Question

Two bodies m₁ and m₂ attached to each other by weightless rigid rod of length a can slide along fixed axes which formed angle with horizontal axis (as drawn shows). Find the Lagrangian of this system?

Answer 1

Let the kinetic energy of each block be $T_1$ and $T_2$

Total energy of the system be $T=T_1+T_2$

$T_1=\frac{m_1v2}{2}$

$T_2=\frac{m_2 v^2}{2}$

$x=\frac{l\sin\theta\cos\phi}{2}$

$y=\frac{l\sin\theta\cos\phi}{2}$

Now, taking the differentiation,

$\dot{x}^2+\dot{y}^2+\dot{z}^2=v^2 ....(i)$

$\dot{x}=\frac{l}{2}(\cos\phi \cos\theta\dot{\theta}-\sin\theta\sin\phi \dot{\phi})$

$\dot{y}=\frac{l}{2}(\sin\phi \cos\theta\dot{\theta}-\sin\theta\cos\phi \dot{\phi})$

$\dot{z}=-\frac{l}{2}\sin\theta \dot{\theta}$

Now, substituting the values in (i)

$\Rightarrow \dot{x}^2+\dot{y}^2+\dot{z}^2=\frac{l^2}{4}(\cos^2\phi \cos^2\theta\dot{\theta}^2+\sin^2\phi \sin^2\theta\dot{\phi}^2+\sin^2\phi \cos^2\theta\dot{\theta}^2-\sin^2\theta\cos^2\phi \dot{\phi}^2+\sin^2\theta\dot{\theta}^2)$

$v^2=\frac{l^2}{4}(\dot{\theta}^2+\sin^2\theta\dot{\phi}^2)$

Now, substituting the value of $v^2=ma^2\dot{\psi}^2+\frac{ml^2}{4}(\dot{\theta}^2+\sin^2\theta\dot{\phi}^2)$