# Rotational Kinematics-Bead Constrained in a Hoop Rotating

1. Sep 27, 2008

### Darkalyan

1. The problem statement, all variables and given/known data

2. Relevant equations
a=w^2*r

3. 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?

Last edited by a moderator: May 3, 2017
2. Sep 27, 2008

### ak1948

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.

3. Sep 27, 2008

### Darkalyan

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?