Minimum Angular Velocity

Question

A bead is free to slide on a vertical circular frame of radius $R$ comes to equilibrium when $cosθ = g/Rω²$.

The minimum value of angular velocity comes out to be $sqrt{g/R}$, which we can find out by balancing  Gravitational and centripetal force with Normal reaction to bead from the frame.

Why can't the angular velocity have values between 0 and $sqrt{g/R}$?

BioPhysicist · Answer

The angular velocity can definitely have values in that range. It's just that you then lose that equilibrium position you reference.

If $omega<sqrt{g/R}$ then you only have two equilibrium positions. One at $theta=0$ (the bottom of the ring) and the other at $theta=pi$ (the top of the ring).

If $omegageqsqrt{g/R}$ then you gain a third equilibrium that you mention where the relation holds of $costheta=gR/omega$

R.W. Bird · Answer

If your formula was correct, then the maximum $cosθ = 1 = g/(Rω^2)$. However, it is not correct.  Assuming $θ$ is measured up from the (downward) vertical to the radius going to the bead, then:  $N(cosθ) -mg =0$  and $N(sinθ) = m(Rω^2)$.
Solve the first equation for $N$ and put it into the second.  Then: $(mg/cosθ)sinθ = m(Rω^2)$  and $(cosθ)/( sinθ) = g/(Rω^2)$.  The angular velocity is not limited.

Minimum Angular Velocity

2 Answers

Add your own answers!

Ask a Question