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!

Box and friction on a merry-go-round

  1. May 5, 2017 #1
    1. The problem statement, all variables and given/known data
    A box of mass m is set on a merry-go-round of radius r with a linear speed v. The friction coefficient between surfaces is μ.
    2. Relevant equations
    ac = v²/r
    3. The attempt at a solution
    I imagined this problem because I'm trying to understand more deeply circular motions. I have many ideas (more questions maybe, tell me cause I'm probably wrong) such as : the merry-go-round is trying to set the box in motion, so the friction force is pushing the box in the opposite direction. The problem is I don't understand the action of centripetal acceleration in this case even if I place myself in the turning frame. I know the circular motion is a state that requires an acceleration perpendicular to the speed in order to make only change its direction but I don't really understand what's going on here.
     
  2. jcsd
  3. May 5, 2017 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello Baptiste, :welcome:
    I see that there is no question in the problem statement :smile:

    To help you understand what is going on here: According to Newton we need a force to make an object follow a circular trajectory. If you place yourself on the merry-go-round you need to hold on to something or else you will move away from the axis of rotation and ultimately fall off. The box can only depend on friction for an inward force and friction is maximized to ##\mu##N with N the normal force ##mg##. So if ##\omega^2 r## (or ##v^2/r## ) exceeds ##\mu g## it will start sliding outwards.
     
  4. May 5, 2017 #3

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    The box always has linear momentum in a direction tangent to the circle (perpendicular to the radius line from the center to the box).
    The frictional force is in the direction pointing toward the center.
    You may find this video interesting

    It is not friction, but there is a force acting on the rotating object, and they do a nice working out with the vectors.
     
  5. May 5, 2017 #4
    Hi, thank you very much for your welcome and very quick answers !

    I understood BvU; friction force must play the role of a string that pulls the mass toward the center at every moment. I'm currently reading a chapter about non-inertial frames. I understand these principles in the case of a linear speed (the example given was a train on a straight way). The fictitious force that the observer can see in the train accelerating is actually opposite to the real acceleration. If I keep this idea in the case of the merry-go-round the fictitious acceleration the person could feel must go outward (as you said). But I don't have any intuition about the way to explain why the box is going outward from an inertial frame outside the merry-go-round. I see that my question is pretty stupid but I don't feel it.

    Thanks for you video scottdave it was incredible, really another way to explain circular motions.
     
  6. May 5, 2017 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Very good indeed, because in the fairground frame the box is not going 'outward' at all !

    In the 'fairground frame' (the outside world):
    If the friction is suddenly 'switched off', then -- completely in accordance with Newton -- the box will continue with the speed it has at that moment along a straight line -- a tangent to the circle it has followed until then

    It is going outward only in the sense that the total distance to the axis of rotation increases. The perpendicular distance stays the same (figure 3 in the link, but I suspect there will be nice movies to be found as well with a little googling).
     
  7. May 5, 2017 #6

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    As soon as it starts slipping, it would 'want' to go in a straight line, perpendicular to the axis. But friction has not disappeared, just become less (kinetic friction is always less than static friction), so there would still be a certain amount of frictional force, which is going to be opposite direction from the relative sliding motion of the 2 surfaces. An observer on the merry go round, probably would observe it to move backwards and outward, until it leaves the platform.
     
  8. May 5, 2017 #7
    Hi again, I found a fantastic video explaining what happens from both frames : . But there's still something I don't understand. At 2 min 22 he's mentioning the tension in the string acting as two forces : one vertically balancing the weight and another horizontally that plays the role of the centripetal force. The only important force here is acting inward, but why does the ball move away from its vertical ? I think it has something in common with inertia but I can't go further. I also watched some experiments showing the angle between the vertical and the string increases when the rotation speed increases too. That's what you mentioned when you said that the distance to the axis was increasing (but I didn't understand what "perpendicular distance" means".

    Many many thanks.
     
  9. May 5, 2017 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    1. String forces the ball to follow a circular trajectory. So it has to pull towards the axis of rotation.
    2. String also disallows the ball to fall down to earth. So it has to pull upwards.
    3. Strings can only exercise forces on balls (or anything else) in a direction along the string.
    Two forces in different directions add up as vectors -- thus determining the direction of the string.
    You don't need a merry-go-round for these experiments: a small weight on a string you hold in your hand shows the same phenomena.

    Number 3 is very important to be aware of.
    ------

    Perpendicular distance is the distance from the axis of rotation to the straight line of the trajectory the box follows after the string is broken: the tangent to the circle it was describing before. Top view:
    upload_2017-5-5_23-21-15.png

    R stays the same, even when the box flies further away from the axis of rotation in the direction it had at the moment the line snapped
     
  10. May 6, 2017 #9
    Wow, oh my god. I, I don't have words. It was so fundamental. Thank you very much !
     
  11. May 6, 2017 #10
    It just to say to OP that it is a beginning, further it will be more interesting and interesting
     
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: Box and friction on a merry-go-round
  1. Merry-go-round problem (Replies: 1)

  2. Merry Go Round Problem (Replies: 1)

Loading...