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Center of gravity of a stack of coins

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A penny has a mass of 2.5 g and is 1.5 mm thick; a nickel has a mass of 5.7 g and is 1.9 mm thick.

    If you make a stack of coins on a table, starting with five nickels and finishing with four pennies, how far above the tabletop is the center of gravity of the stack?

    2. Relevant equations

    Y center of gravity = (y1m1+y2m2+y3m3 etc) / (m1+m2+m3 etc)

    3. The attempt at a solution

    Ycg = (5*(5.7*1.9)+4*(2.5*1.5)) / ((5*5.7) +(4*2.5))
    Ycg = 1.80 mm

    This answer doesn't make sense because the entire stack is roughly 15 mm high... What am I doing wrong?
     
  2. jcsd
  3. Nov 11, 2009 #2

    hage567

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    Homework Helper

    I'm just looking at this quickly so hopefully I'm not missing something, but the equation says Y_cg = (y1m1 + y2m2+...)/(m1+m2+...), those y's are coordinates. So the way you've done it, each coin has its centre of mass at the same place (so all the nickels are at (0, 1.9) and all the pennies are at (0, 1.5). Draw it out, so you can see the positions in the stack.
     
  4. Nov 29, 2009 #3
    Yep that was it :D Awesome! Thanks for the help.
     
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