Recent content by TimmyD1

Another way of looking at the solution is to study what is usually called the energy function $$h(q_1...\dot{q}_n,\dot{q}_1...\dot{q}_n,t)=\sum_j\dot{q}_j\frac{\partial L}{\partial \dot{q}_j}-L$$ The energy function is derived as one term from the expression for $$\frac{dL}{dt}$$when...

So I thought more about this $$\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}=mb^2\ddot{\theta}=\dfrac{\partial L}{\partial\theta}$$ Multiply both sides with $$\dot{\theta}$$ gives $$\frac{1}{2}mb^2\frac{d\dot{\theta}^2}{dt}=\dfrac{\partial L}{\partial...

I could not solve the differential equation and I think finding the acceleration of the system will be enough for me! Amazing guidance by the way!

Alright! Before I tackle this, do we have any initial conditions?

So with the help of MATLAB, I've found that $$\frac{1}{mb^2}\int\frac{\partial L}{\partial \theta}dt\approx 0.83719t+C$$ What next!?

So the acceleration would simply be $$\ddot{\theta}=\frac{\dfrac{\partial L}{\partial \theta}}{mb^2}$$ now I would assume to integrate the expression to get velocity but not sure what the bounds would be...

Sorry it's a typo that I just realized, $$x=\sqrt{(1.25b-b\cos\theta)^2+(b\sin\theta)^2}-0.25b$$ if we can ignore the notation and geometric point of view for now, it just bothers me why I can't get the same answer using lagrangian. I think that using $$v=r\omega$$ relation messes it up...

I ran into the same integral using the lagrangian you defined

The exercise is to use lagrangian formalism, I have solved it using energy principle but using the lagrangian i ran into problems.

Hello! I have some problem getting the correct answer for (b). My FBD: For part (a) my lagrangian is $$L=T-V\iff L=\frac{1}{2}m(b\dot{\theta})^2+mg(b-b\cos\theta)-\frac{1}{2}k\boldsymbol{x}^2,\ where\ \boldsymbol{x}=\sqrt{(1.25b-b)^2+(b\sin\theta)^2}-(1.25b-0.25b)$$ Hence my equation of...