24 May 2025
Coordinate Systems, Virtual Work, and D'Alembert's Principle
Coordinate transformations, degrees of freedom, generalized velocities, virtual work, and D'Alembert's equation.
Coordinate systems
Cartesian coordinates $(x,y,z)$ use fixed, mutually perpendicular unit vectors. Two common curvilinear transformations are cylindrical coordinates $(\rho,\phi,z)$,
\[x=\rho\cos\phi, \qquad y=\rho\sin\phi, \qquad z=z,\]and spherical coordinates $(r,\theta,\phi)$,
\[x=r\sin\theta\cos\phi, \qquad y=r\sin\theta\sin\phi, \qquad z=r\cos\theta.\]Their standard coordinate ranges are
\[\rho\geq0,\quad 0\leq\phi<2\pi; \qquad r\geq0,\quad 0\leq\theta\leq\pi,\quad 0\leq\phi<2\pi.\]The inverse relations include
\[\rho=\sqrt{x^2+y^2}, \qquad \phi=\operatorname{atan2}(y,x),\]and
\[r=\sqrt{x^2+y^2+z^2}, \qquad \theta=\cos^{-1}\!\left(\frac zr\right).\]At $\rho=0$ the azimuth $\phi$ is undefined; at $r=0$ both spherical angles are undefined. These are coordinate singularities, not physical singularities. Differentiating the position vector gives
\[\mathbf v =\dot\rho\,\hat{\boldsymbol\rho} +\rho\dot\phi\,\hat{\boldsymbol\phi} +\dot z\,\hat{\mathbf z}\]in cylindrical coordinates, and
\[\mathbf v =\dot r\,\hat{\mathbf r} +r\dot\theta\,\hat{\boldsymbol\theta} +r\sin\theta\dot\phi\,\hat{\boldsymbol\phi}\]in spherical coordinates. The factors $1$, $r$, and $r\sin\theta$ arise because the curvilinear basis vectors and arc lengths depend on position.
Degrees of freedom and generalized velocities
A system of $N$ point particles requires $3N$ Cartesian coordinates before constraints are imposed. If $k$ independent holonomic constraints
\[f_\alpha(\mathbf r_1,\ldots,\mathbf r_N,t)=0, \qquad \alpha=1,\ldots,k,\]have rank $k$, the system has
\[\boxed{s=3N-k}\]degrees of freedom. Choose independent generalized coordinates $q_1,\ldots,q_s$ and write
\[\mathbf r_i=\mathbf r_i(q_1,\ldots,q_s,t).\]The chain rule gives the actual velocity,
\[\boxed{ \mathbf v_i =\sum_{j=1}^{s}\frac{\partial\mathbf r_i}{\partial q_j}\dot q_j +\frac{\partial\mathbf r_i}{\partial t}}.\]The $q_j$ need not be lengths: they may be angles or other independent parameters. Their time derivatives $\dot q_j$ are generalized velocities.
Virtual displacement and virtual work
A virtual displacement compares neighbouring allowed configurations at the same instant, so $\delta t=0$. Hence
\[\boxed{ \delta\mathbf r_i =\sum_{j=1}^{s}\frac{\partial\mathbf r_i}{\partial q_j}\delta q_j}.\]It must also satisfy the linearized constraints,
\[\sum_i\nabla_i f_\alpha\cdot\delta\mathbf r_i=0.\]This differs from an actual displacement during a time $dt$, for which an additional $(\partial f_\alpha/\partial t)dt$ term can occur.
At static equilibrium, ideal constraint forces do no virtual work. If $\mathbf F_i$ denotes the applied force, the principle of virtual work is
\[\delta W=\sum_i\mathbf F_i\cdot\delta\mathbf r_i=0.\]Substitution of the generalized displacement gives
\[\delta W=\sum_jQ_j\delta q_j, \qquad \boxed{Q_j=\sum_i\mathbf F_i\cdot \frac{\partial\mathbf r_i}{\partial q_j}}.\]$Q_j$ has the units of energy divided by the units of $q_j$. Thus it is a force for a length coordinate and a torque for an angular coordinate.
D’Alembert’s principle
Let $\mathbf R_i$ be ideal constraint forces. Newton’s equations are
\[\mathbf F_i+\mathbf R_i=m_i\mathbf a_i.\]Multiplication by allowed virtual displacements and summation eliminates the constraint forces because $\sum_i\mathbf R_i\cdot\delta\mathbf r_i=0$. Therefore
\[\boxed{ \sum_i(\mathbf F_i-m_i\mathbf a_i)\cdot\delta\mathbf r_i=0}.\]To express the inertial term in generalized coordinates, start from
\[T=\frac12\sum_i m_i\mathbf v_i^2.\]The coordinate transformation implies
\[\frac{\partial\mathbf v_i}{\partial\dot q_j} =\frac{\partial\mathbf r_i}{\partial q_j}, \qquad \frac{d}{dt}\left(\frac{\partial\mathbf r_i}{\partial q_j}\right) =\frac{\partial\mathbf v_i}{\partial q_j}.\]Consequently,
\[\frac{\partial T}{\partial\dot q_j} =\sum_i m_i\mathbf v_i\cdot\frac{\partial\mathbf r_i}{\partial q_j},\]and differentiation with respect to time gives
\[\frac{d}{dt}\frac{\partial T}{\partial\dot q_j} -\frac{\partial T}{\partial q_j} =\sum_i m_i\mathbf a_i\cdot\frac{\partial\mathbf r_i}{\partial q_j}.\]D’Alembert’s principle therefore becomes
\[\sum_j\left[ Q_j-\frac{d}{dt}\frac{\partial T}{\partial\dot q_j} +\frac{\partial T}{\partial q_j} \right]\delta q_j=0.\]Because the independent $\delta q_j$ are arbitrary,
\[\boxed{ \frac{d}{dt}\frac{\partial T}{\partial\dot q_j} -\frac{\partial T}{\partial q_j}=Q_j}.\]This is D’Alembert’s equation in generalized coordinates and is the direct bridge from Newton’s laws to Lagrange’s equations.
Solved Problems
1. Bead on a uniformly rotating smooth rod
A bead of mass $m$ slides without friction on a straight horizontal rod that rotates with prescribed constant angular speed $\Omega$ about one end. Obtain its radial equation directly from D’Alembert’s principle.
The constraint is time-dependent:
\[\mathbf r=r\,\hat{\mathbf e}_r, \qquad \phi=\Omega t.\]The rotating polar unit vectors obey
\[\dot{\hat{\mathbf e}}_r=\Omega\hat{\mathbf e}_\phi, \qquad \dot{\hat{\mathbf e}}_\phi=-\Omega\hat{\mathbf e}_r.\]Therefore
\[\mathbf v=\dot r\,\hat{\mathbf e}_r+\Omega r\,\hat{\mathbf e}_\phi,\]and then
\[\mathbf a=(\ddot r-\Omega^2r)\hat{\mathbf e}_r +2\Omega\dot r\,\hat{\mathbf e}_\phi.\]At fixed time an allowed virtual displacement is along the rod,
\[\delta\mathbf r=\delta r\,\hat{\mathbf e}_r.\]The smooth rod’s reaction is perpendicular to this displacement, so it does no virtual work. There is no applied force along the rod. D’Alembert’s principle gives
\[(-m\mathbf a)\cdot\delta\mathbf r =-m(\ddot r-\Omega^2r)\delta r=0.\]Since $\delta r$ is arbitrary,
\[\boxed{\ddot r-\Omega^2r=0}, \qquad r=C_1e^{\Omega t}+C_2e^{-\Omega t}.\]The outward term $m\Omega^2r$ is not an additional real force in the inertial frame; it is the radial part of the acceleration required by the rotating constraint. The transverse reaction supplies $2m\Omega\dot r\,\hat{\mathbf e}_\phi$.
2. Atwood machine in one generalized coordinate
Two masses $m_1$ and $m_2>m_1$ are joined by a light inextensible string over an ideal pulley. Let $x$ increase when $m_2$ moves downward; the same $x$ makes $m_1$ move upward. The system has one degree of freedom.
For an allowed virtual change $\delta x$, gravity does virtual work
\[\delta W=m_2g\,\delta x-m_1g\,\delta x,\]so the generalized force is
\[Q_x=(m_2-m_1)g.\]Both masses have speed $\dot x$, hence
\[T=\frac12(m_1+m_2)\dot x^2.\]D’Alembert’s equation gives
\[\frac{d}{dt}\frac{\partial T}{\partial\dot x} -\frac{\partial T}{\partial x}=Q_x,\]or
\[(m_1+m_2)\ddot x=(m_2-m_1)g.\]Therefore
\[\boxed{a=\ddot x=\frac{m_2-m_1}{m_1+m_2}g}.\]From $T-m_1g=m_1a$ for the rising mass,
\[\boxed{T=\frac{2m_1m_2}{m_1+m_2}g}.\]Both expressions have the correct limits: $a=0$ when $m_1=m_2$, while $a\to g$ when $m_2/m_1\to\infty$.
Descriptive Questions
- Distinguish actual displacement from virtual displacement for a rheonomous constraint, and state precisely why $\delta t=0$ in the latter.
- Derive the generalized-velocity formula from $\mathbf r_i=\mathbf r_i(q_1,\ldots,q_s,t)$ and interpret its explicit-time term.
- Explain the rank condition behind $s=3N-k$ and why dependent constraint equations must not be counted separately.
- Derive D’Alembert’s equation in generalized coordinates and identify the condition under which ideal constraint forces disappear from it.
Numerical Problems
- At an instant a particle has cylindrical data $\rho=0.40\ \mathrm m$, $\dot\rho=0.30\ \mathrm{m\,s^{-1}}$, $\dot\phi=2.0\ \mathrm{s^{-1}}$, and $\dot z=-0.20\ \mathrm{m\,s^{-1}}$. Find its speed. Answer: $0.8775\ \mathrm{m\,s^{-1}}$.
- A particle has $r=2.0\ \mathrm m$, $\theta=\pi/3$, $\dot r=0.40\ \mathrm{m\,s^{-1}}$, $\dot\theta=0.20\ \mathrm{s^{-1}}$, and $\dot\phi=0.50\ \mathrm{s^{-1}}$. Find its speed in spherical coordinates. Answer: $1.034\ \mathrm{m\,s^{-1}}$.
- In an ideal Atwood machine, $m_1=2.0\ \mathrm{kg}$ and $m_2=3.0\ \mathrm{kg}$. Find the acceleration and string tension using $g=9.81\ \mathrm{m\,s^{-2}}$. Answer: $1.962\ \mathrm{m\,s^{-2}}$ toward the $m_2$ side, $23.544\ \mathrm N$.
- A bead moves without friction on the vertical parabola $y=ax^2$, where $a=0.50\ \mathrm{m^{-1}}$. At $x=0.40\ \mathrm m$ it has $\dot x=0.60\ \mathrm{m\,s^{-1}}$. From D’Alembert’s equation, find $\ddot x$ for $g=9.81\ \mathrm{m\,s^{-2}}$. Answer: $-3.507\ \mathrm{m\,s^{-2}}$.
Maxima verification: rotating-rod residuals and all problem values.
References
- D’Alembert’s principle — Wikipedia
- H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Chapter 1, Pearson (2002).
- L. D. Landau and E. M. Lifshitz, Mechanics, 3rd ed., Sections 1–5, Butterworth-Heinemann (1976).
- J. R. Taylor, Classical Mechanics, Chapter 7, University Science Books (2005).
Discussion