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Homework Help: Frozen Lake , Uniform Stick.

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data

    A straight uniform stick of L= 1m and mass m = 1 KG lies on a surface of a frozen lake.Someone kicks the stick at one end,imparting an impulse of 2Ns normal to the stick.Describe the subsequent motion of the stick.

    2. Relevant equations

    I=mΔV



    3. The attempt at a solution

    Obviously the first thing we have to do is find the linear velocity at the end of the stick ( actually all molecules of the stick will be moving with that velocity). so I=m(V1-V0)

    <=> 2=1*V1 , V0=0 since it was at rest.Therefore V1=2m/s which will remain constant due to lack of friction.The stick will be sliding forward since there are no forces to counteract with the initial force.But this force also created a torque which causes the object to rotate.

    Should i assume that since there is no friction , the stick is rolling right from the beginning ? How can i find the angular velocity ?

    I believe that since the kick is an external force and thus an external torque then linear and angular momentum are not conserved .

    Any ideas ?

    The force of the kick also caused the object to rotate
     
  2. jcsd
  3. Apr 26, 2012 #2
    Well i don't think it's rolling otherwise Vcm= wR <=> w= 2/0.5 = 4 rad/s which is not the right answer . Hmm,so we have an object that is slipping and rotating and there is no friction ... :S Grrr !
     
    Last edited: Apr 26, 2012
  4. Apr 26, 2012 #3
    No one ? :(
     
  5. Apr 26, 2012 #4

    tiny-tim

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    Hi ZxcvbnM2000! :wink:

    (erm :redface: … NEVER reply to your own question, it takes you off the "no replies" list!)

    just use impulse = change in momentum

    and torque of impulse = change in angular momentum (about the same point) …

    what do you get? :smile:

    (and i'm off to bed :zzz:)
     
  6. Apr 26, 2012 #5
    I understand that Impulse = change in momentum

    but why does torque of impulse = change in angular momentum ?

    Impulse is not a force it is Ft :S So how can it create a torque . Can you explain this physically ?

    Let me check if what u say is making sense unit-wise .

    R*mat = Iω Kg*m*m /s = Kg*m*m*rad/s which is true since rad/s = 1/s since ω is also measured in hertz ... hmm

    i still don't get it though.

    1) The stick is stationary . I kick its end . Will it begin rotating and slipping at the same time or not ? . I am lost :S

    Assuming that i understood everything let me proceed to the calculations..

    2=1*V so Vcm = 2 m/s

    mVcmR = Iω Ι = 1/12 ΜL^2 so Icm = 1/12 therefore 1*2*0.5*12 = ω

    So Vcm = ωR is not valid in this case because if it were true then the stick would be rolling , right ?

    Damn rotational and translation is soooo confusing .

    Anyway , have a good night and thanks for your time :)
     
  7. Apr 27, 2012 #6

    tiny-tim

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    Hi ZxcvbnM2000! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)
    yes that's all correct :smile:

    (except that your mVcmR should be the torque, which in this case of course is the same)

    so it's rotating at 12 rad/s, and moving at only 2 m/s

    (the question doesn't ask for it, but can you find the centre of rotation?)
    i don't understand what you mean by rolling … there's nothing for it to roll against :confused:
    I dω/dt = torque = force.distance

    so Iω = ∫ torque dt

    = ∫ (distance x force) dt

    = distance x (∫ force dt)

    = distance x impulse ! :smile:

    (eg, in this case, if you applied the same impulse slowly, eg 0.5 N for 4 s (total 2 N.s),

    then for 4 s the angular acceleration would be dω/dt = torque/I = 0.5 R/I (a constant),

    and so the total increase in angular velocity is ω = torque*time/I = 0.5 IR times 4 = 2 IR)
     
  8. Apr 27, 2012 #7
    Well i already said the moment of inertia is ML/12 so i assumed that it is the centre of mass of the stick . Can i use it as a general rule and say that for any body which is free to move any force applied to it ( besides those who cross the axis of rotation of the CoM) will cause it to rotate about its CoM.

    In terms of energies : Einitial = 1/2 ( Iω +mVcm) .

    Why isn't Vcm = ωR ? Does this means that a particle at the end of the stick will cover more distance than the CoM in the same time t ?

    :)
     
    Last edited: Apr 27, 2012
  9. Apr 27, 2012 #8

    tiny-tim

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    a body rotates about its centre of rotation

    (the clue's in the name! :biggrin:)

    the formulas torque = Iα and torque of impulse = I(∆ω) only (and always) work about the center of mass and the centre of rotation …

    in this case it isn't obvious where the centre of rotation is, so we use I and torque about the centre of mass …

    that doesn't mean it's rotating about the centre of mass! :wink:
     
  10. Apr 27, 2012 #9
    That's very useful thank you !

    Lastly, can you tell me whether what i said about Vcm = ωR is true or nonsense ?

    That's my last question ever...





    about rotation :p
     
  11. Apr 27, 2012 #10

    tiny-tim

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    the c.o.m. moves in a straight line, and the stick revolves around that moving point (so yes, the rest of the stick moves further than the c.o.m.)

    if VP is the velocity of a point P at distance R from the c.o.m., then

    |VP - Vcm = ωR …​

    (in 3D: VP - Vcm| = ω × R)

    so Vcm = ωR only if P is stationary :wink:
     
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