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Homework Help: Wheel attached to a spring with teeth

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data
    FpoyM09.png


    2. Relevant equations
    F = ma
    Torque

    3. The attempt at a solution
    All I want to ask is for this problem is: is there friction at ALL points along the path of the wheel because it wasn't clear from the problem. Thanks
     
  2. jcsd
  3. Jan 21, 2013 #2
    You have to assume no friction at x > l, and some very large friction at x < l, which causes the wheel to roll rather than slip.
     
  4. Jan 21, 2013 #3

    haruspex

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    There is no friction for x > l; effectively unlimited friction for x < l.
    (It will be interesting calculating what happens when it first meets the track.)
     
  5. Jan 21, 2013 #4
    So there is a finite friction force for x < l?
     
  6. Jan 21, 2013 #5

    haruspex

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    For x < l, just treat it as rolling. The 'friction' force is whatever is required to achieve that. (At the x=l, the force will be instantaneously infinite.) Be careful with the transition to rolling. (Treat it as an impact.)
     
  7. Jan 21, 2013 #6
    Hi haurspex thanks for responding. I was wondering if I was on the right track. First I used the fact that since there was no friction for x > l, the only force is the spring force which is conservative so [itex]v_{0} = \sqrt{\frac{k}{m}}b[/itex] where [itex]v_{0}[/itex] is the speed of the wheel just before hitting the tracks. We can set up a conservation of angular momentum equation (with respect to the origin fixed to the corner of the back wall let's say) for the moments instantaneously before and after the collision to get [itex]mv_{0}R = mv_{f}R + I\omega = 2mRv_{f}[/itex] using the fact that the wheel is rolling without slipping instantaneously after the collision. Next I stated that at the farthest point, x, that the wheel gets to, it will be instantaneously at rest so that [itex]-\frac{1}{2}mv_{f}^{2} = \int_{l}^{x}kxdx - fdx [/itex] leaving us with, after plugging back in known quantities, [itex]\frac{1}{4}kb^{2} = k(l^{2} - x^{2}) + 2f(x - l)[/itex]. I need to find a way to get friction from here but I don't want to go further if I've made mistakes. Thanks.
     
    Last edited: Jan 21, 2013
  8. Jan 21, 2013 #7

    haruspex

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    fdx? It's rolling. What energy is lost to friction?
     
  9. Jan 22, 2013 #8
    Huh? But friction is doing work as the wheel passes through the region so don't we have to take that into account? In one of the worked examples in Kleppner there is a wheel rolling without slipping down an incline and he takes into account the work done by friction both for force and torque.
     
  10. Jan 22, 2013 #9
    The wheel is rolling. The point on the wheel where friction applies has zero velocity. What is the work of friction on it?
     
  11. Jan 22, 2013 #10

    WannabeNewton

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    Friction will convert all the loss of energy due to translation into a gain of energy in rotation. That's what Kleppner was saying with that example when he calculated the work done by torque due to friction and the translational work which included friction. It looks like your goal was to solve for friction using the work done by torque and eliminate it from that other expression which is what Kleppner does in his example. However haruspex I'm not sure if you can just toss out the friction term completely when calculating the work done by pure translation because friction is converting that energy into rotational energy so friction does no work on the total combined rotational and translational motion but if you are calculating the work done by translational and rotational separately then you have to take them into account and then combine them to make it vanish.
     
    Last edited: Jan 22, 2013
  12. Jan 22, 2013 #11

    haruspex

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    When rolling, the force of friction can certainly be involved in conversion of energy, but it cannot be said to be doing work. This is because, as voko points out, the force does not move the object it is acting on. At any given instant, it acts on the point of contact, and the point of contact is stationary. An analogous situation is use of a fulcrum.
     
  13. Jan 22, 2013 #12

    WannabeNewton

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    Yes but it does no work when combining the work due to rotation and translation together not when considering them separately.
     
  14. Jan 22, 2013 #13

    haruspex

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    I gave due consideration to that view and concluded it was wrong. Work is ∫F.dx, where dx is the distance yielded to the force (a vector), and '.' is the dot product. When rolling, dx is zero.
    Anyway, in the present context, PhizKid had the work done by friction as a loss to the system.
     
  15. Jan 22, 2013 #14

    WannabeNewton

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    What phiz kid wrote down refers to the translation of the center of mass (and as phiz kid stated he was going to find the work due to rotation after in order to combine the two results together). If what you said was true then the inclusion of the -fL term by Kleppner in this example (the one alluded to by phiz kid) would be incorrect:

    http://img829.imageshack.us/img829/7642/kleppnerexample.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  16. Jan 22, 2013 #15

    haruspex

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    The work is being done gravity, not by the friction.
     
  17. Jan 22, 2013 #16

    WannabeNewton

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    I'm talking about the -fL in (Wsinb - f)L and he labels friction by f. The expression refers to the work done due to the translation of the center of mass as he derives a few pages before. He then considers the work done due to the torque as phiz kid stated and eventually combines the two together to eliminate the friction.
     
  18. Jan 22, 2013 #17

    haruspex

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    Then (Wsinb - f) is the magnitude of the force doing the work. The fact that f happens to be involved in the expression does not mean friction is doing work.
    Take my fulcrum example. A fulcrum is placed under a light, rigid horizontal beam length L, distance D from a load at one end of the beam. I move the other end down a distance x. The vertical force on the fulcrum is F. Work done on the load = xFD/(L-D). Does that mean the fulcrum did work?
     
  19. Jan 22, 2013 #18

    WannabeNewton

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    I'm not claiming friction does any work I'm saying you have to first include its negative expression in the translational and then take into account its positive expression in the rotational to get rid of it in order to get the right final equation not throw out the fdr term from the very start because that refers to the translation of the center of mass.
     
  20. Jan 22, 2013 #19
    I made a typo in my last post with the bounds of my integral (since the bounds should be the displacement from the un-stretched length) but I fixed it here. I did it the way WBN said to do it and the answer I got agreed with the one listed on the University of Chicago's homework page for the same class. I did [itex]-\frac{1}{4}mv_{0}^{2} = \int_{0}^{l - x}(kx - f)dx = (\frac{1}{2}kx^{2} - fx)|^{l - x}_{0} = \frac{1}{2}k(l - x)^{2} - f(l - x)[/itex] and [itex]-\frac{1}{2}I\omega^{2} = -1/4mv_{0}^{2} = \int_{\theta _{1}}^{\theta_{2} }fRd\theta = fR\Delta \theta = f(l - x)[/itex] so combining the two together and using that [itex]v_{0} = -\sqrt{\frac{k}{m}}b[/itex] (I chose the negative solution for the square root because the wheel is moving backwards immediately before it hits the track) I get that [itex]\frac{1}{2}b^{2} = (l - x)^{2}[/itex] giving [itex]x = l - \frac{1}{\sqrt{2}}b[/itex].

    haruspex, I had one more question regarding the way we treated the hitting of the track as a collision: what if we had no track but were told that the region x < l just happened to have a very large friction in order to make rolling without slipping happen but there was nothing but regular ground - no track or anything. Would we still treat it as an impact because in that instantaneous moment pure translation motion is being converted into rotational + translation motion so we can invoke a conservation of angular momentum equation for that infinitesimal period of time for which the collision occurs because physically I don't see an actual collision in contrast with the previous case of physically hitting the gear tracks.
     
    Last edited: Jan 22, 2013
  21. Jan 22, 2013 #20

    haruspex

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    Yes, but the difficulty with describing it as just a highly frictional surface is that it would have to have an infinite coefficient of friction, which may be felt to be too great a violation of reality. Stating it as a gear track removes such concerns.
     
  22. Jan 23, 2013 #21
    Yes we do treat it as a collision regardless?
     
  23. Jan 23, 2013 #22

    haruspex

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    Yes.
     
  24. Jan 23, 2013 #23

    ehild

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    In case the track is a common rough surface the wheel can slip and rotate on it. There is kinetic friction: it acts against translation and accelerates rotation up to the instant when linear velocity of the CM becomes equal to ωR. That takes some time and the effect of the external spring force can not be ignored.

    ehild
     
  25. Jan 27, 2013 #24

    ehild

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    Assuming no impact but very high kinetic/static friction on the track, we can set up an equation for the change of velocity of the CM, and an other for its angular velocity. The wheel slips when entering the track, so it keeps on slipping till pure rolling is set up. The spring is relaxed at x=l, so there is no spring force at the beginning and we ignore it during the whole process, assuming that rolling is set up in a very short time.

    The friction acts against linear velocity and accelerates rotation.

    mdv/dt=-Fk=> dv/dt=-μg
    Idω/dt=FkR=> dω/dt= μg/R

    So v=Vi-μgt, ω=(μg/R)t

    The rolling condition is that v=Rω: Vi-μgt=μgt, it happens after t=Vi/(2μg) time slipping. During that time, the velocity decreased to Vf=Vi/2 and the wheel covered the distance Δs=(Vi+Vf)/2*Vi/(2μg)=0.75Vi2/(μg).
    Note that the final velocity is the same we got with conservation of angular momentum during impact.

    Assume m=1 kg, μ=1, Vi=1 m/s. The force of kinetic friction is Fk=10 N. Rolling is set up in t=0.05 s.
    The distance travelled without spring would be Δs=0.0375 m. The spring force at such displacement would b Fs=kΔs. If k = 100 N/m, Fs=3.75 N. That can not be really ignored with respect to friction.

    How rough the surface has to be that the spring force stay less than 0.1 Fk?
    3/4 Vi2 /(μg)*k<0.1(μmg). With the data assumed, μ has to be greater than 2.7 But a gear track can supply that big friction.

    If Vi=0.1 m/s, t=0.005 s, Δs=0.000375, and the maximum spring force is 0.0375 N, that can be really ignored with respect to the force of friction.

    ehild
     
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