Bead on a loop in a non-inertial frame

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SUMMARY

The discussion focuses on analyzing the equilibrium points of a bead of mass m on a rotating loop of radius R, with a constant angular velocity ω, using non-inertial frames of reference. Key equations include the effective force equation, which incorporates centrifugal and Coriolis forces, as well as the conditions for equilibrium where angular acceleration and velocity are zero. The participant expresses difficulty in applying non-inertial frame concepts to derive stationary points for the angle θ, despite being comfortable with Lagrangian mechanics. The conclusion emphasizes that solving for equilibrium positions does not require differential equations.

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  • Understanding of non-inertial frames of reference
  • Familiarity with centrifugal and Coriolis forces
  • Knowledge of equilibrium conditions in rotational dynamics
  • Basic proficiency in Lagrangian mechanics
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Homework Statement


Consider a bead of mass m on a loop of radius R. The loop is rotating at a constant rate ω. Using a non-inertial frames of reference, find all equilibrium points of the bead on the loop.


Homework Equations


\vec{F_e}=m\vec{a_r}
\vec{F_e}=\vec{F}-m\vec{\ddot{R_f}}-m\vec{\dot{\omega}}×\vec{r}-m\vec{\omega}×(\vec{\omega}×\vec{r})-2m\vec{\omega}×\vec{v_r}
Where \vec{F} is the sum of the forces acting on a particle as measured in a fixed inertial system, -m\vec{\ddot{R_f}} and -m\vec{\dot{\omega}}×\vec{v_r} are the translational and angular accelerations, respectively, of the moving coordinate system relative to the fixed system.
\vec{v_r} is the velocity relative to the rotating axes
-m\vec{\omega}×(\vec{\omega}×\vec{r})
is the centrifugal force, and
-2m\vec{\omega}×\vec{v_r}
is the Coriolis force.


The Attempt at a Solution


The loop itself isn't moving from any fixed frame, it is simply rotating, so there is no translational velocity. The loop is rotating at a constant rate, so there is no angular acceleration, meaning the second and third terms disappear. The forces acting on the bead in the fixed frame are gravity and the normal force. After plugging those in for the first term, you can divide everything by m to find the acceleration of the bead. From there, you are supposedly able to solve the resultant differential equations in θ, the angle the bead makes with the vertical. After that point, you can find the stationary points for θ, such that the angular velocity and acceleration are 0, I think, which gives you the stationary points. I can do this with a Lagrangian without much difficulty, but I cannot for the life of me figure out how to do this with non-inertial frames. I am awful with them. Any help would be greatly appreciated.
 
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If you just need to find the equilibrium positions, you don't need to solve any differential equations. If the bead remains at an equilibrium position, what are the values of ##\vec{a_r}## and ##\vec{v_r}##?
 

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