## Homework Statement

[PLAIN]http://img508.imageshack.us/img508/6861/21296472.jpg [Broken]

I tried to get a_rad and if i didn't make any mistake, then a_rad = 15.79 m/s. I need to find angle when the bean is at vertical equilibrium.

## Homework Equations

circumference= 2(pi)R

## The Attempt at a Solution

v= 2(pi)R/0.25 sec = .2pi/0.25= .89m/sec

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PhanthomJay
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This problem is a bit more complex than that. You have to look at the forces acting on the bead at angle beta, its weight and the normal force, where the net force in the vert direction is 0, and the net force in the horiz direction is the horiz component of the normal force, as caused by the centripetal acceleration, w^2(R), where R is not the radius of the circle.

This problem is a bit more complex than that. You have to look at the forces acting on the bead at angle beta
Will this be the direction of a_rad?

PhanthomJay
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Will this be the direction of a_rad?
NO, the hoop is not rotating about its center point, it is rotating about the vertical axis, that is, out of and into the plane, as if you were spinning a top.

NO, the hoop is not rotating about its center point, it is rotating about the vertical axis, that is, out of and into the plane, as if you were spinning a top.

PhanthomJay
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Yes.

Yes.
Just to check is this right?

PhanthomJay
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Just to check is this right?
No. Firstly, r is not 0.1; as you noted before, the radial acceleration points to the right from the bead, so that is the value of "r" to use, the distance from the bead to the vertical axis in the horizontal direction. Also, it is better to use the angular speed instead of the out of plane tangential speed, that is , a_radial = w^2(r), where r = R*(____??___), and w is 4 rev/s, or w = 8 pi rad/s

No. Firstly, r is not 0.1; as you noted before, the radial acceleration points to the right from the bead, so that is the value of "r" to use, the distance from the bead to the vertical axis in the horizontal direction. Also, it is better to use the angular speed instead of the out of plane tangential speed, that is , a_radial = w^2(r), where r = R*(____??___), and w is 4 rev/s, or w = 8 pi rad/s
Ok, I got r=(sinB)R , since the hoop is spinning, the bead will be going up. As it goes up r will will be changing with respect to angle.

where you get this?

PhanthomJay
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Ok, I got r=(sinB)R , since the hoop is spinning, the bead will be going up. As it goes up r will will be changing with respect to angle.
Yes, if the rotation speed was increased or decreased, then r would change with respect to the angle.
Since a_radial= v^2/r, and v=wr, then a_radial = w^r^2/r = w^2(r) (where w = omega, the angular speed in rad/sec)

Since a_radial= v^2/r, and v=wr, then a_radial = w^r^2/r = w^2(r) (where w = omega, the angular speed in rad/sec)
This is confusing me. v= wr, shouldn't speed be just 4rev/s ?

edit: nvm the v= rev/sec , 1 rev= 2(pi)R, and therfore v = 4(2piR) / sec = 8(pi)R / sec

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PhanthomJay
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No, w= rev/sec= 8(pi)radians/sec, then v=wr = 8(pi)r/ sec, where r = RsinB

No, w= rev/sec= 8(pi)radians/sec, then v=wr = 8(pi)r/ sec, where r = RsinB
I see it.

Edit: Got it!

angle = 81.07 if I did this right.

[PLAIN]http://img176.imageshack.us/img176/2347/32500903.jpg [Broken]

Picture for who any1 wants to see what I'm doing.

r=(sinB)R , R = 0.1m

T=4rev/sec

V= 4(1 rev)/sec = 4(2(pi)r)/sec

a_rad= v^2/r , (8pi*r)^2/r = (8pi)^2 * r

sinB= Fx / x rearrange to get Fx, Fx= xsinB

cosB= (mg)/x rearrange to get x, x= mg/cosB

Pluged in what I know.

(8pi)^2rm = sinB(mg)/cosB

(8pi)^2r / g = sinB/cosB m cancle out

(8pi)^2(sinBR) /g = sinB/cosB replace r, r = r=(sinB)R

(8pi)^2R / g= 1/cosB

B= cos^-1 [ g/ (8pi)^2R ] = 81.07

Thanks a lot for you help PhanthomJay.

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PhanthomJay
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That's the biggest thanks I ever got,
You're Welcome!

Did you get parts b and c?

That's the biggest thanks I ever got,
You're Welcome!

Did you get parts b and c?
You really did help me out a lot.

Yup finished b and c.