Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we have a differential equation of the from

$L = \frac{1}{2}((I_1 + I_2 + m_1r_1^2 + m_2q_2^2)\dot{q}_1^2 + m_2\dot{q}_2^2) - a_g(m_1r_1 + m_2q_2) \sin q_1$.

and we want to find $g = \frac{\operatorname{d}}{\operatorname{d}t} \frac{\partial L}{\partial \dot q_1}$, and $h = \frac{\partial L}{\partial q_1}$.

For the first term I get $g = (I_1 + I_2 + m_1r_1^2 + m_2q_2^2)\ddot{q}_1$ which I'm pretty sure is correct. However, I believe the second term should be $h = 2m_2q_2\dot{q_1}\dot{q_2} + a_g(m_1r_1 + m_2q_2) \cos q_1$. I see where the cosine term appears from, but not $2m_2q_2\dot{q_1}\dot{q_2}$. Can anybody help me out?

share|improve this question
    
I'm not so sure about your $g$, since $\frac{d}{dt}q_2^2\dot q_1 = 2\dot q_2 q_2\dot q_1 + q_2^2 \ddot q_1$. I think. –  Myself Apr 20 '11 at 17:45
    
@Myself: You're missing a $q_2$ in the first term. –  joriki Apr 20 '11 at 17:49
    
@Joriki: right, thanks! I made an edit to avoid confusion. –  Myself Apr 20 '11 at 17:50
add comment

1 Answer 1

up vote 2 down vote accepted

The term $2m_2q_2\dot{q_1}\dot{q_2}$ that you're missing doesn't come from $h$ but from $g$ -- you forgot to take into account the variation of $q_2$ with time:

$$\frac{\mathrm d}{\mathrm d t}m_2q_2^2\dot{q_1}=m_2q_2^2\ddot{q_1}+2m_2q_2\dot{q_1}\dot{q_2}\;.$$

share|improve this answer
1  
Ah! You Sir are a gentleman and a scholar! Thank you. –  user9842 Apr 20 '11 at 18:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.