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!

Rolling conceptual question

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    mta6fb.jpg


    2. Relevant equations



    3. The attempt at a solution

    My answer is a because for ball B it loses kinetic energy when it drops into the dip and comes back up.

    I'm having a hard time picturing ball b when it goes down the dip.. does it drop down and then jumps up back on the track? or unless it is stuck onto the track the whole time and slows down when it climbs back up onto the track after the dip?
     
  2. jcsd
  3. Apr 30, 2013 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    How? Where does the KE go?
     
  4. Apr 30, 2013 #3
    I was suggested by one of my tutors that it loses kinetic energy due from impact?

    Ok i'm thinking if conservative forces are the only forces present here.. that should mean that both balls are accelerating at the same rate and will both have the same final velocity at the bottom of the ramp, however my theory is that ball A will not continue accelerating but ball B will when it reaches the dip so it will have a greater velocity and will have taken less time?
     
  5. May 1, 2013 #4

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Humm Ok loosing KE by impact is possible. I had ignored the sharp edges to the dip in B.

    It's possible the ball will leave the track as it enters the dip (think fast car going over a cliff), then it hits the bottom of the dip somewhere and bounces. The right hand side of the dip appears near vertical so the ball could/will become trapped in the dip (if it rolls up the vertical right hand edge it has no horizontal velocity)... but in that case the answer is rather easy. Too easy?

    If the ball makes it to the end of the track then.. Take a look at the change in PE. It's the same for A and B. If no KE were lost bouncing in the dip then they would both arrive at the end with the same KE (eg same velocity). If some KE is lost by that means then Vb < Va. There is only one answer with that option.
     
    Last edited: May 1, 2013
  6. May 1, 2013 #5
    Hmm, what other cases could we explore other than the one discussed? that the ball stays on the track when it enters the dip and accelerates and gains velocity but still loses velocity on its way up to the vertical right hand edge? I think in any case ball B does lose kinetic energy correct?
     
  7. May 1, 2013 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    There is not enough information to answer the question. Yes, the diagram shows sharp edges, but it's not generally a good idea to trust that. It doesn't look like an arc of a circle, yet the 'r < R' suggests it is supposed to be. It looks like half an ellipse, but if the ends of the dip are truly vertical there's an excellent chance it will never make it past the dip.
    If there is a sudden change in slope, as it would appear there is, then certainly some KE is lost. But it is not possible to say on that basis alone which will take the shorter time. A dip of the right shape (think brachistochrone) would allow the ball to take less time, and the loss in post-dip speed might not make up for that. Even if the dip's edges are smooth enough to avoid energy loss, the time comparison could not be resolved without further details.

    I assume this is the same teacher who set the shoddy question about the cylinder, the taut string and the dropped mass.
     
  8. May 1, 2013 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    If you assume the ball never leaves the surface and always rolls without slipping, then I think there is a definite answer to the question.
     
  9. May 1, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I disagree. A small dip can mean it arrives sooner; a larger dip, later.
     
  10. May 1, 2013 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    As the ball is rolling down the dip, what is the direction of the acceleration of the center of the ball? Does that acceleration have a horizontal component?
     
  11. May 1, 2013 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It must do, since its horizontal velocity will be greatest at the bottom of the dip. You're aware of the brachistochrone, right?
     
  12. May 2, 2013 #11

    TSny

    User Avatar
    Homework Helper
    Gold Member

    If the dip has the shape stated in the problem, then as the ball rolls down from point A to point B, the center of the ball will have a tangential component of acceleration and a centripetal component. See attached diagram. It seems to me that the net acceleration must have a horizontal component to the right between A and B. Thus the ball gains in horizontal speed between A and B and therefore arrives at B before the other ball arrives at B'.
     

    Attached Files:

  13. May 2, 2013 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It's very difficult to be sure what the exact shape is supposed to be. If it is a semicircle or semiellipse, as it appears, then the ball must become airborne. Even if it makes it up the other side it will be rising vertically, so cannot make it past the dip. Your argument will work if we assume (a) that the entry to the dip is sufficiently smooth that rolling contact is maintained and (b) that the profile is symmetric.
     
  14. May 2, 2013 #13

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Suppose we assume..

    a) the track has smooth corners so the ball stays on the track (eg like a roller coaster)
    b) track lengths are the same
    d) we ignore friction and drag
    c) we can apply conservation of energy.

    then..

    The PE at the start is converted to KE at the end. Since the two tracks start and end at the same height the change in PE is the same so the final KE must be the same. So I think the velocity must end up the same.

    The time taken can be different because the average velocity is different. The averge velocity is higher in track b. Specifically during the dip. It's the same as a) at all other points.

    That gives answer d.
     
  15. May 2, 2013 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    As TSny points out, you need that the average horizontal velocity is higher. That isn't quite so clear to me, but it's probably true.
     
  16. May 2, 2013 #15

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I agree. The sharp corners ruin the problem.
     
  17. May 2, 2013 #16
    I think we should just ignore the sharp edges since it is given that the balls roll without slipping. This also means that the balls don't go airborne. Ignoring air drag is almost sure for most problems. Since the track lengths are the same and they are inclined at equal angles, by conservation of energy vA=vB.

    The awesome logic that "net acceleration must have a horizontal component to the right between A and B. Thus the ball gains in horizontal speed between A and B and therefore arrives at B before the other ball arrives at B'. Has already been given by TSny.

    Note that during the whole motion vB≥vA. Which I think was the point of the problem.
     
    Last edited: May 2, 2013
  18. May 2, 2013 #17

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I agree that's probably the right answer, given the adjustments mentioned to the statement of the problem. But the horizontal velocity argument is not quite as straightforward as it may seem. On ascending, the horizontal acceleration is negative. In fact, if you ignore the section of path beyond the dip, so the ball only has to reach the far side of the dip without becoming airborne, then a very steep ascent can make the B route take longer.
    The argument can be completed by requiring that the ball resumes horizontal travel without becoming airborne. Then, by a time reversal argument, for any given horizontal interval (x, x+dx) in the dip, whether on ascent or descent, route B is faster.
     
  19. May 3, 2013 #18
    If we ignore the possibility that the ball might go airborne and take the other assumptions, the solution as I understand it is-

    "When the ball rolls in the dip there is a component of acceleration tangential to the curve acting on the ball. This component increases the velocity of the ball till the lowest point in the dip and then decreases it till it retains its original value. This means that for all points in the dip velocity of the ball is greater than the initial velocity of the ball.
    So the average speed of the ball increases.

    Time taken=Average Speed of the ball/Total path length

    Since the path lengths are given to be same, time taken in B is less that in A."

    Another argument could be that since its a dip the potential energy of the ball is converted into kinetic energy. So the speed of the ball is always more inside the dip. In this argument the shape of the dip can be anything provided that it lies below the track, and doesn't result in the ball going airborne or slipping. Am I overlooking something?
     
  20. May 3, 2013 #19

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes - you're overlooking that although the speed is higher that does not prove the horizontal component is higher. The same flaw applies to the first argument, as it stands, in relation to the ascent. As I said, it is possible to construct a dip profile in which:
    - it does not go airborne before reaching the far side of the dip
    - it takes longer through the dip than flat
    But, by requiring it resumes horizontal travel without becoming airborne it is possible to complete the argument by considering time reversal of the ascent, turning it into a descent.
     
  21. May 4, 2013 #20

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    I'm not convinced that's necessary because they appear to have arranged for the path length to be the same in A and B. The dip makes that part of B longer so they reduced the length of horizontal track after the dip to compensate. All that matters is the velocity along the track because the track length is the same.
     
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: Rolling conceptual question
  1. Conceptual question (Replies: 3)

Loading...