Answer to Question #168036 in Classical Mechanics for Aysu

Question #168036

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?


1
Expert's answer
2021-03-03T11:25:54-0500

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)"

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