1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Acceleration of the ring

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data

    A ring is moving on a vertical rod attached to a spring. Find the velocity of the ring when spring becomes horizontal. The spring is ideal with spring constant 400N/m. Mass of ring is 10kg. Natural length of spring is 4m. Initially the ring is at rest as shown in figure.(attachment)



    2. Relevant equations



    3. The attempt at a solution
    I want to solve the problem using Newtons laws and not by conservation of energy method. I have no idea on where should I start. I have made the following equation:

    Fcosθ + mg = ma ........where F is the force by the spring
    ........ θ is the angle between spring and rod
    ........ a is the acceleration of the ring
     

    Attached Files:

  2. jcsd
  3. Mar 23, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Next you need some geometry to relate theta to the length of the spring, and an equation relating the length of the spring to F.
     
  4. Mar 23, 2014 #3
    thanks for the reply Haruspex:smile:
    From hooks law:
    F=k*x ......where x is the length of spring.

    from geometry:

    cosθ=h/x

    so x=h/cosθ

    so F=k*h/cosθ
     
  5. Mar 23, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, that would be where x is the extension of the spring.
    This way is going to get messy because h and x are both variable. Let the horizontal distance be D (=4m) and write the height and spring extension in terms of D and theta.
     
  6. Mar 23, 2014 #5
    So i should write it as:

    F=k(D/sinθ -3) ..... D =horizontal distance

    so putting this value in first equation:

    ma=Fcosθ + mg

    ma=k(D/sinθ-3)cosθ + mg

    dv/dt = k/m(D/sinθ-3)cosθ + g

    [itex]\frac{dv}{dt}[/itex]=[itex]\frac{k}{m}[/itex]([itex]\frac{D}{sinθ}[/itex]-3)+g
     
    Last edited: Mar 23, 2014
  7. Mar 25, 2014 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not 3. The relaxed length is also D.
    You dropped the cos.
     
  8. Mar 25, 2014 #7
    o,made silly mistakes.


    [itex]\frac{dv}{dt}[/itex]=[itex]\frac{k}{m}[/itex]([itex]\frac{D}{sinθ}[/itex]-D)cosθ + g

    now what to do after this?Should we consider a small change in θ and integrate it over the "θ" range to find the velocity??
     
  9. Mar 25, 2014 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You need to express the vertical height as a function of theta, and thereby express dv/dt in terms of theta.
     
  10. Mar 25, 2014 #9
    sorry but i am not getting what you saying?Can you please explain a little.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted