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Two blocks connected by springs moving in a circle

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Block A and block B are on a frictionless table as shown in Figure 2. Spring 1 connects block A to a frictionless peg at 0 and spring 2 connects block A and block B. Their masses are respectively mA=0:45 kg and mB=0:32 kg. When the blocks are in uniform circular motion about 0, the springs have lengths of lA=0.80 m and lB=0.50 m. The springs are ideal and massless, and the linear speed of block B is 2.8 m/s. If the distance that spring 2 stretches is 0.070 m, determine the spring constant of spring 2.


    2. Relevant equations
    F=ma
    a=v2/r
    F=kx

    3. The attempt at a solution
    y-direction is perpendicular to surface and x-direction is parallel to surface
    rA=lA, rB=lA+lB
    T1=Tension is spring 1, T2=Tension in spring 2
    NA=Normal acting on block A, NB=Normal acting on block B
    xB=0.07m (The extension of spring 2)

    For block A (From FBD)
    Resolving (y-direction): FAy=mAaAy=NA-mAg=0

    Resolving (x-direction): FAx=mAaAx=T1-T2

    ⇒(mAvA2)/rA=T1-T2


    For block B (From FBD)
    Resolving (y-direction): FBy=mBaBy=NB-mBg=0

    Resolving (x-direction): FBx=mBaBx=T2

    ⇒(mBvB2)/rB=T2


    And this is where I have become stuck, because I know I can't simply say T2=kxB as this would give the spring constant for a single spring that has extended to a total length of 1.3m. And continuations from this point do not seem to go anywhere e.g. substituting centripetal force of block B into the equation for block A, does not help as far as I can see.
     
  2. jcsd
  3. Oct 18, 2015 #2

    haruspex

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    why would it do that? T2 is the tension, xB is the extension.
     
  4. Oct 18, 2015 #3
    I thought of it originally, but then I thought it couldn't be because it doesn't require me to use anything from block A, and the block B would be under the same tension at that distance with a single spring, at least I think so. But surely it can't be as simple as to just equate (mBvB2)/rB with kxB?
     
  5. Oct 18, 2015 #4
    I think I managed to convince myself that (mBvB2)/rB=kxB was wrong because it seemed too easy and didn't involve values from spring 1 or block A.
     
  6. Oct 18, 2015 #5

    haruspex

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    I don't see how string 2 could tell the difference if string 1 and block A were replaced by an inextensible massless string of the same (current) length.
     
  7. Oct 19, 2015 #6
    Thank you for clearing that up for me in my mind. I've been doing this question for the past three days and I kept thinking to myself that the answer couldn't be as simple as I thought it was.
     
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