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Falling mass

  1. Mar 17, 2005 #1
    have a look at the diagram. Mass on top is called Mass 1 and the mass below is caled Mass 2. The spring has spring constant k. At time t=0 the string is cut and the system falls freely. Neglect air resistance

    1) Determine the position of the centre of mass at time t>0. The position should be given as distance Ycm froim the CM position at t=0

    Nothing spciel about this. Imagin the whole thing as a big body and the mass falls.
    [tex] d_{CM} = v_{1} t + \frac{1}{2} g t^2 [/tex]
    and since v1 =0
    [tex] d_{CM} = \frac{1}{2} gt^2 [/tex]

    2) Determine the acceleration on mass 1 and mass 2 immediately after the string has just been cut
    to start with for mass 2
    kx = mg (1)
    for mass 1 T = kx + mg
    but once the string is cut T = 0
    kx + mg = 0
    mg + mg = 0 from 1
    2mg = 0
    this is the net force 2mg = ma, thus a = 2g
    (is there a flaw in this logic?)

    for the mass 2
    kx = mg but once string is cut kx = 0 since nothing pulls up
    thus mg = ma = 0 a = 0??
    once again whats the flaw with this logic??

    We now refer to the motion for t>0 to a reference frame that has origin in the CM. Denote this frame the position of the lower mass 2 by z.

    [B} determine z as a function 0of t. Consider for values of t such that 2 has not hit the floor. [/B]
    Now for the this part im a bit confused.
    Certainly z = L/2 + something
    what is this something
    the text gives the something to be mg/2k cos(root 2k/m) t
    not quite sure how they got that part
    Thank you in advance for ANY help!!
     

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    Last edited: Mar 17, 2005
  2. jcsd
  3. Mar 17, 2005 #2

    Doc Al

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    Staff: Mentor

    OK.
    You mean to say that before the string is cut, the net force on the top mass is: T - 2mg. And once the string is cut, the net force becomes just: -2mg.
    Looks good to me.

    Prior to the string being cut, the net force on the lower mass is: kx - mg. Once the string is cut, what happens to the spring force? Hint: What's the extension of the spring at that moment?

    [Note: I merged your two threads under the new title; I will delete your second post.]
     
    Last edited: Mar 17, 2005
  4. Mar 17, 2005 #3
    for mass 2

    ma = kx - mg
    but kx = T - mg from the first part
    ma = T - mg - mg = T - 2mg
    but T = 2mg
    ma = 2mg - 2mg = 0
    a = 0
    is this good enough??
     
  5. Mar 17, 2005 #4

    Doc Al

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    It's not clear whether you are talking about before the string is cut or after. (T = 0, after the string is cut.)

    Answer my question: What is the spring force just after the string is cut? (What was it just before the string was cut?)
     
  6. Mar 17, 2005 #5
    the spring force just after the string is cut is -mg for the top mass and +mg for the bottom mass since the spring will try to acheive equilibrium
     
  7. Mar 17, 2005 #6

    Doc Al

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    Exactly right. Just after the string is cut, the net force (on each mass) is -2mg on the top mass and 0 on the bottom mass. Which makes sense, since the net force on the entire system (masses plus spring) is just the weight of the system = -2mg.
     
  8. Mar 17, 2005 #7
    ok that makes sense now
    can you help with the third part that i added in post #1

    thanks for the help so far!! i appreciate it!
     
  9. Mar 17, 2005 #8

    Doc Al

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    The two masses execute simple harmonic motion about the center of mass (which is exactly in the middle, of course). Hints: (1) The masses start with maximum amplitude. (2) Treat each mass as being at the end of its own spring (What's the spring constant of half a spring?)
     
  10. Mar 17, 2005 #9
    for half a spring the spring constant is 2k??

    because half th spring is going to exert a force of kx should be 2k x/2


    so the SHM for each mass would be

    [tex] m \frac{d^2 y}{dt^2} + 2ky = F cos( \omega t) [/tex]
    solving the homogenous DE [tex] y = e^{ \lambda t} [/tex]
    [tex] \lambda = i \sqrt{(\frac{2k}{m})}[/tex]
    Using Mr Euler's Identity [tex] Re(y) = Cos \sqrt{\frac{2k}{m}} t[/tex]

    and i learnt this is the ODE course so i know taht Re(y) is also a soltuion to the non homogenous equation

    i see how they got hte cos part
    but how did they get the mg/2k term??
     
    Last edited: Mar 17, 2005
  11. Mar 17, 2005 #10
    i'm really baffled :confused: certainly [tex] y = cos \sqrt{\frac{2k}{m}} [/tex] is the position of the mass if there was no freefall involved. If there is a freefall not only is there that acceleration but also another acceleration taht pulls it downwards

    that mg/2k seems to have been derived from the earlier partr of this question since Fnet = mg - 2kx but how does that related to the position of the mass itself when in freefall??
     
  12. Mar 17, 2005 #11
    In post #9 you have equation for SHM. Right hand side should be mg for this case. You have already found the complementory function (solution to the homogen DE). Find also the particular solution. That will look like gm/2k. Combine both to get the complete solution. You may have to use boundary conditons at the end.
     
  13. Mar 18, 2005 #12

    Doc Al

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    Right!
    Excellent!
    What's the amplitude of the motion? Initially, the stretch in the spring is X = mg/k. So L + mg/k is the initial total separation between the two masses; so the amplitude of the motion of each mass about its equilibrium point is mg/2k.
     
  14. Mar 18, 2005 #13
    i tink i finally solved it

    for the oscillation only

    [tex] x(t) = A cos(w)t [/tex]
    where A is some amplitude
    this mass starts off from L/2 position with respect tot he CM so that is the initial condition
    [tex] z = \frac{L}{2} + \frac{mg}{2k} Cos \sqrt{\frac{2k}{m}} t [/tex]
     
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