1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Frozen Lake , Uniform Stick.
  1. Frozen Pendulum (Replies: 6)

  2. Pressure in a Lake (Replies: 3)

Loading...