Block sliding down a moving plane whose angle of inclination changes

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SUMMARY

The discussion focuses on the dynamics of a bar of soap on a frictionless square plate that pivots at a constant angular velocity, ω. The equation of motion derived is \(\ddot{x} - \omega^2 x = -g \sin(\omega t)\), where x represents the distance from the pivot to the soap. The Lagrangian formulation is utilized, leading to the identification of kinetic and potential energy components. The missing term \(-\omega^2 x\) is attributed to the changing angle of inclination over time, which affects the motion of the soap.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with angular motion and angular velocity
  • Knowledge of basic differential equations
  • Concept of potential and kinetic energy in physics
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equation in classical mechanics
  • Explore the effects of angular motion on linear dynamics
  • Learn about the relationship between angular velocity and linear displacement
  • Investigate the implications of non-inertial reference frames in physics
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Students of physics, particularly those studying classical mechanics, as well as educators and researchers interested in the dynamics of systems with changing angles of inclination.

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Homework Statement



A bar of soap of mass m is at rest on a frictionless square plate of side-length D rests on a horizontal table. At t=0 I start to raise one edge of the plate the plate pivots about the opposite edge. The angular velocity about the pivot is a constant ω Show that the equation of motion for the soap is of the form [itex]\ddot{x} - \omega ^{2} x = - g sin( \omega t )[/itex]. Solve this for x(t) given that . In this equation x is the distance from the pivot to the soap (i.e. measured up the slope).

Homework Equations





The Attempt at a Solution



We will find that the potential energy is [itex]V = mgx sin( \theta )[/itex], and the kinetic energy is just [itex]T = \frac{1}{2} m \ddot{x}^{2}[/itex]. Then our Lagrangian is just [itex]L = \frac{1}{2} m \ddot{x}^{2} - mgx sin( \theta )[/itex] and since we know omega is constant, then θ = ωt, so our Lagrangian becomes [itex]L = \frac{1}{2} m \ddot{x}^{2} - mgx sin( \omega t)[/itex].

But using the Euler-Lagrange equation yields [itex]\ddot{x} = -g sin( \theta )[/itex].

Where did the [itex]- \omega ^{2} x[/itex] come from? i want to say that it has something to do with the fact that the angle is changing over time...but i can't see it right now. any help would be great. thanks in advance.
 
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You forgot the kinetic energy from motion perpendicular to the direction of x. As you know the motion of the slope, you can relate this to x and t.
 

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