Sci Advisor HW Helper P: 3,149 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 $R \sin \theta$. This makes the component of centripetal force on the bead tangential to the hoop $m \omega^2 R \sin \theta \cos \theta$. The component of the gravitational force tangential to the hoop is $m g\sin \theta$. Draw some pictures to verify the above and then proceed.