How to calculate acceleration of blobs in any degree pendulum?

AI Thread Summary
The discussion focuses on developing a method to calculate the acceleration of blobs in multi-degree pendulums using Lagrangian mechanics. The user presents equations for the system's configuration and Lagrangian, seeking validation of their approach. Key equations of motion are derived, leading to expressions for angular accelerations in terms of angles and angular velocities. The user requests confirmation of their calculations, particularly for a simple pendulum case, before proceeding to more complex scenarios. The conversation emphasizes the importance of accurately deriving the equations for n-point masses in pendulum systems.
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I want to create a method to calculate acceleration of blobs in any degree pendulum (double, triple and more). I have this equation but I am not sure if it is correct, or how to extract acceleration from it.

unknown.png


[Mentor Note -- this is a new thread start to correct errors in the previous 2 thread starts]
 
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In the configuration as attached sketch for 2D n-fold pendulum shows
(x_0,y_0)=(0,0)
(x_i,y_i)=(x_{i-1},y_{1-1})+(l_i \cos\theta_i, l_i \sin\theta_i)=(\sum_{j=1}^i l_j\cos\theta_j, \sum_{j=1}^i l_j\sin\theta_j)
(\dot{x_i},\dot{y_i})=(\sum_{j=1}^i l_j (-\sin\theta_j) \dot{\theta_j}, \sum_{j=1}^i l_j\cos\theta_j \dot{\theta_j})

Lagrangean of the system is
L=\sum_{i=1}^n \frac{m_i}{2}(\dot{x_i}^2+\dot{y_i}^2)+g\sum_{i=1}^n m_i x_i
with gravity force applying to x positive direction. Express L as ##L(\theta_1, \theta_2,...,\theta_n, \dot{\theta_1}, \dot{\theta_2}, ... , \dot{\theta_n})##

L=\sum_{i=1}^n \frac{m_i}{2}[\{\sum_{j=1}^i l_j (-\sin\theta_j) \dot{\theta_j}\}^2+\{\sum_{j=1}^i l_j \cos\theta_j\ \dot{\theta_j}\}^2]+g\sum_{i=1}^n m_i \sum_{j=1}^i l_j\cos\theta_j

We get equation of motion or Lagrangean equation from this Lagrangean, e.g.

\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta_k}}=\frac{d}{dt}\sum_{i=k}^n m_i[\{\sum_{j=k}^i l_j (-\sin\theta_j) \dot{\theta_j}\}l_k (-\sin\theta_k)+\{\sum_{j=k}^i l_j \cos\theta_j\ \dot{\theta_j}\}l_k \cos\theta_k]
=\sum_{i=k}^n m_i[\{\sum_{j=k}^i l_j (-\cos\theta_j) \dot{\theta_j}^2\}l_k (-\sin\theta_k)+\{\sum_{j=k}^i l_j (-\sin\theta_j\ \dot{\theta_j}^2\}l_k \cos\theta_k]
+\sum_{i=k}^n m_i[\{\sum_{j=k}^i l_j (-\sin\theta_j) \ddot{\theta_j}\}l_k (-\sin\theta_k)+\{\sum_{j=k}^i l_j \cos\theta_j\ \ddot{\theta_j}\}l_k \cos\theta_k]
+\sum_{i=k}^n m_i[\{\sum_{j=k}^i l_j (-\sin\theta_j) \dot{\theta_j}\}l_k (-\cos\theta_k)\dot{\theta_k}+\{\sum_{j=k}^i l_j \cos\theta_j\ \dot{\theta_j}\}l_k(- \sin\theta_k) \dot{\theta_k}]
equals to
\frac{\partial L}{\partial \theta_k}=\sum_{i=k}^n m_i[\{\sum_{j=k}^i l_j (-\sin\theta_j) \dot{\theta_j}\}l_k (-\cos\theta_k)\dot{\theta_k}+\{\sum_{j=k}^i l_j \cos\theta_j\ \dot{\theta_j}\}l_k (-\sin\theta_k)\dot{\theta_k}]+g\sum_{i=k}^n m_i \ l_k(-\sin\theta_k)

So the resulted equation is
\sum_{i=k}^n m_i[\{\sum_{j=k}^i l_j (-\cos\theta_j) \dot{\theta_j}^2\} (-\sin\theta_k)+\{\sum_{j=k}^i l_j (-\sin\theta_j)\ \dot{\theta_j}^2\} \cos\theta_k<br /> +\{\sum_{j=k}^i l_j (-\sin\theta_j) \ddot{\theta_j}\} (-\sin\theta_k)+\{\sum_{j=k}^i l_j \cos\theta_j\ \ddot{\theta_j}\}\cos\theta_k<br /> -g (-\sin\theta_k) ]=0
or
\sum_{i=k}^n m_i[\ \sum_{j=k}^i l_j \sin(\theta_k-\theta_j) \dot{\theta_j}^2<br /> +\sum_{j=k}^i l_j \cos(\theta_k-\theta_j )\ddot{\theta_j} +g \sin\theta_k\ \ ]=0

where k=1,2,...,n. Coefficients of g have trigonometric function though formula OP does not have it.  

These n equations give you formula of ##\ddot{\theta_k}## as functions of ##\theta_i##s and ##\dot{\theta_i}##s, which you are looking for.
 

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I found this equation:
Równania _wahadła2 (2).png


It comes from this page: https://www.researchgate.net/publication/336868500_Equations_of_Motion_Formulation_of_a_Pendulum_Containing_N-point_Masses and I transformed them like this:

Równanie _wahadła1.png

Did I do it right?
 
How about checking it for n=1, a simple pendulum.
You should get
l_1 \ddot{\theta_1}+g \sin\theta_1=0.

If it's OK proceed to check it for n=2. You should get ##\ddot{\theta_1},\ddot{\theta_2}## as function of ##\theta_1,\theta_2,\dot{\theta_1},\dot{\theta_2}## which is calculated from (19) and (20) of the paper you quote.
 
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