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Uniform Circular Motion

  1. Feb 6, 2006 #1
    here is the problem:
    ---
    Two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 1 cm from the center and the outer block is 6 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.75, and the string is taut.

    What is the maximum angular frequency such that neither block slides?
    ---
    the sum of all forces on each block = mass * centripetal acceleration.
    f1/f2 = friction on the outer/inner blocks, respectively
    m1/m2 = mass of the outer/inner blocks, respectively
    R1/R2 = radius of the outer/inner blocks, respectively
    T = tension

    so, f1 + T = m1 * a; f2 + T = m2 * a, which can also be written as...
    f1 + T = m1 * ω^2 * R1; f2 + T = m2 * ω^2 * R2

    solving for ω^2, i got...
    ω^2 = (F1 + T)/(m1 * R1);
    ω^2 = (F2 + T)/(m2 * R2).

    that's how far i've gone, and i'm pretty sure i've gone the wrong direction.
    will anyone please direct me to a right approach?
    thanks.
     
  2. jcsd
  3. Feb 7, 2006 #2
    It should be [tex] f_2 -T =m_2 a_2 [/tex]
    the direction of the friction force of the inner one is opposite to that of the tension force.

    Find the relation between T and mg, and you can find [tex] \omega[/tex]
     
    Last edited: Feb 7, 2006
  4. Feb 7, 2006 #3

    Fermat

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    At block1, f1 and T are both acting in the same direction, radially inwards, so (f1 + T) is correct. But at block2, f2 is acting radially inwards but T is acting radially outwards!

    Apart from the wrong sign, your approach looks ok so far.
     
  5. Feb 7, 2006 #4
    oh, i see.
    how can i relate T and mg? i'm not given a lot of information for the problem.
    instead, can i add the two equations to get rid of T?
    doing that, i got...
    F1 + F2 = ω^2 * (m1*R1 + m2*R2)
    i'm stuck there again...
     
  6. Feb 7, 2006 #5

    Fermat

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    You've got rid of the T, that's right.

    Now, can you relate the friction forces with the coefft of friction and the normal reaction of the the blocks ?
     
  7. Feb 7, 2006 #6
    well, the friction forces = fric. coeff. * the normal force.
    the normal force, however, depends on the mass of the block, doesn't it?
     
    Last edited: Feb 7, 2006
  8. Feb 7, 2006 #7

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    The turntable is horizontal (I expect!) so the normal reaction is the weight of each mass = mg. (Acting vertically downwards)

    At the moment you don't need to know m1 or m2, simply that they are equal masses.
    They will cancel out.
     
  9. Feb 7, 2006 #8
    yes, the turntable is horizontal =)
    for ω^2, i got...
    0.75(m1+m2)g
    --------------
    m1*R1 + m2*R2
    maybe i forgot my algebra, but is it possible to cancel m1 and m2 there?
     
  10. Feb 7, 2006 #9

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    m1 = m2!

    just cancel out top and bottom :)
     
  11. Feb 7, 2006 #10
    omg !!
    bleh ! >_<
    i'm an idiot.
    it's 2 am, after all... ^^

    thanks, guys.

    edit:

    i got 1.5mg = m(R1+R2)ω^2, got 1.4491 for ω, but it's still wrong!
    did i do something wrong?

    edit:

    bleh. i had to change the radius from cm to m.
     
    Last edited: Feb 7, 2006
  12. Feb 12, 2007 #11
    okay so i have the same exact problem, and i was able to find the angular frequency, however i am having trouble finding the tension! i tried plugging back in the angular frequency to the equations that we previously determined and nothing is working!!! can you tell me what you did in order to find the T?
     
  13. Feb 12, 2007 #12

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    If you got the angular frequency OK, then you should be able to get the tension by using one of the earlier equations. You should check your working. If you can't find the error, then post your working and someone will check it out.
     
  14. Feb 12, 2007 #13
    so from the previos problems we found that f1 + T=m1*w^2*R1 and f2-T= m2*w^2*R2. there were two unknowns in the equations, tension and the angular frequency. i know that to solve for the angular frequency we had to solve for w^2 get rid of the tension and simplify. so now to solve for the tension shouldn't i be able to use those previous equations to find T? i'm probably doing it wrong since everything i do comes up with a wrong answer. but what i did was solve for f so that f=m1*w^2*R1-T and f=m2*w^2*R2+T then i set those equal to each other and solved for T so that T= (m1*w^2R1-m2*w^2*R2)/2 and that still doesnt work...can you help me derive the correct equation to solve for the tension?
     
  15. Feb 13, 2007 #14

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    I don't see anything wrong with your working, except that R1 and R2 should be the other way arouind. i.e. T = (1/2)w²(m2R2 - m1R1)

    1) do you have the value for m1/2 ?
    2) do you have the answer ?
     
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