Question #145943
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.
1
Expert's answer
2020-11-23T10:32:50-0500

As per the given question,


x1=l1sinϕ1x_1=l_1\sin\phi_1

y1=l1cosϕ1y_1=-l_1\cos\phi_1

x2=l1sinϕ1+l2sinϕ2x_2=l_1\sin\phi_1+l_2\sin\phi_2

y2=(l1cosϕ1+l2cosϕ2)y_2=-(l_1\cos\phi_1+l_2\cos\phi_2),

Now, taking the differentiation ,

dx1dt=l1cosϕ1dϕ1dt\frac{dx_1}{dt}=l_1\cos\phi_1 \frac{d\phi_1}{dt}


dy1dt=l1sinϕ1dϕ1dt\frac{dy_1}{dt}=l_1\sin\phi_1 \frac{d\phi_1}{dt}


y2dt=l1sinϕ1dϕ1dt+l2sinϕ2dϕ2dt\frac{y_2}{dt}=l_1\sin\phi_1\frac{d\phi_1}{dt}+l_2\sin\phi_2\frac{d\phi_2}{dt}


x2dt=(l1cosϕ1dϕ1dt+l2cosϕ2dϕ2dt)\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=TV=m1v122+m2v222VL=T-V=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}-V


T=m1((x1dt)2+(y1dt)2)2+m1((x2dt)2+(y2dt)2)2T=\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=(m1+m2)l122(dϕ1dt)2+(m2)l222(dϕ2dt)2+m2l1l2dϕ1dtdϕ2dtcos(ϕ1ϕ2)+(m1+m2)gl1cosϕ1+m2gl2cosϕ1L=\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


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