 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2012/paper-1/1/a/solution

Use the figure's upward-positive vertical coordinate. The support is at

(x, 0)

and the bob at

(x + l sin ϕ, - l cos ϕ)

. Thus

v_{2}^{2} = (\overset{x}{˙} + l \dot{ϕ} cos ϕ)^{2} + (l \dot{ϕ} sin ϕ)^{2} = \overset{x}{˙}^{2} + 2 l \overset{x}{˙} \dot{ϕ} cos ϕ + l^{2} \dot{ϕ}^{2} .

The potential is

V = - m_{2} g l cos ϕ

, up to an irrelevant constant. Combining both kinetic energies gives

L = \frac{1}{2} (m_{1} + m_{2}) \overset{x}{˙}^{2} + \frac{1}{2} m_{2} l^{2} \dot{ϕ}^{2} + m_{2} l \overset{x}{˙} \dot{ϕ} cos ϕ + m_{2} g l cos ϕ .

The cross term expresses the moving support's contribution to the bob velocity; it would be lost by treating the pivot as fixed.

 New to topics? Read the docs here!