Rotational Kinematics-Bead Constrained in a Hoop Rotating

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SUMMARY

The discussion focuses on the analysis of a bead constrained in a rotating hoop, specifically addressing the relationship between angular acceleration and gravitational forces. The key equation derived is θ = arccos(-g/(Rw²)), where 'g' represents gravitational acceleration, 'R' is the radius of the hoop, and 'w' is the angular velocity. The initial confusion regarding dimensional consistency in the equations was clarified, leading to the correct formulation of the problem. The final solution accurately reflects the balance of forces acting on the bead.

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Darkalyan
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Homework Statement


http://docs.google.com/Doc?id=d277r7r_58d3chgqfj


Homework Equations


a=w^2*r


The Attempt at a Solution



a=w^2*r
w^2*r*cos(theta)=-mg

theta=arcos(-mg/(Rw^2))

I'm pretty sure that's right, because the vertical component of the acceleration has to equal the vertical component due to gravity. However, I wasn't sure if acceleration was simply w^2*R or if I had to do something more complicated to find it. Is my reasoning/answer correct?
 
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That's pretty close. You can see it's not quite right, though, because your second equation in your attempt at a solution part is dimensionally inconsistant. You have a force equal to an acceleration.
 
oooh. didn't catch that. thank you. so then i just add a m to the left side of the equation, which makes the m's cancel out, and the final answer is theta=arcos(-g/(Rw^2)) Is that right now?
 

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