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

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.
  • #1
cheater1
34
0

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



a_rad= v^2/R = 4(pi)^2R/T^2
circumference= 2(pi)R

The Attempt at a Solution



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

a_rad= 4pi^2(0.1)/.25= 15.79 m/s^2
 
Last edited by a moderator:
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.

so that means the a_rad is pointing right from the bead?
 
  • #6
cheater1 said:
so that means the a_rad is pointing right from the bead?
Yes.
 
  • #7
PhanthomJay said:
Yes.

Just to check is this right?
a_rad= 4pi^2(0.1)/.25= 15.79 m/s^2
 
  • #8
cheater1 said:
Just to check is this right?
a_rad= 4pi^2(0.1)/.25= 15.79 m/s^2
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
 
  • #9
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.

a_radial = w^2(r)
where you get this?
 
  • #10
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.
where you get (a_radial =w^2r)?
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)
 
  • #11
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:
  • #12
No, w= rev/sec= 8(pi)radians/sec, then v=wr = 8(pi)r/ sec, where r = RsinB
 
  • #13
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:
  • #14
That's the biggest thanks I ever got,
You're Welcome!

Did you get parts b and c?
 
  • #15
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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1. What is a radial acceleration problem?

A radial acceleration problem is a physics problem that involves calculating the acceleration of an object moving in a circular path, also known as radial motion. This type of problem is common in fields such as astronomy, where objects orbit around a central mass, or in mechanics, where objects move along a curved path.

2. How do you calculate radial acceleration?

The formula for radial acceleration is a = v^2/r, where a is the acceleration, v is the velocity, and r is the radius of the circular path. This formula can be derived from Newton's second law of motion, F = ma, where the force (F) is equal to the mass (m) multiplied by the acceleration (a).

3. What is the difference between radial acceleration and tangential acceleration?

Radial acceleration is the acceleration of an object moving in a circular path, while tangential acceleration is the acceleration of an object moving in a straight line. In other words, radial acceleration is directed towards the center of the circle, while tangential acceleration is directed tangent to the circle.

4. How can I apply the concept of radial acceleration in real life?

Radial acceleration is a fundamental concept in many fields of science, including astronomy, mechanics, and engineering. It is used to understand the motion of objects in circular paths, such as planets orbiting around a star, cars navigating a curved road, or roller coasters moving along a track. It also has practical applications in designing and analyzing circular motion devices, such as centrifuges or particle accelerators.

5. What are some common mistakes when solving a radial acceleration problem?

One common mistake is confusing radial acceleration with tangential acceleration, as they have different formulas and directions. Another mistake is not accounting for the radius of the circular path, which is a crucial component in the formula for radial acceleration. It is also important to use consistent units and pay attention to the directions of vectors, as they can affect the final answer. Additionally, not considering factors such as friction or air resistance can lead to inaccurate solutions.

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