Lagrange 102

Classical Mechanics Level 3

A bead of mass m m slides without losses along a wire in the shape of the curve y = x 2 y = x^2 . The gravitational acceleration g g is in the negative y y direction. Gravitational potential energy is measured with respect to y = 0 y = 0 .

What is the Lagrangian for the particle's motion?

Note: In the equations on the right, x x represents the particle's horizontal position

1 2 m x ˙ 2 − m g x \large{\frac{1}{2} m \dot{x}^2 - m g x } 1 2 m x ˙ 2 − m g x 2 \large{\frac{1}{2} m \dot{x}^2 - m g x^2 } 1 2 m ( 1 + 4 x 2 ) x ˙ 2 − m g x 2 \large{\frac{1}{2} m (1 + 4 x^2) \, \dot{x}^2 - m g x^2} 1 2 m ( 1 + 4 x 2 ) x ˙ 2 − m g x \large{\frac{1}{2} m (1 + 4 x^2) \, \dot{x}^2 - m g x }

Steven Chase by Steven Chase , 37, USA

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2 solutions

Steven Chase
Jul 14, 2018

Coordinates and velocity:

x = x y = x 2 x ˙ = x ˙ y ˙ = 2 x x ˙ \large{x = x \\ y = x^2 \\ \dot{x} = \dot{x} \\ \dot{y} = 2 x \dot{x}}

Kinetic Energy:

v 2 = x ˙ 2 + y ˙ 2 = ( 1 + 4 x 2 ) x ˙ 2 E = 1 2 m v 2 = 1 2 m ( 1 + 4 x 2 ) x ˙ 2 \large{v^2 = \dot{x}^2 + \dot{y}^2 = (1 + 4 x^2) \dot{x}^2 \\ E = \frac{1}{2} m v^2 = \frac{1}{2} m (1 + 4 x^2) \dot{x}^2 }

Potential Energy:

U = m g y = m g x 2 \large{U = m g y = m g x^2 }

Lagrangian:

L = E − U = 1 2 m ( 1 + 4 x 2 ) x ˙ 2 − m g x 2 \large{L = E - U = \frac{1}{2} m (1 + 4 x^2) \dot{x}^2 - m g x^2 }

3 Helpful 0 Interesting 1 Brilliant 0 Confused

We have y = x 2 y = x^{2} y ˙ = 2 x x ˙ \dot{y} = 2x\dot{x} and v = x ˙ i ^ + y ˙ j ^ v = \dot{x}\hat{i} + \dot{y}\hat{j} v = x ˙ i ^ + 2 x x ˙ j ^ v = \dot{x}\hat{i} + 2x\dot{x}\hat{j} and height from reference level = y y

L = T - U

1 2 m v 2 − m g h \frac{1}{2}mv^{2} - mgh 1 2 m ( x ˙ 2 + 4 x 2 x ˙ 2 ) − m g y \frac{1}{2}m(\dot{x}^{2} +4x^{2}\dot{x}^{2}) - mgy 1 2 m ( 1 + 4 x 2 ) x ˙ 2 − m g x 2 \frac{1}{2}m(1 +4x^{2})\dot{x}^{2} - mgx^{2}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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