UPSC 2021 Maths Optional Paper 2 Q6c — Step-by-Step Solution

15 marks · Section B

Lagrange's equations · Mechanics & Fluid Dynamics · asked 9× in 13 yrs · Read the full method →

Question

Obtain Lagrangian equation for two particles of unequal masses connected by inextensible string passing over a smooth pulley.

Single generalised coordinate $x$ (downward displacement of $m_1$ ); both KE and PE in terms of $\dot x$ and $x$ ; standard EL gives acceleration.

Setup. Mass $m_1$ on one side, $m_2$ on other side. String inextensible (total length constant). Pulley smooth (no friction).

Let $x$ = downward displacement of $m_1$ from some reference. Then $m_2$ is displaced upward by $x$ (since string is inextensible).

Both masses move with speed $\dot x$ : $T=\dfrac{1}{2}m_1\dot x^2+\dfrac{1}{2}m_2\dot x^2=\dfrac{1}{2}(m_1+m_2)\dot x^2$ .

Take $y$ -up. If $m_1$ descends by $x$ , its height decreases by $x$ ; if $m_2$ ascends by $x$ , its height increases by $x$ .

$V=-m_1 g x+m_2 g x=(m_2-m_1)gx$ .

(Reference $V=0$ at $x=0$ .)

$L=T-V=\dfrac{1}{2}(m_1+m_2)\dot x^2-(m_2-m_1)gx=\dfrac{1}{2}(m_1+m_2)\dot x^2+(m_1-m_2)gx$ .

$\partial L/\partial\dot x=(m_1+m_2)\dot x$ , $d/dt$ : $(m_1+m_2)\ddot x$ .

$\partial L/\partial x=(m_1-m_2)g$ .

EL: $(m_1+m_2)\ddot x-(m_1-m_2)g=0$ ,

\ddot x=\dfrac{(m_1-m_2)g}{m_1+m_2}.

\boxed{\;L=\dfrac{1}{2}(m_1+m_2)\dot x^2+(m_1-m_2)gx;\;\;\ddot x=\dfrac{(m_1-m_2)g}{m_1+m_2}.\;}