Hamiltons Principle on Moving Cart with Fixed Beam

[teknodude]

Homework Statement

-Use Hamiltons Principle

Homework Equations

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

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.

 

[mark726]

Thank you for sharing your work on Hamilton's Principle. Your approach seems sound, and I agree with your reasoning for using w,xx instead of (w,xx-q,xx) in the potential energy term for the beam. The q term represents the displacement of the base, which is already accounted for in the kinetic energy term. However, it is always a good idea to double check for any sign errors in your equations. Good luck with your calculations!