Question

A bead lies on a frictionless hoop of radius R that rotates around a vertical diameter with constant angular frequency W. what should W be so that the bead maintains the same position on the hoop at angle $\theta$ with respect to the verticle? what is the special value W and why is it special?

Answer 1

$\omega =\sqrt{\frac{g}{Rcos\theta }}$

The horizontal component of the normal force provided by the ring on the bead keeps it moving in its circular path. It is directed perpendicular to the tangential velocity of the bead. The vertical component of the normal force has to equal the weight of the bead.

N is normal surface

m is mass

g is acceleration due to gravity

$N_v=mg$

$N_{cos\theta }=mg$

$N=\frac{mg}{cos\theta }$

$N_H=Nsin\theta$

$N_H=\frac{mg}{cos\theta }sin\theta$

Now we can begin working with the centripetal force equation

$F_{c}iscentripitalforce$

$\omega isrotationalvelocity$

$RisradiusofcircularpathRsin\theta$

$\sum F_c=m{\omega }^{2}R$

$\frac{mg}{cos\theta }sin=m{\omega }^{2}Rsin\theta$

$\omega =\sqrt{\frac{g}{cos\theta }}$