Physics midterm question help !!!?

Question

A single bead can slide with negligible friction on a wire that is bent into a circular loop of radius 15 cm. The circle is always in a verticle plane and rotates steadily about its vertical diameter. The position of the bead is described by the angle θ that the radial line, from the center of the loop to the bead, makes with the vertical. (the circle is rotating to the right)
If if the period of the rotation is .450 s, at what angle(s) up from the bottom of the circle can the bead stay motionless relative to the turning circle? (b) Repeat the problem taking the period of the circle's rotation as 0/850 s.

i think that in the Y direction we have
Y direction>>>T*sin(omega)=mg
x direction >>>T*con(omega)=(m*v^2)/r
where v=(2*PI*r)/T
from this you get
tan(omega)=(g*T^2)/(4*PI^2)
ang omega is 2.88. but it seems like a small angle, Im not sure

Answer 1

The other answer posted in response to this question is wrong, except in a minor technicality that I will explain at the end. Here is the right answer:

When the bead is at an angle theta from the bottom of the loop, there are two forces acting on the bead. One is the force of gravity, which points downward with magnitude mg. The other is the normal force of the wire, N, which points toward the center of the loop.

When the bead is at its equilibrium position, it has no net vertical acceleration (ay=0). Its horizontal acceleration is the acceleration due to motion in a circle about the axis of the loop's rotation: ax=r*omega^2, but we must be careful defining r. Lets define R as the radius of the loop of wire. Then r=R*sin(theta) is the radius of the circular motion of the bead. The bead is always a distance R from the center of the loop, but the radius that matters here is the distance from the axis of rotation to the bead, r.

The way you can see this is to draw a picture of the loop from the side. Pick a point for the bead, and label the angle theta from the bottom of the loop to the bead. You can form a right triangle by drawing a line from the center of the circle to the bead of length R, then drawing a line from the bead horiontally over to the axis of rotation, then drawing a line straight up back to the center of the loop. The length of the horizontal side of the triangle is r=R*sin(theta). If you now picture the diagram from the top, you'll see this is also the radius of the bead's rotation about the axis.

Now we use Newton's second law: the sum of the forces in a given direction is equal to the mass times the net accelertion in that direction. F=ma.

In equilibrium, the vertical component of N minus the downward pull of gravity equals the vertical acceleration, which is zero:
Ny-m*g=m*ay=0
so Ny=m*g.
The horizontal component of N is equal to the centrepital "force":
Nx=m*ax=m*r*omega^2

It is also true that
Nx=N*sin(theta)
and Ny= N*cos(theta)
since these are just the horizontal and vertical components of the normal force.

If you plug these in for Nx and Ny, and plug in r=R*sin(theta). You get that
N*cos(theta)=mg
N*sin(theta)=m*R*sin(theta)*omega^2

Solve the first equation for N and plug it back into the second equation. After simplifying, you get that
cos(theta)=g/R/omega^2
so theta=arccos(g/R/omega^2).

Omega is the angular speed of the loop.
omega=2*pi/T, where T is the period.

With your numbers in part a, omega=14.0 rad/s.
theta=arccos[(9.8 m/s^2)/(0.15 m)/(14.0 /s)^2] =arccos(0.33) =70.5 degrees.

Now for a fairly pedantic point: It is true that a bead in the 6 o'clock position on a wire experiences no forces to move it away from the bottom; however, this is not a stable state. If the bead moves even a tiny bit away from exactly the bottom, the bead will be accelerated by the rotation of the loop. The bead's innertia will carry it outward and the normal force of the loop on the bead will push it upward, until it reaches the point where the centrepital acceleration is just the right amount to be supplied by the horizontal component normal force. This will be at some angle theta above the bottom, with theta between zero and 90 degrees.

Answer 2

I hope you didn't copy this question word for word from a textbook. It is very flawed.

If the bead can slide without friction, what's to get the bead off the 6 o'clock position of the circular wire? At that position omega = 0 degrees, the only force acting on the bead is its weight W = mg. As there is no non vertical force on the bead, due to lack of friction, there is nothing to push it off the 6 o'clock position and up the side of the wire. We need a horizontal component of the force on the bead for this to happen.

OK, I'm going to add friction force F = kN = k mg cos(omega) to the net forces on the bead...even though you specified it is negligible. Therefore, the bead will stabilize at some angle omega where the friction force and the weight component along the wire are equal but opposite. Thus, W sin(omega) = k mg cos(omega) Then sin(omega)/cos(omega) = tan(omega) = k mg/mg = k. Then omega = arctan(k).

This, by the way confirms my statement about needing friction for this to happen. Note what the omega is when there is no friction, that is, when k = 0.

Re the periods...they have no bearing on the problem if we assume k is a sliding friction coefficient that is fixed no matter what the rotation rate is.

Addendum: Look at your additional info, T sin(omega) = mg. If T is a period, which is in seconds according to your problem, then T sin(omega) must also be in seconds since the sine is dimentionless. So how do you rationalize T sin(omage) --> seconds, while mg --> kg m/sec^2? You can't...you can't equate seconds to force (weight).

If T is torque, you still have a units inconsistency between the RHS and the LHS of the equals sign. In this case the LHS is Newtons meter and the RHS is in Newtons. Further, to have torque, one needs a force actng over a distance r. That is, T = F*r and there we are again, we need a force other than weight acting on the bead to get it off the 6 o'clock position. And that force would be friction is a real case.

I have no clue where you're getting this question from...but it is still flawed, even with your additional information. With a frictionless interface between the wire and the bead, there is still no force to get the bead off omega = 0.

Recommend you withdraw this question, research it some more, and then resubmit it with all the necessary assumptions and givens. Also, make sure you define your variables...like is T period or torque?