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Homework Help: Center of mass of necklace

  1. Mar 16, 2005 #1
    Been trying this one using:
    x1m1 = x2m2

    I guess I just maybe am not using the right mass or length for center of mass..

    Here's the question:

    A 0.19 kg meter stick balances at its center. If a necklace is suspended from one end of the stick, the balance point moves 16.3 cm toward that end. What is the mass of the necklace?
  2. jcsd
  3. Mar 16, 2005 #2
    You need to use calculus, not physics. Look up center of mass of a rod on google.

    I don't remember the formulas, that was like two months ago.
  4. Mar 16, 2005 #3
    don't mislead him, no calculas is needed in this problem, the center of mass of a rod is at its mid-point....
    here is some hint:
    draw the graph first, and find the torque for LHS and RHS... the torque of the necklace is straight forward, but the torque for the rod is a bit tricky...
  5. Mar 16, 2005 #4

    Andrew Mason

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    The stick has a mass/length of .19 kg/m or 1.9 g/cm.=[itex]\rho[/itex] The torque when the necklace is added and the balance point is moved is 0:

    [tex]M_nd_1 + \rho d_1^2/2 - \rho d_2^2/2 = 0[/tex]

    [itex]d_1 = 33.7[/itex] and [itex] d_2 = 66.3 cm[/itex]

    (you can think of the torque from one side of the stick as a point mass sitting half way between the end and the balance point on a massless stick).

  6. Mar 16, 2005 #5
    There's no need to divide the rod into 2 pieces and to calculate the torque of each of them.

    This problem is as simple as 1,2,3. :zzz:


    Write the equation of the torques around the new ballance point:


    That's it!
    Enjoy physics!
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