Question

A square hoop ABCD is made of fine smooth wire. The hoop is horizontal and rotating with a constant angular velocity ω about a vertical axis through A. A small bead can slide on the wire. Which forces are acting on a bead? I know there's no friction since a wire is smooth and there is gravitational force.

Answer 1

Hmm... I think I may be able to help... Lemme think on it for a sec.

Answer 2

That would be Centripetal force I believe... because it is moving in a circular motion with a gravitational force and no friction from the smooth wire...

Answer 3

Can you write a term for centripetal force?

Answer 4

The forces that acts on bead is gravity, normal force equal in magnitude to gravity and centripetal force. I'm trying to write an expression for centripetal force. What given do you have?

Answer 5

I have given that the length of a side of the square is 2a. I wrote expression for acceleration which is \[\frac{d^2\vec r}{dt^2}=-(2 \omega \dot y+2a \omega^2) \vec i+(\ddot y-\omega^2y)\vec j\]. I will post text of a problem to see what is y and how did I get expression for acceleration. You will see that I have to show that \[\ddot y-\omega^2 y=0\] so I think that there's no active component of any force in the direction of \[\vec j\]

Answer 6

In an inertial frame \[\frac{d^2 \vec r}{dt^2}=(\frac{d^2 \vec r}{dt^2})'+2 \vec \omega \times (\frac{d \vec r}{dt})'+\vec \omega \times (\vec \omega \times \vec r)\] where ' refers to rotating frame so when I write Newton's second law I have \[m\frac{d^2 \vec r}{dt^2}=\vec F-2m \vec \omega \times (\frac{d \vec r}{dt})'-m\vec \omega \times (\vec \omega \times \vec r)\] So, now when we know this, I think that only force I have to add is gravitational. Do you agree?