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Mouse and ladder

  1. Aug 4, 2009 #1
    1.Statement: A mouse starts climbing up a ladder with uniform speed u wrt ladder and at the same time ladder is released to fall from rest. After what time the mouse is again at its initial position wrt the ground if
    a) mass of mouse is negligible wrt ladder
    b) mass of mouse be m and that of ladder be M


    PS:initially the mouse is at the bottom of the ladder and ladder is held vertical and then released at t=0
     
  2. jcsd
  3. Aug 4, 2009 #2
    i was able to think for the first part...
    as the mouse have acceleration=g (downward)
    n initial velocity =u (which is uniform actually) so in order to have displacement=0
    we can write..0=ut-1/2gt^2 so we have t=2u/g
    but what about 2nd part??
     
  4. Aug 4, 2009 #3
    please help someone :(
     
  5. Aug 4, 2009 #4

    tiny-tim

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    Welcome to PF!

    Hi phykeep! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    (and don't be so impatient… it can take hours to get an answer here!!)

    hmm … right result, wrong reason! :biggrin:

    You should really use the acceleration on the whole mouse-ladder system.

    In part a), the mouse can be taken to have zero mass, so effectively g only acts on the ladder.

    The u in ut - 1/2gt2 is not necessarily the same u as they gave you … it happens to be in a), though you have to show that, and it isn't in b)!

    Hint: call the height of the mouse above the ground z, and the height of the ladder above the ground h.

    So you know that the "speed" of (z - h) is u. :smile:
     
  6. Aug 4, 2009 #5
    heyee tiny-tim..thanks a lot for ur reply...
    yep i got ur point for part b but the problem is that velocity of mouse with respect to ground will become u+gt...(m i right??correct me if i m wrong)
    so 0=(u+gt)t-1/2gt2
    but this will give negative values for t...
    and i could not understand the reasoning for part a..i mean is my answer for that part correct???
     
  7. Aug 4, 2009 #6
    Re: Welcome to PF!

    Just out of curiosity, what are the final answers? These look like really interesting questions and I want to solve them for myself.
     
  8. Aug 4, 2009 #7
    Re: Welcome to PF!

    well even i dont know the final answers... :(
     
  9. Aug 4, 2009 #8
    sorry i made a mistake...in post #5
    actually the answer for the part b is t=2u/3g...
    because the velocity of mouse with respect to groung is u-gt...
    but what about first part??
     
  10. Aug 4, 2009 #9
    Wow, this is a beautiful question. Who knew you don't need to use any force to move at a constant velocity up a falling ladder?

    You are mistaken in your analysis. The velocity term in this equation:
    [tex]y(t)=y_0+vt+\tfrac{1}{2}at^2[/tex]
    relates to the INITIAL velocity. Plugging in the velocity of the ladder as a function of time into the initial velocity, is just wrong. You are already counting in the effect of gravity by plugging it in to the acceleration term in the above equation.

    Try approaching the problem like this:
    What is the velocity of the mouse wrt (I think I like this abbreviation!) the ladder?
    What is the acceleration of the mouse wrt the ladder?
    What is the acceleration of the ladder wrt to the ground?

    From these three, you can deduce the initial velocity of the mouse wrt the ground, and its acceleration wrt to the ground, at which point, you can plug those into the above equation, and solve for [tex]y(t)=y_0[/tex]

    For the second question, you must first do a force analysis of the two masses, to see what the acceleration of the ladder is wrt the ground (Remember that since the mouse is advancing up the ladder with a constant velocity, it is at equilibrium wrt the ladder.)

    Just an FYI, I got the same result for both cases, [tex]t=\frac{2u}{g}[/tex] which was really surprising and beautiful. :) Though I may be wrong, so I'd like to get a second opinion.
     
    Last edited: Aug 4, 2009
  11. Aug 4, 2009 #10

    tiny-tim

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    Yes, but you don't understand why, and nor will the examiner. :wink:
    uhh? whatever happened to m and M? :confused:

    Use z and h and good ol' Newton's second law for the whole mouse-ladder system.

    Try both parts again. :smile:
     
  12. Aug 4, 2009 #11
    Well I think i have a good understanding of the question by seeing ur opinion..
    please let me give my explanation .i.e. t=2u/g...so that i get to know whatever i m thinking is right or wrong..
    for first part mouse have zero acceleration wrt ladder thats why its acceleration=g(downwards) wrt to ground..(since ladder's acceleration=g(downward)
    now for 2nd part...since mouse is moving with constant velocity wrt to ladder..it is in equilibrium wrt to ladder...so have zero acceleration wrt to ladder...
    but ladder have a downward acceleration=g..therefore mouse have acceleration=g wrt to ground..
    n as far as initial velocity of mouse is concerned it is equal to u wrt to ground..since intial velocity of ladder is equal to zero...
    correct if i m wrong anywhere ...
    or kindly post ur way of solving the equations..(if i m wrong somewhere).....
     
  13. Aug 4, 2009 #12
    Yep, that's how I solved it, but I'd like tiny-tim's input on this as well.
     
  14. Aug 4, 2009 #13
    yep...right now i m trying to solve it from ur point of u as well...
    its very cool to have many number of approaches to the same question..
     
  15. Aug 4, 2009 #14

    kuruman

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    If I can add my two bits here: The answer to part (b) can be conceptually understood better if one considers that both ladder and mouse are in free-fall, therefore "weightless". In other words, we are transforming to a non-inertial frame that accelerates down with acceleration g. The problem then can be solved in two steps.

    Step 1
    A mouse of mass m is on a platform of mass M. The platform is on a horizontal frictionless floor. The mouse starts walking across the platform with velocity u relative to the ladder. Find an expression for the position of the mouse relative to its initial position as a function of time. Check that your expression reduces to u*t as the mass of the platform becomes very large. Hint: By considering what happens to the CM, Newton's 3rd Law is automatically taken care of.

    Step 2
    Transform back to the lab frame by adding a term
    [tex]-\frac{1}{2}gt^{2} [/tex]
    to your answer in Step1. Set this expression equal to zero and solve for the time. Check that your expression reduces to your answer in part (a) as the mass of the platform becomes very large.
     
  16. Aug 4, 2009 #15
    hmm...this one is also good...wow..3rd kind of explanation...cool!
     
  17. Aug 4, 2009 #16
    tiny-tim,RoyalCat,kuruman thanks for ur help :)
     
  18. Aug 4, 2009 #17

    kuruman

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    You're welcome and good luck.
     
  19. Aug 4, 2009 #18

    tiny-tim

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    Hi phykeep! :smile:

    (been out … just got back)

    Yes, kuruman's method is fine, but you should be able to do it the direct way also.

    Use F = ma for the mouse-ladder combo (and ignore internal forces) …

    that's (M + m)g = Mh'' + mz'' …

    combine that with z'' - h'' = ut. :wink:
     
  20. Aug 4, 2009 #19
    yes i did it in that way also...


    but dont u think the above equation..which u wrote...is dimensionally incorrect..z" is double derivative of z..anyways but thanks for that solution too..i got the answer... :)
     
  21. Aug 5, 2009 #20

    tiny-tim

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    oops!

    oops! :redface: got carried away with those dashes! :rolleyes:

    yes, i should have written z - h = ut

    (and then call the initial velocities v and u+v)
     
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