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
$$\frac{1}{2}mb^2\frac{d\dot{\theta}^2}{dt}=\dfrac{\partial L}{\partial...
So the acceleration would simply be
$$\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$$ relation messes it up...
Hello! I have some problem getting the correct answer for (b).
My FBD:
For part (a) my lagrangian is
$$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...