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Nov22-05, 11:22 PM
Sci Advisor
HW Helper
P: 3,144
If you measure the angle from the vertical line connecting the center of the hoop to the lowest point on the hoop, then the radius of the "orbit" is [itex]R \sin \theta[/itex]. This makes the component of centripetal force on the bead tangential to the hoop [itex]m \omega^2 R \sin \theta \cos \theta[/itex]. The component of the gravitational force tangential to the hoop is [itex]m g\sin \theta[/itex].

Draw some pictures to verify the above and then proceed.