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• Users: baw
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1. ### Formula for the energy of elastic deformation

Lets say we applied ##\sigma_1## at first and got ##\epsilon_1## as well as some ##\epsilon_2## and ##\epsilon_3##. The (specific) work done is ##\frac{\sigma_1^2}{2E}##. If we now apply ##\sigma_2## we already have some initial strain, so the plot ##\sigma_2(\epsilon_2)## moves downward by...
2. ### Formula for the energy of elastic deformation

In every book I checked, the energy (per unit mass) of elastic deformation is derived as follows: ## \int \sigma_1 d \epsilon_1 = \frac{\sigma_1 \epsilon_1}{2} ## and then, authors (e.g. Timoshenko & Goodier) sum up such terms and substitute ##\epsilon ## from generalised Hooke's law i.e. ##...
3. ### The definition of generalised momentum

Ok, thank you
4. ### The definition of generalised momentum

Ok, that's right, but are ##\frac{dL}{dq}## and ##\frac{\partial L}{\partial q}## exactly same things like F and F?
5. ### The definition of generalised momentum

That's not exactly what I meant. I'm wondering how could Lagrange, deriving his equation, decide which of ##\frac{dL}{dq}## or ##\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}## to choose. I guess the only thing he could compare it to ( in terms of consistency ) was Newton's or d'Alebert's...
6. ### The definition of generalised momentum

I mean the second: ##\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}## or ##\frac{dL}{dq}##. There must be some physical reason that makes the idea of ##\frac{dL}{dq}## incorrect and ##\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}## correct, that could be found befeore we e.g. solve the same...
7. ### The definition of generalised momentum

aren't joules and Newton-meters same thing? When we integrate torque (Nm) with respect to the angle (dimensionless) we get work (J).
8. ### The definition of generalised momentum

I noticed that approaches I mentioned give different results, but my point is: how do we know which of them is correct? or maby... how did Lagrange know that?
9. ### The definition of generalised momentum

Why, in lagrangian mechanics, do we calculate: ##\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}## to get the (generalised) momentum change in time instead of ##\frac{d T}{dq}##? (T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external...