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Centripital Force and Springs Combined

  1. Nov 13, 2014 #1
    image.jpg
    1. The problem statement, all variables and given/known data
    (Attached is an image depicting the situation)

    A ball of mass m = 1.50kg is hung from a spring attached to a shaft. The length of the spring with the mass hanging from it is Lo = 50.0cm. The shaft then starts to rotate such that the spring stretches to a length L = 65.0cm. The spring constant of the spring is k=1450N/m.

    A) At what angle theta does the rotating spring make with the vertical
    B) How fast is the ball moving
    C) How much work had to be done on the shaft to get the ball moving at the angle and speed given in parts a and b

    2. Relevant equations
    Springs : Fs = -kx
    Centripital Force : Fc = mv²/r
    3. The attempt at a solution

    First thing I did was solve for the length of the spring at equilibrium (assuming the spring is massless). First I solve for the change in distance due to the mass attached.

    mg = kx
    mg/k = x = (1.50kg)(9.81m/s²)/(1450N/m)
    x = 1.015cm

    Length at Equilibrium
    50.0cm - 1.105cm = 48.99cm

    Now that I know the equilibrium length I will find the total force being applied to the spring since the length is known.

    Change in length from equilibrium

    65.0cm - 48.99cm = 16.01cm = .1601m

    The total force acting on this spring is

    F = kx = (1450N/m)(.1601m) = 232.1N

    I don't even know if I'm on the right track with this. I have a free body diagram that I will post, but again I have no idea if I'm on the right track.

    If I've made any mistakes please let me know.
     
    Last edited by a moderator: Nov 14, 2014
  2. jcsd
  3. Nov 13, 2014 #2
    I don't know how to attach the file of my free body diagram
     
  4. Nov 14, 2014 #3
    So far you are on the right track.

    Chet
     
  5. Nov 14, 2014 #4
    Hello Chestermiller. In order to solve for angle theta Atleast one more component of the force must be known. Weight is straight down and I'm guessing the centripital force must make up rest of the net force. Isn't centripetal force is purely horizontal?
     
  6. Nov 14, 2014 #5
    Please write out your force balances on the mass in the x and y directions.

    Chet
     
  7. Nov 14, 2014 #6
    image.jpg
    Here is my attempt at solving for the centripetal force. I don't think its correct but it's all I know how to do
     
  8. Nov 14, 2014 #7
    How come your force balance equations don't have the components of the spring force acting on the mass in the horizontal and vertical directions?

    Chet
     
  9. Nov 14, 2014 #8
    I thought that the spring force is the net force? That's how I was able to find the net force. Please clarify
     
  10. Nov 14, 2014 #9

    BvU

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    It would be wise to see Fc as a resultant force, i.e. the net force. It is the sum of the two forces that work on the mass: one from gravity, the other from the spring.
    The resultant force is what makes the mass follow a circular trajectory.
    You calculate its magnitude (you need that for B.), but in A) the exercise asks for an angle...
     
  11. Nov 14, 2014 #10
    The spring force is not the net force. The net force is the vector sum of the spring force and the weight. Now please, write out the force balances in the horizontal and vertical directions in terms of the horizontal and vertical components of the spring force, and the weight of the mass.

    Chet
     
    Last edited: Nov 14, 2014
  12. Nov 14, 2014 #11
    Sounds good. Only there's one problem. I don't know the direction of the spring force, only its magnitude.

    What I know :
    Magnitude and direction of the weight
    Magnitude of the spring force
    Don't know:
    Direction of spring force
    Magnitude and direction of resultant centripital force.

    What can we do?
    Is there a trick?
     
  13. Nov 14, 2014 #12
    Wait hold on I think I got something
     
  14. Nov 14, 2014 #13
    No, there's no trick. You express the components of the spring force in terms of sine and cosine of theta. Then you write the force balances. Then you solve for theta.

    Chet
     
  15. Nov 14, 2014 #14
    Chet[/QUOTE]
    Something else needs to be known.
    The sum of forces in the y direction must be 0 so that we can have an equation.

    (Fs means force of spring)
    ΣFy = Fscosθ - mg
    ΣFx =Fssinθ

    If the sum of forces in the y direction = 0 then we have our equation. Otherwise i don't kneo
     
  16. Nov 14, 2014 #15
    Well, then you have your equation for the y direction. That gives you theta.

    What is your force balance equation for the x direction?

    Chet
     
  17. Nov 14, 2014 #16
  18. Nov 14, 2014 #17
  19. Nov 14, 2014 #18
    First attachment is solving for the angle which I found to be 86.4 degrees. (It really doesn't look that way in the picture I hope I didn't do it wrong). Second attachment is the sum of forces in direction x which is also the magnitude of the centripital force
     
  20. Nov 16, 2014 #19

    BvU

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    Looks good. Now on to b) and c) ?
     
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