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!

Homework Help: A ball on a string

  1. Feb 6, 2015 #1
    1. The problem statement, all variables and given/known data
    (Im translating this from another language, so if it doesn't make any sense, please let me know and ill try to do a better job.)
    A metal ball (mass=120g) is hanging from a string (lenght=85cm). The ball is struck horizontally. When the angle of the swing reaches 125 degrees the ball starts to deviate from its circular path. What was the ball's initial velocity?
    m=0.120kg, r=0.85m, θ=125°

    2. Relevant equations
    I don't know what to put here. :l

    3. The attempt at a solution
    Here's what i came up with: Conservation of energy. Eki=Epf
    I figured the max height, the ball reaches should be h=r+r*sin(35°) ⇒0.85m + 0.85m*sin(35°)=1.3375m
    1/2 mv2=mgh
    v=√(2gh)
    v=√(2*9.81*1.3375)
    v=5.1227 m/s ≈ 5.1m/s

    There, the answer i came up with. The answer section at the end of the book says the correct answer is 5.6m/s.

    What am i missing?

    P.S. I was told, that i should mention that im studying physics without a teacher.
     
  2. jcsd
  3. Feb 6, 2015 #2

    Nathanael

    User Avatar
    Homework Helper

    It says that the ball begins to deviate from it's circular path at an angle of 125°

    But this does not mean the ball's velocity will be zero.


    Edited because I said something wrong (I said it would not be at it's highest point). Stephen's post made me realize
     
    Last edited: Feb 6, 2015
  4. Feb 6, 2015 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    If we call x the horizontal axis and y the vertical axis, when the ball deviates from a circular path it is not motionless. It's y velocity is zero but it has a non-zero x velocity. The total energy of the ball at that point is more than mgh since it has some kinetic energy.
     
  5. Feb 6, 2015 #4

    NascentOxygen

    User Avatar

    Staff: Mentor


    Have you sketched the ball's path, including and beyond that 125°?
     
  6. Feb 7, 2015 #5
    Damn, i was afraid that might be it. Dont really know what to do with that information
    Up to that 125°, but not beyond.

    So, i guess i'll try to figure out the ball's x-velocity at the top and add it to the E equation.
     
  7. Feb 7, 2015 #6
    Im thinking of a force diagram at the 125° position (gravity ( acting down ) and centripetal ( acting outward ) ) to produce a resultant force that is tangential to the path radius. (this seems to be the critical factor)
    You can figure the centripetal force required to end up with a tangential resultant, then the velocity required to produce this force.
    From that you have the final KE, add to the PE at that point, gives the total KE at the start.
    From that you can get the velocity.
     
  8. Feb 7, 2015 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're confusing centripetal with centrifugal.
    Centrifugal force "acts outwards", but it is classed as a fictitious force because it only exists in the frame of the (accelerating) object. If you take an inertial reference from there is no such force. Instead, there is centripetal force, but that is a resultant force, not an acting force.
    To rephrase your remark in the inertial frame:
    To keep the ball moving in a circle, the sum of the acting forces (gravity and tension in the string) must have a radial component equal to the centripetal force. Tension cannot supply a force outwards along the radius, so when gravity's inward radial component exceeds the centripetal force the circular path cannot be maintained.
    Your form works if you replace 'centripetal' with 'centrifugal'.
     
  9. Feb 7, 2015 #8

    rcgldr

    User Avatar
    Homework Helper

    A nit pick, that should be greater than or equal, the answer is probably looking for the point when it's equal, since that's the lower bound for when the ball no longer travels in circular path.

    lost an update to this post - that should have been: a follow up ... greater than or equal to the minimal force to keep the string in tension.
     
    Last edited: Feb 9, 2015
  10. Feb 7, 2015 #9

    NascentOxygen

    User Avatar

    Staff: Mentor

    And I thought my hint was such a good one. :oldgrumpy:
     
  11. Feb 7, 2015 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't mind nit picks so long as there's a nit to pick. I stand by the exact wording of what I wrote.:)
     
  12. Feb 9, 2015 #11

    rcgldr

    User Avatar
    Homework Helper

    My previous post lost an update (I'll refresh the page and make sure my updates take place from now on if there's a laggy response from the forum). Also nit pick was the wrong term, "follow up" would have been a better term for what I intended to post. I was pretty much stating the obvious, that the ball stops following a circular path just after the moment the tension in the string goes to zero (and the angle is greater than 90° ). When this happens, the path of the ball transitions into a parabola, and it's still moving vertically (upwards) and horizontally (towards the pivot point of the string).
     
    Last edited: Feb 9, 2015
  13. Feb 9, 2015 #12
    It was, i just hadn't had the time to act on it by the time i answered :p

    So, in the beginning, there is tension T and gravity G. ∑F=T+G=man, which is the total force acting on the ball, resulting in an inward acceleration (an). Right?
    At the point θ=125°, the string goes slack and T=0 and the only force acting on the ball is gravity. ∑F=G=ma=mg. At this point it still has some velocity v2, up and inward, which i can't figure out.
    I tried this:
    at θ=125°, G=man
    mg=mv22/r
    v2=√(gr)
    v2=2.8876 m/s

    Doing the energy calculation with this included (Eki=Epf+Ekf) i got 5.8805m/s. Then i realised G=man can't be right, G is acting straight down. I should calculate this with the x-component of T? but T=0? an isnt the total acceleration? This exercise is going to give me an aneurysm.
     
  14. Feb 9, 2015 #13

    rcgldr

    User Avatar
    Homework Helper

    Hints, the tension goes to zero when v^2 / r = g sin(35°). The change in height = .85 x (1 + sin(35°)) meters.
     
  15. Feb 9, 2015 #14
    thats not a hint, thats the solution :s
    Based on that, I did the calculations and got to 5.6m/s, but i dont know why. I need someone to explain this "v^2 / r = g sin(35°)". What is it and where does it come from?

    multiply it by g, you get an/g=sin(35)
    from what i can tell sin(35) should rather equal g/an

    multiply it by m and you get man=G sin(35), which seems to me like an x-component of G, which it doesnt have?
     
  16. Feb 9, 2015 #15

    rcgldr

    User Avatar
    Homework Helper

    Sorry for the hint / solution. You already determine that if the tension in the string is non-zero, then the ball travels in a circular path regardless of the force of gravity, and if the tension transitions to zero, then the only force acting on the ball is gravity, and the the ball's path transitions to a parabola (free fall). So the issue was determining the equation for tension versus velocity and angle. This is where I couldn't think of a hint other than to write out the equation. If I had a drawing tool, I could have created a diagram showing the forces involved.

    For general form of the equation, tension in string = m (v^2/r - g sin(θ)), where θ is the angle from horizontal. For θ = -90° (straight down), the tension is m (v^2/r - g sin(-90°)) = m(v^2/r + g). The problem is asking you to solve for tension = 0 = (v^2/r - g sin(35°)) .

    g sin(θ) is the radial component of gravity, the component of gravity in the direction of the string. You have a right triangle where the gravity vector is the hypotenuse (oriented vertical), and the magnitude of the radial component is the hypotenuse x sin(angle opposite string) . g sin(θ) is similar in concept to a box sliding on a frictionless inclined plane.
     
    Last edited: Feb 9, 2015
  17. Feb 9, 2015 #16
    If its any help, by my process, the velocity at 125° = 2.187 m/s
    (use as a check)
     
  18. Feb 9, 2015 #17
    Okay, i think this is starting to make sense now.
    This is how i get there
    Ftotal=T+G=man
    direction defined: T=man- mg sin(θ)
    set T=0=m(v2/r - g sin(θ))
    Right?

    Solve for v, insert into the energy equation, solve for the other v and done.

    So the components of G are what i missed in the first place?
    beware my mad mspaint skills
     

    Attached Files:

  19. Feb 9, 2015 #18
    i ended up with :
    v (at 125°) = sqrt ( ( r * m * g * sine 35° ) / m )
     
  20. Feb 9, 2015 #19
    Yes, dean barry, that's pretty much what i had. The m's cancel out. I already got to the correct answer, now its just a matter of making sure i understand the whys and hows.
     
  21. Feb 9, 2015 #20

    rcgldr

    User Avatar
    Homework Helper

    Your diagran showing that the radial component of g is g sin(θ) is correct.
     
  22. Feb 10, 2015 #21
    OK, see if you can figure this one out.
    (see the attached sketch)
    L = 1.0 m
    A = 30°
    g = 10 (m/s)/s
    Two identical balls (both homogenous spheres), each 10 kg, radius = 0.1 m
    Ball 1 is released from rest and rolls down the incline without slipping, has elastic collision with ball 2, ball 2 gets to A = 30 °
    where the string tension vanishes, calculate the height H.
     

    Attached Files:

  23. Feb 10, 2015 #22
    H=1.750m (1.8m, rounded to 2 significant digits) ? My calculation assumes that all the potential energy of ball1 is transfered to ball2, at least i think it does :f
     
    Last edited: Feb 10, 2015
  24. Feb 12, 2015 #23
    Its not, The PE is shared (not equally) between linear KE and rotational KE as it rolls down to the contact point.
     
  25. Feb 12, 2015 #24
    Alright. H=3.9179m ≈ 3.9m, then?
     
  26. Feb 12, 2015 #25
    Only the linear KE of ball 1 is transferred to ball 2 at impact, but during the fall, 2/7 of the energy translated from the PE gets tied up as rotational energy, leaving 5/7 translated into linear KE.
    (typical KE sharing for a rolling, non slipping, homogenous sphere)
    Whatever velocity you figured for ball 2 at the impact, was also the impact velocity of ball 1 as it was an elastic collision of two equal masses.
    So in summary:
    ball 1 rolls 2.45 vertically gathering 5/7 of the PE as linear KE, the impact leaves ball 1 spinning but stationary, ball 2 sets off at the same linear speed as ball 1 and rotates 120 ° to slack point.
    i got 2.45 m as H
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted