What Is the Angle When the Bean Is at Vertical Equilibrium?

• cheater1
In summary, the conversation discusses a problem involving finding the angle of a bead on a rotating hoop when it is at vertical equilibrium. The solution involves analyzing the forces acting on the bead and using equations such as a_rad = w^2(r) and r = R(sinB) to find the angle. The final answer is 81.07 degrees.
cheater1

Homework Statement

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

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

Last edited by a moderator:
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.

PhanthomJay said:
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?

cheater1 said:
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.

PhanthomJay said:
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.

cheater1 said:
Yes.

PhanthomJay said:
Yes.

Just to check is this right?

cheater1 said:
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

PhanthomJay said:
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?

cheater1 said:
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)

PhanthomJay said:
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 therefore v = 4(2piR) / sec = 8(pi)R / sec

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

PhanthomJay said:
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

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.

Last edited by a moderator:
That's the biggest thanks I ever got,
You're Welcome!

Did you get parts b and c?

PhanthomJay said:
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.

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