# Homework Help: Rotational Physics - Bead on spinning ring

1. Oct 11, 2014

### AgentRedfield

I think I'm not understanding some conceptual part of rotational kinematics because all the questions seem connected. I want to figure it out as best I can so please don't solve it but any hints in the right direction would be really appreciated, thanks!

The Question:
A stiff piece of wire is bent into a circle and mounted (vertically) to rotate as shown. A wooden bead with a hole through its center can slide frictionlessly on that wire. The radius of the wire hoop is 15.0 cm and it rotates steadily with a period of 0.450 s
.

a) At what angle θ will the bead rotate, along with the hoop, without sliding (up or down)? Show that there are two possible angles (solutions) for this hoop size and rotational speed. Include a clear free body diagram of the bead as part of your solution.
b) At what period would the ball not leave the bottom position (θ =0)?
c) At what period would θ =90 degrees? What would that angular speed then be? Explain, in terms of forces on the bead, why one can’t spin the hoop fast enough for the bead to reach θ =90 degrees.
d) If the hoop were twice the radius, what would be the non-zero angle (at which the bead would rotate)?

Relevant Equations:
Fg=mg
a=mv2/r
v=wr
w=(2*pi)/T

Attempt at solving:

a) If the bead is at θ then there is a normal force directed towards the center of the circle and the force of gravity directed downward. Sum of forces in the 'y' gives Ncosθ-Fg=0 and in 'x' is Nsinθ=ma
I think the acceleration is rw2 so lots of substitution gives θ=tan-1 (rw2 /g)
θ=71.47 degrees
This answer seems reasonable but I'm not sure if I skipped something. Also I don't know how to find the second angle.

b) I think Fg=N but if θ is zero it seems like acceleration would be zero so period would be zero which makes no sense.

c) Normal force would be zero and Fg=mv2/r
so g=rw2 and w=2pi/T, after substitution I got:

T=0.777 s

I think I'm missing the conceptual part of c because my answer seems like normal numbers.

d) I imagine this would be solved the same as part a but the question seems to imply it would only have one angle it rotates at, what changed?

Thanks again for your time and help!

2. Oct 11, 2014

### TSny

What is the meaning of r here?

3. Oct 11, 2014

### AgentRedfield

edit: Would it be r*sin(theta), where r is 0.15 m?

Last edited: Oct 11, 2014
4. Oct 11, 2014

### TSny

Note that the $r$ goes back to using $v^2/r$ for the acceleration. So, you just need to consider the meaning of $r$ as used in $v^2/r$. Then you can decide with confidence if $r = (.15 \rm{m}) \sin \theta$.

5. Oct 11, 2014

### AgentRedfield

Since I'm saying acceleration is only in the x-direction then the $r$ should be $Rsinθ$ because that is the x-component of $r$.

My Work:
y:
$N \cos \theta = mg$
$N=\frac{mg}{\cos \theta}$
x:
$N \sin \theta = \frac{mv^2}{r} = \frac{mv^2}{R\sin \theta}$
$N = \frac{mv^2}{R}$
$\frac{mg}{\cos \theta} = \frac{mv^2}{R}$
$\cos \theta = \frac{gR}{(\frac{2 \Pi R}{T})^2}$

$\theta = 70.42^{\circ}$

This is really close to my earlier answer. The logic of the steps seem right though. I'm also not sure how to get the second angle it would stay at.

I believe your answer for $\theta$ is correct. I'm not quite sure of the interpretation of part (a) in regard to finding the second angle. Are negative values of $\theta$ allowed? Do they only want stationary positions of the bead that are stable with respect to small disturbances away from the stationary positions?