1

In question : Why is the d'Alembert's Principle formulated in terms of virtual displacements rather than real displacements in time? there is a response :

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Perhaps a simple example is called for.

Example. In 2D consider a point mass $m$ with position ${\bf r}=(x,y)$ that is constrained to move on a frictionless vertical rod, which in turn has pre-determined horizontal position $$x=f(t),\tag{1}$$ where $f$ is a given function of time $t$. In other words, eq. (1) is a holonomic constraint, and there is one generalized position coordinate $q\equiv y$, i.e. one degree of freedom. The constraint force ${\bf F}^{(c)}$ is horizontal. The virtual displacements $\delta q \equiv\delta y$ are by definition vertical with $\delta t=0$, leading to that the constraint force ${\bf F}^{(c)}$ does no virtual work$^1$ $$ {\bf F}^{(c)} \cdot \delta {\bf r}~=~0.\tag{2} $$ However if we allow $\delta t\neq 0$, then eq. (2) will no longer hold, and this implies that we can no longer derive d'Alembert's principle from Newton's 2nd law.

--

$^1$It is tempting to call eq. (2) the Principle of virtual work, but strictly speaking, the principle of virtual work is just d'Alembert's principle for a static system. For d'Alembert's principle, see also this and this related Phys.SE posts.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

$\boldsymbol {My}$ $\boldsymbol {question}$ is if $\delta t\neq 0$ why will eq. (2) no longer hold?

  • „ eq. (1) is a holonomic constraint“ I don’t think so , holonomic constraint is something like this $x^2+y^2=L^2$ for example – Eli Mar 27 '20 at 15:06
  • So in your example you don’t have constraint force at all – Eli Mar 27 '20 at 15:13
  • @Eli - the response I quoted has a constraint force, so are you suggesting that the response I quoted is incorrect ? In any case what is the exact difference between the virtual time =0 and an actual displacement in time behaviour in this case ? –  Mar 27 '20 at 15:21
  • equation (2) is a dot product between the constraint force and the virtuelle displacement which is not a function of time – Eli Mar 27 '20 at 15:30
  • @Eli If (2) doesn't hold, what would the revised equation be for a differential increase in time ? –  Mar 27 '20 at 15:34
  • Hi @Little Cheese. I updated the original answer. Does it still need further explanation? – Qmechanic Mar 27 '20 at 18:30
  • @Qmechanic Not sure. Eli says your eq (1) is not holonomic, but your words say it is ? I have read most of the 250 or so answers on D'Alemberts principle but still don't get it ! It looks like all thats being done is to absorb the constraints into the generalised coordinates - but this is confused by the virtual / normal differential movements (or at least it has confused me!). –  Mar 27 '20 at 19:09
  • @Little Cheese: It is holonomic by definition. I updated the answer. How about now? – Qmechanic Mar 27 '20 at 19:38
  • @Qmechanic. Apologies, I still don't understand why "if we allow δt≠0, then eq. (2) will no longer hold" (apart from the obvious - its a change from the definition). Is there an easy example to explain what goes wrong? Goldstein says enigmatically, but with no explanation : "No variation in time ...since only then is the virtual displacement perpendicular to the force of constraint if the constraint itself is changing in time". –  Mar 28 '20 at 13:58
  • @Little Cheese: I updated the answer. How about now? – Qmechanic Mar 28 '20 at 14:58
  • @Qmechanic. Very neat & simple demonstration of what goes wrong, thanks. –  Mar 28 '20 at 15:53
  • @Little Cheese: Do you mind if I close your post as a duplicate? – Qmechanic Mar 28 '20 at 15:56
  • @Qmechanic - yup go ahead and mark it as a duplicate. (A minor point in (6) how does ∂r/∂q become just ey - it looks like there should be another multiplier present, but it doesn't affect the answer?) –  Mar 28 '20 at 16:05
  • @Little Cheese: I updated the answer. – Qmechanic Mar 28 '20 at 16:38

0 Answers0