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Calculate the spring constant k

  1. Jan 25, 2010 #1
    hi,
    i am preparing for the test and have the following question.
    as usual i don't see where my mistake is.
    an image is attached.

    A thin uniform rod has mass M = 0.5 kg and length L= 0.55 m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The rod is released from an angle θ1= 63.0o, and moves through its horizontal position at (B) and up to (C) where it stops with θ2 = 105.0o, and then falls back down. Friction at the pivot is negligible. Calculate the speed of the CM at (B).

    this i found. 1.023 m/s

    The spring in (A) has a length of 0.11 m and at (B) a length of 0.14 m. Calculate the spring constant k.

    ok so i decided to use a and b to calculate.
    i decided to use the height of the rod in b as hight zero.
    this means that in a the system has two energies. the potential and the spring.
    the equation of energy conservation is:
    mgh+0.5kx^2=0.5mv^2
    thus
    -0.5*9.81*0.55/2*cos(63)+0.5*k*0.03^2=0.5*0.5*1.023^2
    but it is wrong.

    thanks
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2010 #2
    ok.
    so i managed so solve it but with a and c and not a and b.
    this means of course that i calculate the kinetic energy in b wrong.
    can u help and explain how it should be done?
    just 0.5mv^2 or just 0.5Iw^2 or their sum?
    what do i do when

    thanks a lot
     
  4. Feb 7, 2010 #3
    would it be fair to say
    that it has only rotational energy becuase it is fixed at one end?

    thanks
     
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