mass $m$ on a horizontal spring experiences stiffness $k$ and drag $-c\dot x$. Choose displacement $x$ from the spring's relaxed position. Then $T=m\dot x^2/2$, $V=kx^2/2$, $L=T-V$, and $Q_x=-c\dot x$. The equation is

To solve it, try $x=e^{st}$. Then $\dot x=se^{st}$ and $\ddot x=s^2e^{st}$; division by the nonzero exponential gives $ms^2+cs+k=0$. Thus

When $c^2<4mk$, set $\gamma=c/(2m)$ and $\omega_d=\sqrt{k/m-\gamma^2}$. Real motion is

For $m=1\,\mathrm{kg}$, $k=5\,\mathrm{N\,m^{-1}}$, $c=2\,\mathrm{kg\,s^{-1}}$, one has $\gamma=1\,\mathrm{s^{-1}}$ and $\omega_d=2\,\mathrm{s^{-1}}$. If $x(0)=0.10\,\mathrm m$ and $\dot x(0)=0$, then $A=0.10\,\mathrm m$ and $-\gamma A+\omega_dB=0$, giving $B=0.05\,\mathrm m$.

At $c^2=4mk$, the repeated root gives $(A+Bt)e^{-ct/(2m)}$. Above that value the two roots are real and negative. These regimes describe oscillatory decay, critical damping, and non-oscillatory decay. In all three,

Example 2 · Tangential resistance on a pendulum

A bob of mass $m$ moves on a fixed circle of radius $a$, with angle $\theta$ from the downward vertical. Suppose air resistance along the tangent is $F_t=-bv_t=-ba\dot\theta$. Virtual work is $F_ta\delta\theta$, so $Q_\theta=-ba^2\dot\theta$.

Using $T=ma^2\dot\theta^2/2$, $V=mga(1-\cos\theta)$ gives

Dividing by $ma^2$ gives $\ddot\theta+(b/m)\dot\theta+(g/a)\sin\theta=0$. The drag coefficient multiplying angular velocity is not $b$ alone in the torque equation. Its factor $a^2$ comes from converting angular velocity to linear speed and force to torque.

Here $\mathcal R=ba^2\dot\theta^2/2$ and $\dot E=-ba^2\dot\theta^2$. For small angles, the damped-oscillator solution applies with $\gamma=b/(2m)$ and undamped frequency $\sqrt{g/a}$.

Example 3 · A periodically driven oscillator

Let the spring mass experience an external force $F_0\cos\Omega t$ in addition to viscous drag. Its Lagrangian remains $m\dot x^2/2-kx^2/2$, with $Q_x=F_0\cos\Omega t-c\dot x$. Thus

Seek a persistent response $x=A\cos(\Omega t-\delta)$. Write it as $a\cos\Omega t+b\sin\Omega t$. Substitution and separate comparison of cosine and sine coefficients gives

Multiply the first equation by $a$, the second by $b$, and add to get $(k-m\Omega^2)(a^2+b^2)=F_0a$. Alternatively solve the pair directly, with $D=(k-m\Omega^2)^2+c^2\Omega^2$:

Choose the phase with $\cos\delta=a/A$ and $\sin\delta=b/A$ to preserve its quadrant. At $\Omega=\sqrt{k/m}$ and $c>0$, $A=F_0/(c\Omega)$ and the displacement lags the force by $\pi/2$. The full motion also includes the decaying homogeneous solution from Example 1. The exact power balance is $\dot E=F_0\dot x\cos\Omega t-c\dot x^2$.

Example 4 · Sliding on a rough incline

A mass slides down a plane inclined at $\alpha$. Choose $q$ down the plane. Then $T=m\dot q^2/2$, $V=-mgq\sin\alpha$, and $L=m\dot q^2/2+mgq\sin\alpha$. During downward sliding, $Q_q=-\mu_fmg\cos\alpha$.

For $\alpha=30^\circ$ and $\mu_f=0.20$, $\ddot q=9.8(0.5-0.20\times0.866)=3.20\,\mathrm{m\,s^{-2}}$. For an upward-moving block, the friction sign reverses while $q$ keeps its original downward convention. If the block reaches rest, use the static-friction bound to decide whether it stays there.

5. Energy accounting and the force model

Dissipation does not violate conservation of total energy. The mechanical model omits the microscopic thermal degrees of freedom into which energy flows. Similarly, a prescribed drive represents a source whose own energy is not being tracked. The equation with $Q_i$ makes this modeling boundary explicit.

The principal checks are therefore physical: does the drag oppose velocity, does its power have the right sign, and has each force been counted exactly once? A damped coordinate can be absent from $L$ while its momentum still changes because $Q_i\ne0$. The next chapter makes this qualification part of every conservation statement.

Exercises

Hints and solutions are collected separately.

1. 8.1 Why can dry friction not generally be represented by $V(q)$?

2. 8.2 Find the generalized torque of a horizontal force $F$ acting on a pendulum bob.

3. 8.3 Prove that quadratic Rayleigh damping removes mechanical energy at rate $2\mathcal R$ in stationary coordinates.

4. 8.4 Find the critical damping coefficient for $m=2\,\mathrm{kg}$ and $k=18\,\mathrm{N\,m^{-1}}$.

5. 8.5 A $1\,\mathrm{kg}$ oscillator has $k=10\,\mathrm{N\,m^{-1}}$ and $c=2\,\mathrm{kg\,s^{-1}}$. Find its decay rate and damped frequency.

6. 8.6 For an angular damping torque $-b_\theta\dot\theta$, write the exact damped pendulum equation.

7. 8.7 At resonance $\Omega=\sqrt{k/m}$, find the persistent amplitude for $F_0=6\,\mathrm N$, $c=2\,\mathrm{kg\,s^{-1}}$, $\Omega=3\,\mathrm{s^{-1}}$.

8. 8.8 Write the power balance for a spring driven by an arbitrary force $f(t)$ with viscous damping.

9. 8.9 A block is moving upward on a rough incline. With $q$ positive downward, write its equation during that motion.

10. 8.10 Explain why $\sum_iQ_i\dot q_i$ need not be the full remaining-force power for moving coordinates.

11. 8.11 Why does an autonomous damped Lagrangian not imply constant energy?

12. 8.12 Two dampers act on coordinates through $\mathcal R=\tfrac12c(\dot q_1-\dot q_2)^2$. Find both generalized forces and their total power.