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Homework Help: Cylinder rolling up a step

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data:
    Cylinder of mass m, radius r is rolling without slipping with angular velocity ω and Vcm is ωr.
    Cylinder comes across a step of height R /4. The angular velocity just after the collision is.....
    Assuming cylinder remains in contact and no slipping occurs on the edge of the step.


    Relevant equations: Using cons. Of angular momentum.



    3. The attempt at a solution:
    I conserved the angular momentum about the the edge of the stair
    The relevant equation is:
    3/2 Mr^2(I about the edge)×ω-Mv3r/4 (angular monentum due to Vcm)=
    3/2 Mr^2(after climbing up I remains the same)×ω2 (new angular velocity)- Mv2×r (new v hence v2)
    (NOTE:I hope you'll understand the significance of negative sign.)
    After solving I am not getting the required answer.
    Am I going wrong somewhere? Please feel free to correct any error.
    Thanks.
     
    Last edited: Nov 8, 2013
  2. jcsd
  3. Nov 8, 2013 #2

    haruspex

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    The problem is not adequately specified. looks like you have to assume the step has sufficient friction that the cylinder does not slip during the collision.
    I don't understand the - Mv2×r term. (I assume you meant - Mv2/r.) Or maybe you're not assuming the cylinder maintains rolling contact with the step?
     
  4. Nov 8, 2013 #3

    ehild

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    The angular momentum about of the edge of the stair is not conserved. You ignored the torque of gravity.

    Use conservation of energy.

    ehild
     
  5. Nov 8, 2013 #4
    No, the angular momentum about the step IS conserved because whenever we decide to conserve ang momentum we see if net EXTERNAL torque acting on that point is zero and Mg is internal to that cylinder .(Hence the torque is zero)
    I am sorry that I didnt say anything about the assumptions. .
    I have edited the question.
     
    Last edited: Nov 8, 2013
  6. Nov 8, 2013 #5
    @echild , the energy is NOT conserved because it can be clearly seen that collision is inelastic.
    @haruspex, its not mv^2/r its mv2×r ie new linear velocity of cm.×r
    Actually I ve figured it out and got the answer.
    Firstly taking angular momentum about edge of the step:
    Mv×3r/4 +(ang momentum due to v &ω both are inside the page ie forward pure rolling & its the mistake I made.) Mr^2/2×ω = 3/2 Mr^2×ω(new) (here I took pure rolling about bottomost point)
    ω(new) therefore comes out to be 5ω/6..
    Thanks for your inputs...
     
  7. Nov 8, 2013 #6

    haruspex

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    Well done. And I just realised I was wrong to object to the question, It does say no slipping occurs on the step. However, please try to write equations using superscript and subscript in future - that's why I misread the equation.
     
  8. Nov 9, 2013 #7
    How is angular momentum about the step conserved ? The torque due to gravity(external force not internal) acts on the cylinder while it rolls about the edge of the step .

    I agree with ehild that angular momentum about the edge is not conserved.

    Or is it that the angular impulse of the gravity is ignored during the collision just like impulse of the gravity is ignored during collision ?

    Would some mentor reflect on this issue ?
     
    Last edited: Nov 9, 2013
  9. Nov 9, 2013 #8

    TSny

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    The problem is apparently asking for the angular velocity of the [STRIKE]ball[/STRIKE] cylinder immediately after the collision (but before the cylinder has lifted any distance off the original surface). The cylinder is just beginning to rotate without slipping about the corner of the step. So, the torque due to gravity has not yet had an influence.
     
    Last edited: Nov 9, 2013
  10. Nov 9, 2013 #9
    Do you mean that we are neglecting the angular impulse of the gravity OR the angular impulse of the gravity doesn't act on the cylinder ?
     
  11. Nov 9, 2013 #10

    TSny

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    Gravity will act to slow the motion of the cylinder during the rotation of the cylinder about the point of contact with the step. But, if I'm understanding the question, you are looking for the angular velocity of the cylinder immediately after impact with the step but before the cylinder has undergone any rotation about the point of contact with the step. So, gravity has not yet created an angular impulse. [EDIT: Up until the collision, the force of gravity is cancelled out by the normal force.]

    It's sort of like a ballistic pendulum where you find the speed of the pendulum immediately after the bullet collides with the pendulum but before the pendulum has had time to swing up through any angle.
     
    Last edited: Nov 9, 2013
  12. Nov 9, 2013 #11
    With all due respect ,i beg to differ .What we are looking at is what happens before and after the collision ,but that depends on what happens during the collision .Isn't it ? The force due to gravity acts during the collsion .It definitely provides an angular impulse about the edge of the step .But since the force is non impulsive ,I guess we can neglect the angular impulse of the gravity during the collision.

    I understand your point ,but the problem involves a collision .Just like we neglect the impulse of gravity in a collision while conserving linear momentum in vertical direction ,the same way we are neglecting angular impulse of the gravity about the edge of the step .
     
  13. Nov 9, 2013 #12

    haruspex

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    Impulse = ∫F.dt. Gravity is a limited force, so the integral over an infinitesimal time is zero. The impulse from impact with the step is taken to be an arbitrarily large force acting for an arbitrarily short time. Only its ∫F.dt is known.
    Note that we are given in the question that there is no slipping on the edge of the step. Since the normal 'force' acts as an impulse, IN, there is also a frictional impulse If . This obeys the analogous formula If ≤ μkIN.
     
  14. Nov 9, 2013 #13

    TSny

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    I guess it depends on when the normal force from the horizontal surface goes to zero. If the normal force somehow goes to zero at the beginning of the collision or during the time of the collision, then there would be some impulse due to gravity during the collision. But, as you say, you could still invoke the "impulse approximation" and neglect the gravity impulse compared to the impulse from the edge of the step.

    If the normal force doesn't start to reduce to zero until after the collision is over, then during the collision the gravity impulse would be cancelled by the normal force impulse.

    When I first read this problem, I thought the goal was to find the angular speed of the cylinder after it had rolled up onto the upper surface of the step. Then, of course, the force of gravity would need to be taken into account. Perhaps ehild was thinking the same thing, I'm not sure.
     
  15. Nov 9, 2013 #14

    tiny-tim

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    hi ashankrudola! welcome to pf! :smile:

    yes, you've got it right: but the discussion round you has become a bit confusing, so let's summarise it …

    (i) we can do τP = dLP/dt = d/dt[mvc.o.m x (rc.o.m - rP) + Ic.o.mω] about any stationary point P (eg a step which a body is mounting even if there is slipping)

    (ii) if P is moving (eg if P is the instantaneous point of contact of something rolling along a surface rather than at a point), we can still use that formula
    (a) if P is the c.o.m. or
    (b) if P is the c.o.r. (centre of rotation), and P and the c.o.m. are moving in parallel straight lines ​
    … when it becomes τP = d/dt IPω
    (which also works if P if a stationary c.o.r.)

    (iii) energy is not conserved during a collision (a sudden change, such as rolling round the inside of a polygon, or up a step, is a collision: a gradual change, such as inside a smooth curve is not)

    (iv) energy is conserved after the collision

    (v) the collision can usually be taken to be instantaneous, in which case we should really talk about angular impulse, not angular momentum, and the impulse, or angular impulse, of a non-collision force (such as gravity) during the very short time of the collision is zero
     
  16. Nov 9, 2013 #15
    I am impressed by the amount of interest (confusion?) this question has generated but hats off to all the mentors (Tsny , haruspex and tiny tim ) for explaining it scientifically.
    Yes I took the torque due to gravity to be zero in comparison to the impulsive force at that moment because simply I ve had enough of these questions ...
    Take for example a question involving a block moving linearly towards a ridge and then toppling about it. We need to find the ang velocity just after collision with the ridge...
    Here again we use cons of ang momentum about the ridge simply because torque due to mg (weight) is neglected wrt impulsive force at that instant.Rest has been exquisitely explained by the mentors..
    I ve had my final say...
    PS: Summarisation of the points are a gem to read.
    Hats off specially to tiny tim.
     
    Last edited: Nov 9, 2013
  17. Nov 9, 2013 #16
    Thanks TSny ,haruspex,tiny-tim for your excellent inputs .
     
  18. Nov 9, 2013 #17

    haruspex

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    True, but to answer the question in the OP it is important to know whether slipping occurs. If there's no friction at all at the step then ω is unchanged. In between the two extremes, it gets quite interesting.
    Not sure I follow. Angular impulse is the change in angular momentum, so it is appropriate to involve both.
     
  19. Nov 9, 2013 #18

    tiny-tim

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    the question says "rolling without slipping"
    yes, and we should really talk about the change in angular momentum, rather than angular momentum
     
  20. Nov 9, 2013 #19

    haruspex

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    Yes, I know. I was just clarifying that although, as you say, the equation applies regarding moment before impact, what you equate it to after the impact depends on friction.
     
  21. Nov 9, 2013 #20

    rcgldr

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    At the moment of collision, there's an horizontal impulse imparted to the cylinder at R/4 from the bottom of the cylinder, which would increase the angular momentum and decrease the linear momentum of the cylinder. Momentum is not conserved unless whatever the step is attached to, such as the earth, is included as part of a closed system.
     
  22. Nov 9, 2013 #21

    haruspex

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    Not sure what post that's in reference to. The impulse is not necessarily horizontal (in fact it can't be since there's no slipping), but the line of the impulse passes through the corner of the step. Consequently there's no change to the angular momentum about that point.
     
  23. Nov 10, 2013 #22

    ehild

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    The text of the problem was :

    It is confusing in some points.

    How is collision defined? I might be wrong but usually we speak about collision when two bodies interact for some finite time, and we can use conservation of some quantity (quantities) when determining velocities (angular velocities) after the interaction ceased.

    Initially the cylinder was separated from the step. It hit the step, rolled about it and rolled further on the platform after it "came across" the step. That is what come across means is it not? But my English is poor, I may be wrong.

    The cylinder is in contact with the edge of the step during coming across it so I considered that process as part of the collision.

    At the instant the cylinder hits the step, it experiences an impulsive force from the edge normal to the plane of contact. But its surface moves with respect to the edge so there is also kinetic friction between the step and cylinder, which causes a torque with respect to the centre of the cylinder and stops it rotating about its centre. The result is that the whole cylinder will rotate about the edge as fixed axis. "Assuming cylinder remains in contact and no slipping occurs on the edge of the step." can refer only to that stage of interaction.

    So we have two stages of interaction: There is slipping during the first stage, and no slipping in the second stage. The first stage is the "impulsive part" but the cylinder does not rotate about a fixed axis during this stage.

    Anyway, the angular momentum with respect to the edge conserves during the first stage.

    During the second stage, energy is conserved as no slipping occurs, but the angular momentum is not conserved.

    I considered the whole process of interaction as "collision".

    There was a very similar problem a few years ago, conservation of energy was used to solve it.

    http://www.google.com/url?sa=t&rct=...ql7kahKhgSRkG4Q&bvm=bv.56146854,d.Yms&cad=rjt


    ehild
     
    Last edited: Nov 10, 2013
  24. Nov 10, 2013 #23

    ehild

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    What about elastic collision , for example between a ball and wall. There is a sudden change of velocity...


    ehild
     
  25. Nov 10, 2013 #24

    haruspex

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    Of course, all collisions take some nonzero time. For very short events, we can often ignore those details and treat it as instantaneous. The forces are correspondingly taken to be arbitrarily large. More modest forces, such as gravity, are ignored during the collision. Note that this approach is not going to work when you care about the maximum force.
    In my experience, most mechanics course work uses 'collision' to mean an event of short but unknown duration.
    To 'come across' something can mean to happen upon it, but it's usually only used where the subject is animate. Using it that way here seems a little jocular.
    Right, but you may be missing that the frictional force is itself an impulse. It all happens in the same split second. The normal force is large and unknown, the frictional force is correspondingly large and unknown. What we do know is the impulse delivered by the normal force, IN=∫FN.dt, and the maximum impulse that can be delivered by the frictional force, If=∫Ff.dt=∫μsFN.dt=μs∫FN.dt = μsIN.
    There is one stage and no slipping. Taking the collision to be instantaneous, the transition from rolling on the ground to rolling 'around' the corner of the step is also instantaneous. If you like, you can imagine it spread over a finite time, and you will find there's no need for any slipping.
    Btw, it will only rotate around the step instantaneously. It will immediately lose contact with the step and become airborne.
    That's a very different problem and I hope the diagram is wrong! The solution assumes a smooth transition from rolling at the lower level to rolling at the higher, which will only happen if the radii of curvature of the step are sufficiently large. With a right-angled step, as shown, the question of whether the object succeeds in mounting the step, collisions being elastic, gets really difficult. It could bounce an unlimited number of times before finally making it.
     
  26. Nov 10, 2013 #25

    ehild

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    There must be slipping when the cylinder comes into contact with the step, as it rotates around its CM, so its surface points move with respect to the step. The force of friction stops that relative motion.

    I do not like the word "immediately" in Physics. It is an approximation only, interactions last for some time.


    ehild
     
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