- #1
sharrington3
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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
[tex]\vec{F_e}=m\vec{a_r}[/tex]
[tex]\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}[/tex]
Where [tex]\vec{F}[/tex] is the sum of the forces acting on a particle as measured in a fixed inertial system, [tex]-m\vec{\ddot{R_f}}[/tex] and [tex]-m\vec{\dot{\omega}}×\vec{v_r}[/tex] are the translational and angular accelerations, respectively, of the moving coordinate system relative to the fixed system.
[tex]\vec{v_r}[/tex] is the velocity relative to the rotating axes
[tex]-m\vec{\omega}×(\vec{\omega}×\vec{r})[/tex]
is the centrifugal force, and
[tex]-2m\vec{\omega}×\vec{v_r}[/tex]
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.