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Angular momentum question

  1. Mar 14, 2008 #1
    1. The problem statement, all variables and given/known data

    a cylinder with a given radius [tex]R[/tex] rolls to the right without slipping ([tex]\omega= \frac{v}{R}[/tex]). It hits a nail fixed to the pillar. find [tex]h[/tex], the height of the nail above the floor, in which the cylinder will roll back without slipping.
    [tex]I=\frac{MR^2}{2}[/tex] for the cylinder.
    - The nail applies lateral force only.
    - the collision with the nail is ellastic, there's no loss of kinetic E whatsoever.
    - express [tex]h[/tex] with [tex]R[/tex] only.

    2. Relevant equations
    conservation of linear momentum, conservation of angular momentum.
    [tex]J=r\times P[/tex]
    3. The attempt at a solution
    there is no loss of kinetic energy. so if v is the speed before the collision and u is the speed of the ball after, [tex]v=-u[/tex].
    Thus, there is impact as follows: [tex]2mv=\Delta P[/tex]
    Also, I want the cylinder to roll backwards still with the term of not-slipping - [tex]\omega=\frac{v}{R}[/tex] so it's the same with angular momentum: [tex]2F\mu*R=\Delta J[/tex]
    Now, I was thinking about solving this with [tex]J=(h-R)\times P[/tex], but then I don't know what to do with [tex]F\mu[/tex].

    I'd appreciate help in this :)

  2. jcsd
  3. Mar 14, 2008 #2
    What happens when a rolling object slips?

    It slips when the amount of tangenial force is greater than the maximum friction.

    Also, how high h is affects the amount of tangenial force generated by the impulse during the collision.
  4. Mar 14, 2008 #3
    Do you mean I need to break the F generated from the nail to perpendicular and tangential, and relate to the tangential one? and then I need to do (Ft is the tangential F) [tex]F_t*R=2F\mu*R[/tex]?
    About your second line - do you mean the linear impulse or angular?
  5. Mar 14, 2008 #4


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    Hi ! :smile:

    You are specifically told: "The nail applies lateral force only."

    So you can assume there is only a horizontal force (very unrealistic, I would think :rolleyes:)

    And forget friction.

    The question is asking for the height needed for the cylinder not to want to slip!

    As you say, conservation of energy means that the speed out will be the same as the speed in.

    For the cylinder not to want to slip, the angular momentum, A, must therefore also be the same - but of course in the opposite direction.

    So what height must the nail be to produce a torque which changes the angular momentum by 2A? :smile:
  6. Mar 14, 2008 #5
    That's my problem: I get from that this equation: the nail's torque needs to equal both directions' friction torque: [tex](h-R)F=2F\mu*R[/tex], by the angular momentum conservation.
    from this I get [tex]\frac{F}{F\mu}=\frac{2R}{h-R}[/tex]
    and I remain with a ratio of the forces I don't know how to get.
  7. Mar 14, 2008 #6


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    Ah! No …

    There is no "friction torque", because there is no friction.

    The cylinder is rolling with steady speed - it's not pushing at the floor - the friction is zero.

    The question gives you I, the moment of inertia.

    Use it! :smile:
  8. Mar 14, 2008 #7
    I don't understand - the cylinder is spinning - so there's a torque which generates [tex]\omega[/tex], right? how can you say there isn't (im not talking about force, but torque).

    So you're saying I should do something like this: [tex]2I\omega=F(h-r)[/tex]. I need also to use [tex]\omega=\frac{v}{R}[/tex]. but then what can I use for [tex]F[/tex]?
  9. Mar 14, 2008 #8

    you need friction to roll!!!!!!!!!!!!!!!!!!!!!!!!!!!!! thats how wheels work!!!!!!
  10. Mar 14, 2008 #9
    radagast, how would you separate the force exerted by the nail into its tangenial and centripetal components?

    Also, bear in mind conservation of angular momentum. This will give you a clue as to how much force was exerted on the cylinder during collision.
  11. Mar 14, 2008 #10
    Oreg, I don't understand why I need to separate the nail force, instead of separating the r vector to the center of mass, in which I'll get N=(r-h)*F.
    Conservation of angular momentum is what I started with. I still remain with a missing ratio - F/Fmu ... I don't know where to get more information.
  12. Mar 14, 2008 #11


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    No … that's only the way wheels work if they don't work … :rolleyes:

    Imagine a log rolling on ice - it will keep going at the same speed without any friction, and it will keep rolling without slipping (so long as it started off rolling without slipping, of course). :smile:

    Wheels only need friction with the ground during acceleration or braking (or turning).
    I'm saying there's no friction.

    Yes, there is a torque when the cylinder hits the nail (but not before or after), and that's what you have to calculate.

    What can you use for F? You tell us … what equation do you have that involves F, but doesn't involve angular momentum? :smile:
  13. Mar 14, 2008 #12
    well, I have F=ma, and 2mv=integral{f*dt}. from both I can't get anything... because the acceleration is 0. what am I missing?
  14. Mar 14, 2008 #13


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    You are missing … the full version of Newton's second law.

    It's often referred to as force = mass x acceleration.

    But that's not actually what good ol' Netwon said.

    He said total force = change of momentum.

    If the change of momentum is gradual (not sudden), then the rate of force equals the rate of change of momentum, which is mass x acceleration!

    (That's why force = mass x acceleration is so useful - most forces are graudal.)​

    But if the change of momentum is sudden, then we use the total force (usually called the impulse), instead of the rate of force.

    In this case, the cylinder reverses instantaneously, so we use the impulse (total force), F say, and we put it equal to momentum after minus momentum before.

    So F = … ? :smile:
  15. Mar 14, 2008 #14
    ahh. so [tex]F=2mv[/tex].
    For some reason I got confused thinking the momentum difference is integral of F and not F itself... ~(_8^ D'oh!
    [tex]2I\omega=F(h-r) => 2*\frac{mR^2}{2}*\frac{v}{R}=2mv(h-r)[/tex]

  16. Mar 15, 2008 #15


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    :smile: Looks good! :smile:
  17. Mar 15, 2008 #16
    Thanks tiny-tim and Oerg!
  18. Mar 15, 2008 #17
    it still doesnt make sense to me. How does a wheel slip when its angular momentum has changed by 2a.

    Doesnt a wheel slip when the torque is greater than the torque caused by the friction???
  19. Mar 15, 2008 #18


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    just geometry … point of contact is stationary

    No, it slips when the velocity of its point of contact is non-zero.

    In perfect rolling, the bit of the wheel that's on the ground is always (instantaneously) stationary.

    If it isn't stationary, that means the wheel is slipping along the ground.

    If there's no acceleration and no slipping, there should be no friction.

    This isn't physics … it's geometry! :smile:
  20. Mar 15, 2008 #19
    wait wait wait!!!!

    rolling means it is rotating as it is moving along! So doesnt this mean that the exact same point on the object is not in constant contact with the ground?!
  21. Mar 15, 2008 #20


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    :wink: I'm waiting! :wink:
    Yes, you're right, the point of contact changes all the time.

    But whichever bit of the rim of the wheel is on the ground at any particular moment is stationary!

    Confusing, isn't it? :smile:
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