Rotational Motion with Elastic P. Energy?

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SUMMARY

The discussion focuses on analyzing the equilibrium position of a mass m on a wedge inclined at angle θ, attached to a spring with spring constant k, while the wedge rotates with angular velocity ω. The equilibrium condition is derived using the principles of conservation of energy and circular motion dynamics. The final equation for maximum force (Fmax) is established as Fmax = kx^2 + 2mgLsinθ, highlighting the contributions of both spring force and gravitational potential energy. The discussion emphasizes the necessity of considering rotational kinetic energy in the equilibrium analysis.

PREREQUISITES
  • Understanding of rotational dynamics and angular velocity (ω)
  • Knowledge of spring mechanics and Hooke's Law (spring constant k)
  • Familiarity with conservation of energy principles in physics
  • Basic concepts of circular motion and forces acting on objects in motion
NEXT STEPS
  • Study the effects of rotational kinetic energy on equilibrium systems
  • Explore advanced applications of Hooke's Law in dynamic systems
  • Learn about the forces acting on objects in circular motion
  • Investigate the role of potential energy in mechanical systems
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics, rotational motion, and energy conservation principles. This discussion is beneficial for anyone tackling complex equilibrium problems involving rotational dynamics and spring systems.

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Homework Statement


Consider a small box of mass m sitting on a wedge with an angle θ and fixed to a spring
with a spring constant k and a length in a non-stretched state L. The wedge
rotates with an angular velocity ω around the vertical axis. Find the equilibrium
position of the box and discuss the conditions when such equilibrium is possible and when it is impossible. The box can move only in the direction along the wedge slope and cannot move in the perpendicular direction (e.g. it is on a rail)

2. The attempt at a solution
First of all I don't know what it's asking exactly so I started with Fmax? I.e. when the mass is going up the plane.

I started with Conservation of Energy:
1/2mv^2 = 1/2kx^2 + mgh (general equation using r = L and h = Lsinθ)

v^2/r = ω (circular motion)

vmax^2/r = kx^2/mr + 2gh/r (I divided everything by r so it fits into the Fmax equation using mrω^2 instead of mv^2/r)

In the end I get Fmax = kx^2 + 2mgLsinθ

Could someone help me out please? I'm really lost :(
 
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The problem with your energy equation is that the system has rotational kinetic energy due to the spinning. For the box to be in equilibrium it will be moving in a circle which means there is a force [itex]m\omega^2r[/itex] on the box parallel to the horizontal pointing away from the slope. This force comes from the normal force from the slope and the spring.
 

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