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What is the minimum height?

  1. Feb 2, 2017 #1
    1. The problem statement, all variables and given/known data
    An empty ball , of mass m moment of inertia I = (2m.r²)/3, is rolling across the path shown below :


    there is friction fr from A to C .

    r is the radius of the ball , and R is the radius of the circular part within the path .

    what would be the minimal height h , so that the ball can make a complete lap in the loop (the circular part) ?
    2. Relevant equations
    ∑F = m.a
    ∑τ = I . α = r.I.α / where α is the angular acceleration , I the moment of inertia , and r the radius .

    3. The attempt at a solution
    after some computations I've found :
    [tex]a_{T} = \frac{g.\sin(\alpha)}{\frac{2}{3}r^{2}-1}~~[/tex] ( a_T ; is the translational acceleration)

    [tex]f_{r} = \frac{2m.g.\sin(\alpha).r^{2}}{2r^{2}-3}[/tex]

    I know that the ball must roll on a distance of 2πR( the perimeter of the loop). But I can't figure out how to link between the acceleration and this distance ?

    picture link : https://i.imgsafe.org/38bb5dc5ca.png
  2. jcsd
  3. Feb 2, 2017 #2


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    This statement troubles me. Taken at face value, it is referring to static friction (the ball is rolling, we're told), but if so it is irrelevant what the value is, and it would not be the same on the slope and on the flat (where it would be zero). So I suspect it means rolling resistance, which will act as a constant force opposing forward motion.
    Either way, should not need to get involved in details of accelerations. Use energy considerations.
    Your expression for the acceleration cannot be right anyway. It is dimensionally inconsistent (r2 and 1 being added/subtracted).
  4. Feb 2, 2017 #3
    indeed ,it's a constant friction force
  5. Feb 2, 2017 #4
    I'm not checking the calculations, but your formula for acceleration is dimensionally wrong, probably you missed some term in there?

    By the way, the acceleration while the ball is on the loop is not constant(what you found would just be the acceleration of the ball rolling down the inclined plane), so unless you find an equation describing acceleration with respect to height and integrate over that you can't solve the problem this way, but that's not how I'd solve it.

    You need to understand what are the dynamical conditions such that the ball can make a complete lap, so you should reason on what forces are acting on the body while it is on the loop and note that they share a fundamental property, which allows you to apply an important physical principle. I would suggest you try to solve the problem first using a point-particle and then adjusting it for the ball.
  6. Feb 2, 2017 #5


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    It may be constant, but it is not friction.
  7. Feb 2, 2017 #6
    when the ball is on the loop , there is gravitational force friction , and reaction force , they have also sad that the reaction force was zero on the top of the loop , in addition to that , they give a hint wich says that we have to use newton's law to find out the relation between the velocity (translational velocity to be more precise) and the gravity g . now , what I'am wondering is , do the acceleration in the inclined plane have anything to do with loop , I mean can we study the motion of the ball just inside the loop regardless of the inclined path before ?
  8. Feb 2, 2017 #7


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    Yes. All that matters for the loop is that it enters with sufficient speed. Having found that, you can work backwards to the ramp.
  9. Feb 3, 2017 #8
    but how to find this minimum velocity ?? any hints ?
  10. Feb 3, 2017 #9
    What forces are acting on the body at the top of loop? Try to write down their equations
  11. Feb 3, 2017 #10
    on the top of the loop there is only gravity and friction , they said that we remiss the reaction force , so we have (by applying the sum of the forces , and the sum of the torques) ;

    [tex]m.g-f_{r} = m.a[/tex]

    so : [tex] a = \frac{g}{1+\frac{2}{3}r^{3}}[/tex]

    and then ?
  12. Feb 3, 2017 #11
    You can use energy to get speed (it will be related with initial height)at one point inside de loop and then from that point Newton's equations. Don't forget rolling energy.
  13. Feb 3, 2017 #12
    By using conservation of mechanical energy energy :

    [tex] m.g.h=w(f_{r})+\frac{1}{2}m.v^{2}+\frac{1}{2}I.\omega ^{2}=-f_{r}.\frac{h}{\sin(\alpha)} + \frac{1}{2}m.v^2+\frac{1}{3}.m.v^{2}=-f_{r}.\frac{h}{\sin(\alpha)}+\frac{5}{6}m.v^{2}[/tex]

    [tex]g.h = \frac{5}{6}.v^{2}-f_{r}.\frac{h}{\sin(\alpha)} [/tex]
  14. Feb 3, 2017 #13
    I am not sure how you obtained these, but they are wrong. As was already pointed out, it is of primary importance to check if the units are consistent when you obtain a formula, and in your equations they are not.
    And you can't just neglect the loop vincular reaction because it is what makes the circular motion possible(it is a centripetal force).
    I'd suggest you to post the original text to overcome this and other ambiguities, like the frictional force you are talking about.
    As haruspex already pointed out it may be a static friction, thus making the motion a "pure rolling", or it could be a rolling friction.
    In the first case(which i think is the right one), it would not be relevant since it is static and making no work, so conservation of energy applies; in the second case you would have dissipative forces on the plane but probably not on the loop(i dont think the dynamics of a loop with rolling friction can be solved analytically).

    Anyway, to correctly apply energy conservation you need to account for the centripetal force(which, if you don't remember, depends on velocity). If your body reaches the top with 0 speed, it will not make a complete lap but just fall down under the action of gravity.
  15. Feb 3, 2017 #14
    yes it's a static friction , not the rolling friction
  16. Feb 3, 2017 #15
    Where do I apply the conservation of enegy theorem , in the inclined plan or inside the loop ???
  17. Feb 3, 2017 #16


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    No, rolling friction is static friction. Despite what the question states, if this force is constant then it is not friction at all. Static friction would be zero on the horizontal path, and it does no work, so would not sap any KE. It could be rolling resistance.
    Gravity acts down, fr acts horizontally, so unless you mean this as a vector equation it makes no sense.
    The right hand side appears to be Ira. What beast is that? Did you mean Ia/r?
    As already pointed out several times, that is dimensionally nonsense. You cannot add a dimensionless constant, 1, to an entity that has dimension, such as r3.

    Please, let's take this in easy steps. First, what are the vertical forces acting at the top of the loop? What is the vertical component of acceleration at the top of the loop? What equation does that give?
  18. Feb 4, 2017 #17
    ok , on the top of the loop we have :

    the vertical accleration wich is tangential to the loop : [tex] a_{x} = \frac{v^{2}}{R} [/tex]

    and the normal accelration which points downward : [tex] a_{y} = m.g [/tex]

    is it right ?
  19. Feb 4, 2017 #18
    Am I misunderstanding something? How can vertical acceleration be tangential at the top of the loop?
  20. Feb 4, 2017 #19
    because on each point on the loop , we can break the accleration into two components , one pointing to the center of the loop , and the other tangential to the path , it's like an orbital motion .
  21. Feb 4, 2017 #20
    What does the equation a = v2/r represent?
    Edited due to poor wording.
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