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Pendulum with magnet under

  1. May 31, 2012 #1
    1. The problem statement, all variables and given/known data
    this is an experiment from lab..
    i have pendulum - mass (iron) hanging from a massless cord.
    under the mass there is a magnet.
    the movement of the pendulum is in a small angle.
    i repeat the experiment 5 times - each time in a different distance between the iron mass and the magnet (the magnet is on a jack that can change its height).
    the motion is recorded to the computer (multilab software)

    the goal: to find from the time period (as recorded in the computer) the magnet's force as a function of the distance between the magnet and the iron mass

    2. Relevant equations

    T = 2[itex]\pi[/itex][itex]\sqrt{\frac{l}{g}}[/itex]
    this equation is for there is only mg present..
    im not sure how to add the magnetic force..

    3. The attempt at a solution
    im really stuck..
    ive made a force diagram with the magnetic force.. since its small angles i assumed the magnetic force is straight down (just like mg).. i also got rid of the sine [itex]\theta[/itex] from the same reason..
    im not sure how to use the distance between the magnet and iron mass as a variable..

    any help is appreciated
    thank you
     
  2. jcsd
  3. May 31, 2012 #2
    Hi StasKO!! Welcome to PF! :smile:

    [tex]T = 2\pi \sqrt{\frac{l}{a}}[/tex]

    Where a is the total effective acceleration downwards. So if you are in an elevator accelerating upwards with g', the effective acceleration would be (g+g') instead of g. Elevator going downwards would give (g-g'). Does this give you an idea?
     
  4. May 31, 2012 #3
    Continuing what Infinitum said, to calculate the acceleration due to the magnet ONLY, think about the force exerted on the pendulum bob.
     
  5. May 31, 2012 #4
    first of all thank you for your welcome.. this forums helped me a lot..

    back to the issue..
    so as i understand it, i need to replace the g in the equation to a general 'a' which represents the total acceleration downwards. this 'a' is the sum of g and the acceleration due to the magnet..
    now i can find the acceleration only due to the magnet... and with newton's 2nd law i can find the magnetic force:

    Fmagnetic = miron × adue to magnet

    now if this is correct, then how do i change this equation to be dependent on the distance between the magnet and the iron instead of the acceleration?
     
  6. May 31, 2012 #5
    I believe you are given the magnetic strength/pole magnitude? If so, do you know the formula that gives you the magnetic force between the two(its an inverse square relation!)? From that force, calculate the acceleration and use time equation.
     
  7. May 31, 2012 #6
    no.. i didnt recieve any info about the magnet itself..
    i know that i need (eventually) build a graph in excel.. from the slope of that graph i need to extract the magnetic force..
    so if the graph equation is y=ax+b then y = T^2 and x should be the distance (something along this line).. a, which is the slope should give me the magnetic force..

    the thing is that i need to find this connection between the period and the distance between the magnet and iron.. this is where im stuck :(

    edit: the only thing i know about the magnet its that it attracts the iron...
     
  8. May 31, 2012 #7
    Hmm, Im not completely sure of this. But it does seem to be very interesting. Here's something you can try, for two different pendulum lengths,

    [tex]T_1 = 2\pi \sqrt{\frac{l_1}{g+a_1}}[/tex]

    [tex]T_2 = 2\pi \sqrt{\frac{l_2}{g+a_2}}[/tex]

    Now taking ratio of T and sqrt(l) you get [itex]\sqrt{(g+a_2)/(g+a_1)}[/itex] From here, get a relation between the magnetic forces.
     
  9. May 31, 2012 #8
    well.. the thing is that the pendulum's length is not changing..
    its the magnet that changes its position (only vertically).. the magnet is on a jack..

    it might help to tell you that we had a similar experiment but with no magnet.. only pendulum.. so we measured the period T and length l... we repeated the experiment with different lengths l..
    we squared the equation of period time to get a linear relationship between the period T and length l... we marked - y = T^2 and x=l... the coefficient a=(4*pi^2)/g..
    we built the graph in excel and managed to extract the g to get its value from the slope which is 'a'...

    now its the same procedure but with a magnet (and this time g is given)..
    and now instead of finding g (through the relationship between period and length) i need to find the magnetic force (through the relationship between period and distance of magnet and iron mass)..

    edit: i have the numbers for the different periods with respect to different distances of magnet and iron.. im stuck at finding the connection between them lol
     
  10. May 31, 2012 #9
    Ah, yes. Let the length of pendulum be the same in the above two equations, then. This means taking their ratios will give you

    [tex]\frac{T_1}{T_2} = \sqrt{\frac{g+a_2}{g+a_1}}[/tex]

    Now, you know all the other values, hence you can find a relation between a1 and a2. Let these be done at different heights from magnet r1 and r2. From measurements say,

    [tex]r_1 = kr_2[/tex]
    and from calculations,
    [tex]a_1 = pa_2[/tex]

    For some constants k and p. So a distance of r1 gives acceleration a1, and a distance of (r1/k) gives an acceleration of (a1/p). Observe data's for yet more values of r and T. See what happens to the relations between force and distance and generalize it to obtain a conclusion.

    PS : This is only an idea, which I think would work. I don't see why it shouldn't :rolleyes:
     
    Last edited: May 31, 2012
  11. May 31, 2012 #10
    by 'm' you mean a coefficient and not mass right?

    and instead the a's i can put (magnetic force)/(mass of iron) right?

    i had a thought and not sure if its true..
    is it possibe that instead of the length l of the pendulum i'll put the total length - pendulum + distance to magnet?
     
  12. May 31, 2012 #11
    I edited the m thinking that it might cause confusion :wink: its p now, and yes, a coefficient.

    Yep!

    Nope, the total pendulum length still remains the length from the point of suspension to the center of mass of the bob.
     
  13. May 31, 2012 #12
    im still not sure i really understand your solution...
    on what basis do you say that the ratio between two distances is equal to the ratio between two (corresponding) accelerations?
     
  14. May 31, 2012 #13
    Why not? :confused:
     
  15. May 31, 2012 #14
    because it should be derived from some kind of equation i think..

    can i say that the period T is proportional to the distance r so that T=kr for some constant k?

    then find k by building a linear graph (y=T, x=r) and determining k from the slope of that graph..
    then plug T=kr into the time period equation.. and solve for the magnetic force thus creating a function of the magnetic a force as a function of the distance r..

    does this even makes sense?
     
  16. May 31, 2012 #15
    Actually, the acceleration ratio and distance ratio are equal. I know this example, isn't related in anyway to this problem, but just for intuition,

    $$ \frac{s_1}{s_2}=\frac{\frac{1}{2}a_1t^2}{\frac{1}{2}a_2t^2}=\frac{a_1}{a_2} $$

    Once again, the above example is NOT related to the problem, but it is just meant for intuition.
     
    Last edited: May 31, 2012
  17. Jun 1, 2012 #16
    But my procedure is derived from an equation...

    Hmm this would work, IF the time period was directly proportional to r. Try out experimenting a few times, and see if the relation is a linear one.

    Didn't say that. I derived

    [tex]\frac{T_1}{T_2} = \sqrt{\frac{g+a_2}{g+a_1}}[/tex]

    from which, I found the relation between the two distances by measurement.

    [tex]r_1 = kr_2[/tex]

    and the relation between accelerations was derived from the time period equation. By multiple observations, you can find the relation between r and a.
     
  18. Jun 1, 2012 #17
    ohh ok... now i got it lol

    anyways i think now i have enough info and tools to solve this experiment.. now im gonna sit on my as* and finish this...

    thank you so much for the brain storming.. it is much appreciated (be ware i might be back for more questions lol)
     
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