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0 votes
1 answer
75 views

Derivation of lagrange equation in classical mechanics

I'm currently working on classical mechanics and I am stuck in a part of the derivation of the lagrange equation with generalized coordinates. I just cant figure it out and don't know if it's just ...
Jan Oreel's user avatar
2 votes
5 answers
261 views

Why does $\vec{r}\cdot\dot{\vec{r}}=r\dot{r}$?

Why is $$\vec{r}\cdot\dot{\vec{r}}=r\dot {r}$$ true? Before saying anything, I have seen the proofs using spherical coordinates for $$\dot{\vec {r}}= \dot{r}\vec{u_r}+r\dot{\theta}\vec{u_\theta}+r\sin\...
Ulshy's user avatar
  • 69
0 votes
2 answers
129 views

Does the gradient of potential energy exist independent of coordinates?

Potential energy $U(\vec{r})$ of a conservative force field $\vec{F}$ is defined as a function whose variation between positions $\vec{r}_A$ and $\vec{r}_B$ is the opposite of the work done by the ...
Antonio19932806's user avatar
1 vote
1 answer
33 views

Derivatives of the lagrangian of generalized coordinates [closed]

I know that $$U= \frac{1}{2} \sum_{j,k} A_{jk} q_j q_k \quad \quad T= \frac{1}{2} \sum_{j,k} m_{jk} \dot{q}_j \dot{q}_k $$ and the lagrangian is $$ \frac{\partial U}{\partial q_k} - \frac{d}{dt} \...
rannasquaer's user avatar
0 votes
2 answers
76 views

Translation of coordinates to generalised coordinates

The translation form $r_i$ to $q_j$ language start forms the transformation equation: $r_i=r_i (q_1,q_2,…,q_n,t)$ (assuming $n$ independent coordinates) Since it is carried out by means of the ...
Jack Jack's user avatar
  • 187
0 votes
2 answers
2k views

Derivation of generalized velocities in Lagrangian mechanics

So I know that: $$r_i = r_i(q_1, q_2,q_3,...., q_n, t)$$ Where $r_i$ represent the position of the $i$th part of a dynamical system and the $q_n$ represent the dynamical variables of the system ($n$ =...
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