Answer to Question #145943 in Classical Mechanics for sphume

Question

A planar double pendulum consists of two pendula that
swing in a common plane (see sketch). Find the equations
of motion for the angles ϕ1 and ϕ2 using D’Alembert’s
Principle.

Answer 1

As per the given question,

$x_1=l_1\sin\phi_1$

$y_1=-l_1\cos\phi_1$

$x_2=l_1\sin\phi_1+l_2\sin\phi_2$

$y_2=-(l_1\cos\phi_1+l_2\cos\phi_2)$,

Now, taking the differentiation ,

$\frac{dx_1}{dt}=l_1\cos\phi_1 \frac{d\phi_1}{dt}$

$\frac{dy_1}{dt}=l_1\sin\phi_1 \frac{d\phi_1}{dt}$

$\frac{y_2}{dt}=l_1\sin\phi_1\frac{d\phi_1}{dt}+l_2\sin\phi_2\frac{d\phi_2}{dt}$

$\frac{x_2}{dt}=(l_1\cos\phi_1\frac{d\phi_1}{dt}+l_2\cos\phi_2 \frac{d\phi_2}{dt})$

$L=T-V=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}-V$

$T=\frac{m_1((\frac{x_1}{dt})^2+(\frac{y_1}{dt})^2)}{2}+\frac{m_1((\frac{x_2}{dt})^2+(\frac{y_2}{dt})^2)}{2}$

Hence langrangian equation will be

$L=\frac{(m_1+m_2)l_1^2}{2}(\frac{d\phi_1}{dt})^2+\frac{(m_2)l_2^2}{2}(\frac{d\phi_2}{dt})^2+m_2l_1l_2\frac{d\phi_1}{dt}\frac{d\phi_2}{dt}\cos(\phi_1-\phi_2)+(m_1+m_2)gl_1\cos\phi_1+m_2gl_2\cos\phi_1$