(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

-Use Hamiltons Principle

2. Relevant equations

[tex]\int _{t _{1}}^{t _{2}}({\rm }\delta T -\delta U -\delta V _{E }) dt} [/tex]

3. The attempt at a solution

Here are my energy equations.

So we have kinetic energy from the base and the beam.

[tex]T_{{{\it beam}+{\it base}}}=1/2\,\int _{0}^{l}\!m \left( {\frac {d}{dt}}w \left( t \right) +{\frac {d}{dt}}q_{{1}} \left( t \right) \right) ^{2}{dx}\\\mbox{}+1/2\,M \left( {\frac {d}{dt}}q_{{1}} \left( t \right) \right) ^{2}\]} [/tex]

potential energy from the beam and spring

[tex]U_{{{\it beam}+{\it base}}}=1/2\,\int _{0}^{l}\!{EI(\frac {{d}^{2}{w}}{{{\it dx}}^{2}})^2}{dx}\\\mbox{}+1/2\,K \left( q_{{1}} \left( t \right) \right) ^{2}\]}[/tex]

potential g, so I take into account the foreshortening of the beam.

[tex]V_{{E}}=1/2\,\int _{0}^{l}\!{(mg \left( l-x \right))(\frac { {{\it dw}}}{{{\it dx}}}})^2{dx}\][/tex]

I am somewhat confused with the potential energy term for the beam. I'm trying to reason with myself whether the second derivative of deflection should be just w,xx or (w,xx-q,xx). My argument is that the q is just a function of t, but also it is a general coordinate.

Edit: Solved it. It was just w,xx and there is a sign error in there. Had to take down the image and some info.

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# Homework Help: Hamiltons Principle on Moving Cart with Fixed Beam

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