Velocities & virtual displacement

1. Two different questions about a change in position

Suppose a bead sits on a wire that is being lifted. During a real interval of time the bead may move along the wire while the wire itself rises. Now freeze the clock and ask where else the bead could be placed on the wire at that instant. The second question excludes the rise of the wire because time has not advanced.

These questions lead to actual displacement $d\mathbf r$ and virtual displacement $\delta\mathbf r$. The distinction is the geometrical foundation of virtual work. If it is unclear, the later claim that a moving support can do actual work while doing no virtual work will sound contradictory.

Prerequisites: Chapter 3 and the multivariable chain rule in P3. The dot above a variable always means differentiation along a motion with respect to time.

2. Deriving the generalized velocity relation

For smooth holonomic constraints, let the independent generalized coordinates be related to particle positions by differentiable transformations $\mathbf r_a(q_1,\ldots,q_n,t)$. Along an actual trajectory, each $q_i=q_i(t)$.

A change in each coordinate contributes its own first-order positional change, and explicit time dependence contributes another term:

This is a vector equation, meaning the same chain rule applies separately to the $x,y,z$ components. Each partial derivative holds the other coordinates fixed.

Along the trajectory, $dq_i/dt=\dot q_i$. Therefore

The sum describes motion caused by changing configuration coordinates. The last term describes transport built into a moving coordinate geometry. For a fixed track in stationary coordinates, the last term is zero. For a moving support, it need not vanish even if all generalized velocities are zero.

Each term has units of velocity: $(\text{length}/[q_i])([q_i]/\text{time})$. In Cartesian coordinates with no explicit time, the expression reduces to $(\dot x,\dot y,\dot z)$.

3. Two useful derivative identities

The later Lagrangian method treats $q_i$ and $\dot q_i$ as separate arguments of functions such as kinetic energy. When taking $\partial/\partial\dot q_i$, hold all $q_j$, all other velocities, and time fixed. Along an actual motion they are related, but partial differentiation on the space of arguments is still well-defined.

Identity 1: differentiating with respect to generalized velocity

In the velocity formula, the coefficient $\partial\mathbf r_a/\partial q_j$ depends on coordinates and time, not generalized velocities. Differentiating the sum with respect to $\dot q_i$ selects its $i$th term. The explicit-time term contains no $\dot q_i$. Thus

Identity 2: time derivative of a coordinate response

Apply the total derivative to $\partial\mathbf r_a/\partial q_i$:

Now differentiate the velocity formula with respect to $q_i$, holding the velocities fixed:

For smooth transformations, the mixed partial derivatives agree. Comparing corresponding terms gives

The two identities express the same changing geometry through different derivatives. Both will enter the relation between kinetic energy and generalized inertia in the Lagrangian formulation.

4. Virtual displacement: a fixed-time comparison

Definition. A virtual displacement is an infinitesimal change between neighboring admissible configurations at the same instant, consistent with the constraints. For holonomic coordinates, $\delta t=0$ while the independent $\delta q_i$ may vary.

Apply a fixed-time variation to the transformation. Since there is no elapsed time, the explicit-time contribution is absent:

This is not motion with infinite speed. No speed is assigned: we compare configurations, rather than follow a trajectory. It is also not a finite chord between widely separated points. The expression gives the first-order tangent displacement at the configuration under consideration.

The generalized variations are independent only after we have chosen independent coordinates. The Cartesian components $\delta x,\delta y,\delta z$ generally remain related by constraints.

The constraint test

For $f(\mathbf r,t)=0$, differentiate along an actual motion:

For a virtual comparison at the same time,

The gradient is perpendicular to the constraint surface, so the second equation says that the virtual displacement is tangent to the instantaneous surface. For a moving surface, the actual displacement need not lie in that tangent plane because the surface itself advances.

For many particles, replace the first dot product by the sum $\sum_a\nabla_a f\cdot d\mathbf r_a$, and similarly for virtual displacement. Here $\nabla_a$ differentiates with respect to the position of particle $a$.

5. A moving wire makes the distinction visible

Let a horizontal wire rise with constant speed $U$. Its equation is $y=Ut$. Choose $q=x$. Then $\mathbf r=(q,Ut)$ and

The actual displacement has a vertical part. Every virtual displacement is horizontal because the wire is horizontal at the frozen instant.

The interactive diagram fixes $U=2\,\mathrm{m\,s^{-1}}$ and an illustrative actual horizontal speed of $1\,\mathrm{m\,s^{-1}}$. Move the time slider to choose an instantaneous wire position; vary the virtual comparison independently. Arrow lengths are diagrammatic; the stated components define the example.

For time-independent transformations, the explicit-time term vanishes. Then actual and virtual displacements share the same admissible tangent directions. They remain different concepts: the actual change is selected by a motion, while the virtual variation is a free comparison subject to the instantaneous geometry.

6. Applications

Example 1 · Pendulum velocity and tangent displacement

For a fixed-pivot pendulum, $\mathbf r=(a\sin\theta,-a\cos\theta)$. Differentiating this position with respect to time gives the physical velocity, while varying $\theta$ at fixed time gives the virtual displacement.

Differentiation gives $\mathbf v=(a\cos\theta\dot\theta,a\sin\theta\dot\theta)$. Its squared length is $a^2\dot\theta^2(\cos^2\theta+\sin^2\theta)=a^2\dot\theta^2$, so the speed is $a|\dot\theta|$.

The virtual displacement is $\delta\mathbf r=(a\cos\theta,a\sin\theta)\delta\theta$. Dot it with the radius vector:

It is tangent to the circle. A radial string tension therefore has zero virtual work, the key observation for Chapter 5.

Example 2 · Velocity of a pendulum with a translating pivot

When the pivot translates horizontally at speed $U$, the bob has position $\mathbf r=(Ut+a\sin\theta,-a\cos\theta)$. The two contributions to velocity produce a cross term in the speed squared.

The chain rule gives $\mathbf v=(U+a\cos\theta\dot\theta,a\sin\theta\dot\theta)$. Square both components and add:

Using the trigonometric identity produces $v^2=U^2+2Ua\cos\theta\dot\theta+a^2\dot\theta^2$. The cross term is physically meaningful: support motion and relative motion can reinforce or oppose each other. Setting $U=0$ recovers Example 1. The virtual displacement remains $\delta\mathbf r=(a\cos\theta,a\sin\theta)\delta\theta$, because time is fixed.

Example 3 · Kinetic energy in polar coordinates

Consider a particle of mass $m$ with position $\mathbf r=(r\cos\theta,r\sin\theta)$. Both $r$ and $\theta$ vary, so both contribute to its velocity and kinetic energy.

Differentiate each component:

On squaring and adding, the mixed terms are $-2r\dot r\dot\theta\sin\theta\cos\theta$ and $+2r\dot r\dot\theta\sin\theta\cos\theta$, which cancel. The remaining sums are $\dot r^2(\cos^2\theta+\sin^2\theta)$ and $r^2\dot\theta^2(\sin^2\theta+\cos^2\theta)$. Hence

Radial and tangential contributions add in quadrature because their directions are perpendicular. This kinetic energy will return in Units 2 and 4.

Example 4 · Work done by a moving constraint

A bead on the rising wire $y=Ut$ experiences the normal reaction $\mathbf R=(0,N)$. The work calculation depends on whether the displacement follows the actual motion or compares positions on the wire at fixed time.

Since $\delta\mathbf r=(\delta q,0)$, $\mathbf R\cdot\delta\mathbf r=0$. Since $d\mathbf r=(dq,Udt)$, $\mathbf R\cdot d\mathbf r=NUdt$. The instantaneous power is $NU$.

For $N=4\,\mathrm N$, $U=0.50\,\mathrm{m\,s^{-1}}$, and $dt=0.10\,\mathrm s$, the actual work is $0.20\,\mathrm J$ while virtual work remains zero. There is no contradiction: the two expressions use different displacements.

7. Admissibility and velocity constraints

Do not write $\delta\mathbf r=\mathbf v\delta t$. A virtual comparison has $\delta t=0$ and can still have nonzero $\delta\mathbf r$. Do not call the inertial-frame actual displacement tangent to a moving constraint without checking the explicit-time term. Do not assume $\delta q_i$ are independent if unresolved constraints remain between the $q_i$.

For an affine velocity restriction $\sum_i a_i(q,t)\dot q_i+a_0(q,t)=0$, the standard D’Alembert treatment uses admissible virtual variations satisfying $\sum_i a_i\delta q_i=0$. These are generally not independent variations. This is supporting context only; the course's main calculations use ideal holonomic constraints. General nonlinear velocity constraints require additional care beyond this rule.

8. Practice problems

See Chapter 4 hints and solutions.

1. 4.1 Is a virtual displacement a displacement completed in zero physical time?

2. 4.2 What is held fixed in $\partial\mathbf r/\partial q_i$?

3. 4.3 Can $\dot q=0$ while physical velocity is nonzero?

4. 4.4 For a fixed constraint, are actual and virtual displacement identical concepts?

5. 4.5 Explain why a virtual displacement is tangent to a smooth instantaneous constraint surface.

6. 4.6 Distinguish $dq_i$, $\dot q_i$, and $\delta q_i$.

7. 4.7 Explain why ideal moving constraints may do actual work.

8. 4.8 Derive the generalized velocity formula and both derivative identities in Section 3.

9. 4.9 Derive the actual and virtual differential forms of $f(\mathbf r,t)=0$ and compare them.

10. 4.10 For $\mathbf r=(q,3t)$ in SI units and $\dot q=4\,\mathrm{m\,s^{-1}}$, find velocity, speed, and virtual displacement.

11. 4.11 A pendulum has $a=0.50\,\mathrm m$, $\theta=0$, and $\dot\theta=2\,\mathrm{s^{-1}}$. Find its velocity and the virtual displacement for $\delta\theta=0.01$ rad to first order.

12. 4.12 For a polar particle, $r=2\,\mathrm m$, $\dot r=3\,\mathrm{m\,s^{-1}}$, $\dot\theta=2\,\mathrm{s^{-1}}$, and $m=2\,\mathrm{kg}$. Find speed and kinetic energy.

13. 4.13 Explain actual and virtual displacement using a fixed circle and a rising horizontal wire. Include equations and diagrams.

14. 4.14 Develop the generalized notation and velocity relations for $N$ particles and explain the assumptions.

15. 4.15 For a circle of prescribed radius $a(t)$, show that $\mathbf r\cdot\delta\mathbf r=0$ but $\mathbf r\cdot\mathbf v=a\dot a$.

16. 4.16 Verify both derivative identities directly for the translating-pivot pendulum of Example 2.

9. Tangent variations and transport

The coordinate response and the explicit-time derivative represent distinct contributions to motion. The former describes a change within the instantaneous configuration set; the latter accounts for the motion of that set itself. An actual displacement can contain both. A virtual displacement contains only the first because time is held fixed.

This distinction explains why an ideal reaction can do actual work on a moving support while its virtual work vanishes. It also identifies the variations used in the next chapter: tangent to the instantaneous constraints and independent only when the selected generalized coordinates are independent.