Unit 2 · Hints & solutions

$L=T-V$ is the function whose derivatives organize the equations of motion. Mechanical energy is $E=T+V$ for the ordinary systems here. For a pendulum, kinetic and potential energy exchange while their sum is constant; their difference generally changes.

Angles are dimensionless in SI. Since virtual work is energy, $Q_q$ has units $\mathrm{N\,m}$. Since $p_q\dot q$ is energy, $p_q$ has units $\mathrm{kg\,m^2\,s^{-1}}$: it is an angular-momentum-type quantity.

$\partial_{\dot q}L=m(q)\dot q$, so its total derivative is $m\ddot q+m^{\prime}\dot q^2$. Also $\partial_qL=\tfrac12m^{\prime}\dot q^2-V^{\prime}$. Subtraction gives $m\ddot q+\tfrac12m^{\prime}\dot q^2+V^{\prime}=0$. Here $m(q)$ is an effective inertia coefficient, not a variable-mass particle model.

$T=m\dot y^2/2$, $V=mgy$, and $L=m\dot y^2/2-mgy$. Hence $d(m\dot y)/dt-(-mg)=0$, giving $\ddot y=-g$.

From $L=ma^2\dot\theta^2/2-mga(1-\cos\theta)$, $\ddot\theta+(g/a)\sin\theta=0$. For small angles $\omega=\sqrt{g/a}=\sqrt{10}\,\mathrm{s^{-1}}$. Thus $\tau=2\pi/\omega=1.987\,\mathrm s$.

Take $q$ downward for $5\,\mathrm{kg}$. Then $T=4\dot q^2$, $V=-2gq$, $L=4\dot q^2+2gq$ in SI units. The equation is $8\ddot q=2g$, so $\ddot q=g/4=2.45\,\mathrm{m\,s^{-2}}$.

$L=m(\dot r^2+r^2\dot\theta^2)/2$. The radial equation is $m\ddot r-mr\dot\theta^2=0$. The angular equation is $d(mr^2\dot\theta)/dt=0$, or $r^2\ddot\theta+2r\dot r\dot\theta=0$. These describe a straight line in polar variables.

$V\mapsto V+C$ changes $L$ by $-C$. Neither $\partial_qL$ nor $\partial_{\dot q}L$ changes, so neither does the equation. The numerical energy zero shifts by $C$.

$T=\tfrac12m(1+4c^2x^2)\dot x^2$, $V=mgcx^2$. The total velocity derivative is $m(1+4c^2x^2)\ddot x+8mc^2x\dot x^2$, while $L_x=4mc^2x\dot x^2-2mgcx$. Thus $(1+4c^2x^2)\ddot x+4c^2x\dot x^2+2gcx=0$.

The rod length is built into $x=a\sin\theta,y=-a\cos\theta$, leaving one coordinate. The reaction has zero tangent virtual work and drops from the angular equation. Its geometrical effect remains in $T=ma^2\dot\theta^2/2$. If needed, the reaction is recovered from radial Newton balance after solving the angular motion.

No. $-\partial_yV=-mg$ is already included through $L$. The right-hand side contains only remaining forces. Including it again would incorrectly double the gravitational acceleration.

The equality constraint presumes a taut string. Radial balance gives tension $\mathcal T=ma\dot\theta^2+mg\cos\theta$. If the predicted tension is negative, the string would need to push, which it cannot do. The string goes slack and the circular-path model must be replaced.

Kinetic friction opposes motion on both legs, so its net work over a closed path is negative. A position potential would give zero net work over that path. Its direction also depends on velocity, not only position.

With $\mathbf r=(a\sin\theta,-a\cos\theta)$, $Q_\theta=(F,0)\cdot(a\cos\theta,a\sin\theta)=Fa\cos\theta$. Its value vanishes when the permitted tangent is vertical.

For quadratic $\mathcal R$, $\sum_i\dot q_i\partial_{\dot q_i}\mathcal R=2\mathcal R$. With $Q_i=-\partial_{\dot q_i}\mathcal R$, its power is $\sum_iQ_i\dot q_i=-2\mathcal R$. The equality with mechanical-energy change assumes no other drive and no explicit-time energy input.

The roots of $ms^2+cs+k=0$ merge when $c^2=4mk$. Thus $c_c=2\sqrt{mk}=2\sqrt{36}=12\,\mathrm{kg\,s^{-1}}$.

$\gamma=1\,\mathrm{s^{-1}}$ and $\omega_d=\sqrt{k/m-\gamma^2}=3\,\mathrm{s^{-1}}$. The envelope decays as $e^{-t}$ and its oscillation period is $2\pi/3\,\mathrm s$.

$T=ma^2\dot\theta^2/2$, $V=mga(1-\cos\theta)$, and $Q_\theta=-b_\theta\dot\theta$. Therefore $ma^2\ddot\theta+b_\theta\dot\theta+mga\sin\theta=0$. Do not multiply this already angular coefficient by another $a^2$.

$A=F_0/(c\Omega)=6/(2\times3)=1\,\mathrm m$, within the assumed linear spring model. The phase lag is $\pi/2$. This is the long-time periodic response, not the entire transient motion.

$m\ddot x+kx=f(t)-c\dot x$. Multiplication by $\dot x$ gives $d[m\dot x^2/2+kx^2/2]/dt=f(t)\dot x-c\dot x^2$. The first term can supply or remove energy; the second never supplies it for $c>0$.

$V=-mgq\sin\alpha$, $T=m\dot q^2/2$, and $Q_q=+\mu_fmg\cos\alpha$ while $\dot q<0$. Thus $\ddot q=g(\sin\alpha+\mu_f\cos\alpha)$. This positive acceleration reduces the upward speed until a new friction regime must be chosen.

$\mathbf v_a=\sum_i\partial_{q_i}\mathbf r_a\dot q_i+\partial_t\mathbf r_a$. Dotting with $\mathbf F_a^{\rm rem}$ and summing gives power $\sum_iQ_i\dot q_i+\sum_a\mathbf F_a^{\rm rem}\cdot\partial_t\mathbf r_a$. The second term is absent only for stationary transformations or when its projection vanishes.

Although $L_t=0$, $\dot{\mathcal E}=\sum_iQ_i\dot q_i$. For viscous drag this is negative. Autonomy implies conservation only when the remaining-force power is also zero.

$Q_1=-c(\dot q_1-\dot q_2)$ and $Q_2=+c(\dot q_1-\dot q_2)$. Their power is $-c(\dot q_1-\dot q_2)^2=-2\mathcal R$. Equal velocities produce no relative damping force.

Yes. For a free particle $L=m\dot x^2/2$, $x$ is cyclic but $x=x_0+vt$ can change. The conserved quantity is $p_x=mv$.

$\dot p_i=L_{q_i}+Q_i$. Cyclicity sets $L_{q_i}=0$, leaving $\dot p_i=Q_i$. Momentum is conserved when $Q_i=0$, not from cyclicity alone.

$\dot{\mathcal E}=\sum_i(\ddot q_ip_i+\dot q_i\dot p_i)-\sum_i(L_{q_i}\dot q_i+L_{\dot q_i}\ddot q_i)-L_t$. Since $p_i=L_{\dot q_i}$, this becomes $\sum_i\dot q_i(\dot p_i-L_{q_i})-L_t=\sum_iQ_i\dot q_i-L_t$.

$x$ is absent, and with $Q_x=0$, $p_x=m\dot x$ is conserved. The value of $p_y$ generally changes by $-V^{\prime}(y)$.

$\dot\theta$ decreases by a factor of four. Tangential speed is $v_\theta=r\dot\theta=\ell/(mr)$, so it decreases by a factor of two. Radial speed need not follow either rule.

$\mathcal E=m\dot x^2/2+k(t)x^2/2$. Since $L_t=-\dot kx^2/2$ and $Q=0$, $\dot{\mathcal E}=\dot kx^2/2$. The actuator changing stiffness exchanges energy.

Translation gives $\delta L=\epsilon\sum_aL_{x_a}=0$. With unforced Euler–Lagrange equations, $L_{x_a}=\dot p_{ax}$. Therefore $d(\sum_ap_{ax})/dt=0$. For ordinary Cartesian kinetic energy this is $d(\sum_am_a\dot x_a)/dt=0$.

$\phi$ is cyclic, giving $p_\phi=ma^2\sin^2\theta\dot\phi=L_z$. Gravity generally exerts a nonzero horizontal torque about the pivot, so the full angular-momentum vector is not constant.

$P=m(4v)+2m(-v)=2mv$. Total mass is $3m$, so $\dot X=2v/3$, constant in the isolated system. Internal forces can change both individual velocities.

$dL/dt=\sum_iL_{q_i}\dot q_i+\sum_iL_{\dot q_i}\ddot q_i+L_t$. An autonomous Lagrangian has only the last term zero. The other terms generally do not cancel; for a pendulum, $T-V$ changes as $T+V$ remains fixed.

Rotational invariance gives $0=\sum_a[-y_aL_{x_a}+x_aL_{y_a}-\dot y_ap_{ax}+\dot x_ap_{ay}]$. Replacing $L_x,L_y$ by momentum derivatives combines this into $d\sum_a(x_ap_{ay}-y_ap_{ax})/dt=0$, which is conservation of $L_z$.

No. An autonomous spring oscillator conserves $m\dot x^2/2+kx^2/2$, but $\dot p=-kx$ is generally nonzero. Time independence and spatial translation invariance are different symmetries.

For moving constraints $\mathbf v_a=\partial_t\mathbf r_a$ when the generalized velocities vanish. Then $T=T_0=\sum_am_a|\partial_t\mathbf r_a|^2/2$, which can be positive. Rest in relative coordinates is not necessarily laboratory rest.

From $\mathbf v_a=\sum_i\mathbf e_{ai}\dot q_i+\mathbf w_a$, the squared velocity has quadratic, twice-cross, and constant terms. Multiplying by $m_a/2$ yields $M_{ij}=\sum_am_a\mathbf e_{ai}\cdot\mathbf e_{aj}$, $b_i=\sum_am_a\mathbf e_{ai}\cdot\mathbf w_a$, and $c=\sum_am_a\mathbf w_a^2/2$.

$z^{\mathsf T}Mz=\sum_am_a|\sum_i\mathbf e_{ai}z_i|^2$. Positive masses make this nonnegative. Regular independent coordinates ensure a nonzero $z$ moves at least one particle, making it strictly positive. A singular coordinate patch invalidates that last step.

$\dot qT_{\dot q}=2A\dot q^2+B\dot q$, while $2T=2A\dot q^2+2B\dot q+2C$. They agree only if the lower-degree terms have the required zero contribution, in particular for $B=C=0$.

$\sum_i\dot q_ip_i=\sum_i\dot q_iT_{\dot q_i}=2T_2+T_1$. Subtracting $L=T_2+T_1+T_0-V$ cancels $T_1$ and leaves $T_2-T_0+V$.

$T=m(\dot q^2+U^2)/2$, so $p=m\dot q$. Thus $H=p^2/(2m)-mU^2/2$. The laboratory energy differs by $mU^2$, a constant in this special case.

$M=\operatorname{diag}(m,mr^2)$ has zero determinant at the origin. All angles describe the same position there, so the chart is not regular. This is not a physical zero mass; Cartesian coordinates remove the singularity.

$p_1=m(2\dot q_1+\dot q_2)$ and $p_2=m(\dot q_1+3\dot q_2)$. The matrix is $m\begin{pmatrix}2&1\\1&3\end{pmatrix}$ with determinant $5m^2>0$.

$T_{\dot q_i}=\sum_am_a\mathbf v_a\cdot\mathbf e_{ai}$. Its time derivative is $\sum_am_a(\mathbf a_a\cdot\mathbf e_{ai}+\mathbf v_a\cdot\dot{\mathbf e}_{ai})$. Smooth transformations give $T_{q_i}=\sum_am_a\mathbf v_a\cdot\dot{\mathbf e}_{ai}$. Subtracting leaves $\sum_am_a\mathbf a_a\cdot\partial_{q_i}\mathbf r_a$.

The cross contribution to $T$ is $maU\cos\theta\dot\theta$. Thus $T_1=maU\cos\theta\dot\theta$, $T_2=ma^2\dot\theta^2/2$, and $T_0=mU^2/2$.

No. Symmetry means $M_{ij}=M_{ji}$. In polar coordinates $M_{rr}=m$ and $M_{\theta\theta}=mr^2$, with different dimensions because the coordinates have different dimensions.

$p=m\dot q+B(q)$, so $\dot p=m\ddot q+B^{\prime}\dot q$. Also $L_q=B^{\prime}\dot q-V^{\prime}$. Their difference is $m\ddot q+V^{\prime}$. In one coordinate this linear term is locally a total derivative of an antiderivative of $B$.

With $\mathbf r(q,t)$, velocity includes $\partial_t\mathbf r$. Omitting it changes $T$ and can remove the forcing caused by support acceleration. A relative coordinate is allowed, but it must be related to physical position correctly.

$g\sin\theta_0=-A\cos\theta_0$, giving $\theta_0=-\arctan(A/g)$ on the downward branch. The bob tilts opposite the support acceleration.

$L=ma^2\dot\theta^2/2+ma\dot X\cos\theta\dot\theta+m\dot X^2/2-mga(1-\cos\theta)$. Then $dL_{\dot\theta}/dt=ma^2\ddot\theta+ma\ddot X\cos\theta-ma\dot X\sin\theta\dot\theta$ and $L_\theta=-ma\dot X\sin\theta\dot\theta-mga\sin\theta$. The cross terms cancel, leaving $a\ddot\theta+g\sin\theta=-\ddot X\cos\theta$.

The sum obeys $\ddot s+4s=0$, so $\omega_s=2\,\mathrm{s^{-1}}$. The difference obeys $\ddot d+16d=0$, so $\omega_d=4\,\mathrm{s^{-1}}$.

When $x_1=x_2$, its extension $x_2-x_1$ is zero at all times. Only the wall springs supply restoring forces, so the frequency is $\sqrt{k/m}$.

$x=a\sin\theta\cos\phi$, $y=a\sin\theta\sin\phi$, $z=-a\cos\theta$. The coordinate tangent vectors are perpendicular, with lengths $a$ and $a\sin\theta$. Thus $T=ma^2(\dot\theta^2+\sin^2\theta\dot\phi^2)/2$.

$\Omega^2=9.8/(1\times0.5)=19.6\,\mathrm{s^{-2}}$, giving $\Omega=4.43\,\mathrm{s^{-1}}$. This is the azimuthal angular speed, not a small-oscillation frequency.

$\ddot\theta=\sin\theta(\Omega^2\cos\theta-g/a)$. Thus $\theta=0,\pi$ modulo $2\pi$, or $\cos\theta=g/(a\Omega^2)$. The off-bottom pair exists only for $\Omega^2\geq g/a$, with the pair merging at the bottom at equality.

$\ddot\theta\simeq(\Omega^2-g/a)\theta$. It is oscillatory for $\Omega^2<g/a$, exponential for $\Omega^2>g/a$, and the linear test is inconclusive at equality. The cubic term there is restoring, $-(g/2a)\theta^3$.

The Lagrange derivatives cancel the $\dot X\dot\theta$ terms, leaving only $\ddot X$. Constant-velocity frames are related by a Galilean transformation. An accelerating support produces relative forcing, equivalent to a fictitious force in its frame.

$s=A\cos(\sqrt{k/m}\,t)$ and $d=A\cos(\sqrt{(k+2\kappa)/m}\,t)$. Hence $x_1=(s+d)/2$ and $x_2=(s-d)/2$. Both frequencies are present even though only one mass was initially displaced.

The motor maintaining $\Omega$ can transfer energy through the moving ideal constraint. The autonomous reduced Lagrangian conserves $\mathcal E=ma^2\dot\theta^2/2-ma^2\Omega^2\sin^2\theta/2+V$, which differs from laboratory $T+V$.