1. Limited time only! Sign up for a free 30min personal 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!

Vertical Circle Motion

  1. Sep 10, 2014 #1
    1. The problem statement, all variables and given/known data Suppose a non-uniform circular motion where a particle of mass "m" is attached to a string, which rotates on a vertical plane. Once an initial velocity is provided to the particle at the lowest point of the trajectory, no further forces act on the particle. (Air drag is negligible)
    1. Which is the minimum velocity that the particle requires to reach the highest point of the trajectory?
    2. If initial velocity is two-times the one we calculated in question 1, what would be the velocity in the highest point? Calculate the acceleration too:
    3. Calculate the maximum height the particle can reach if the velocity is half the one calculated in 1.
    4. Calculate the initial velocity if the particle rotates only 2[itex]\Pi[/itex]/3 radians:




    2. Relevant equations
    [itex]1/2[/itex]mv^2bottom=[itex]1/2[/itex]mv^2top+mg2r
    Centripetal acceleration=(v^2)/(r)
    v=ωr


    3. The attempt at a solution
    For no.1 , given that KE is fully transformed into potential energy at the highest point of the circle,KE is 0 in the right side of the equation so: v=2√(gr)

    For number 2, if velocity is 4√(gr) then [itex]1/2[/itex]m(4√(gr))^2=[itex]1/2[/itex]mvtop^2+mg2r →vtop=2√(3gr)

    So I tried to calculate the centripetal acceleration and I got an=(2√(3gr)/r
    But know I don't know how to calculate the tangential acceleration with only the values I know and now I'm stuck on this problem. Pleasse help me!
     
    Last edited: Sep 10, 2014
  2. jcsd
  3. Sep 10, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    If all KE converted to PE, how is it still moving in a circle?
     
  4. Sep 10, 2014 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The speed is minimum at the top of the circle. How is tangential acceleration related to speed?


    ehild
     
  5. Sep 10, 2014 #4
    Well question no. 1 asks for the minimum velocity required to get to the top of the circle, so the particle does not necessarily move in a cirlcle after that point.
     
  6. Sep 10, 2014 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It will not even reach that point along the circle as the string becomes slack before.

    ehild
     
    Last edited: Sep 10, 2014
  7. Sep 10, 2014 #6
    If the velocity vector(tangent to the trajectory) is antiparallel to the tangential acceleration vector, speed decreases.
    But that doesn't answer my question :(
    Sorry but this is my first post and I don't know how to use latex
     
  8. Sep 10, 2014 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Well, an other hint: What is the direction of the resultant force at the top of the circle? And how are acceleration and force related?

    ehild
     
  9. Sep 10, 2014 #8
    Ohh I get it the resultant force points towards the center of the circle, therefore there's no tangencial acceleration right?
     
  10. Sep 10, 2014 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is right!:smile:

    ehild
     
  11. Sep 11, 2014 #10
    Hi, I have another question, in another section of my homework it asks me to calculate the initial velocity if the object traveled to 2pi/3 radians, but how can I do it if I don't know any angular velocity nor acceleration?
     
  12. Sep 11, 2014 #11

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Apply conservation of energy again.

    ehild
     
  13. Sep 11, 2014 #12
    Can I use polar coordinates? so y=rsen120 when the angle is 2pi/3 radians, and so the height at that point would be r+rsen120?
     
  14. Sep 11, 2014 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, the height is R(1+sin(120°)).

    Edit: Oops. It is R(1-cos(120°))

    ehild
     
    Last edited: Sep 11, 2014
  15. Sep 11, 2014 #14
    And considering the circle rotates counterclockwise and at 2∏/3 radians tangential acceleration exists again(parallel to velocity, so it accelerates) , is it really possible for the object to only rotate to 2∏/3?
     
  16. Sep 11, 2014 #15

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The particle moves along the circle. See picture. The height from the lowest point is h=R(1-cos(θ)) .
    It does not matter if theta is measured clockwise or anti-clockwise as cosine is even function.

    ehild
     

    Attached Files:

  17. Sep 11, 2014 #16
    Oh yes sorry my mistake we have to use cos indeed. Another question if you don't mind, will this initial velocity depend on the tension of the string?
     
  18. Sep 11, 2014 #17

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The initial velocity is given. The tension of the string depends on the speed. It has to provide the necessary centripetal force together with the radial component of the weight.

    ehild
     
  19. Sep 11, 2014 #18
    What I don't get is how is it possible to calculate the initial velocity using conservation of energy.
    Because if height=(R(1-costheta)) , is it correct to use the formula:1/2m(velocityatbottom)^2=mg(R(1-costheta))? or am I doing something wrong?
    I don't get how can kinetic energy be 0, if the particle keeps moving along the cirlce after that point
     
  20. Sep 11, 2014 #19

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The problem text said that

    So the particle does not go around its circular path after covering 2pi/3 radians.

    If he particle is fixed to a rigid rotating stick that means the particle stops at that angle, then starts to move backwards.

    If it is connected to a string, it still can have speed after reaching that angle, but the radial component of the weight is not enough for the centripetal force, the string gets slack, the motion will not be circular any more.
    I think the problem maker expect the solution according to the first case.

    ehild
     
  21. Sep 11, 2014 #20
    How are newton's 1st and 3rd laws applied to the forces acting on the particle?
    this is the last question of my homework, and I'm having problem answering it correctly in a few lines. Do I have to consider inertial and non inertial frames separately?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Vertical Circle Motion
  1. Vertical circle motion (Replies: 6)

Loading...