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Homework Help: Speed/energy problem, could someone please check my work

  1. Sep 7, 2007 #1
    1. The problem statement, all variables and given/known data

    when a car has travelled 100m up a 1 in 10 (sine slope) the driver stops and has a drink, he places an empty can of cola on the ground so that it can roll, the can weighs 50g and the radius of rotation k of the can is r

    a, calculate the energy of the can at the top and bottom of the slope and the form it takes

    b. calculate the speed of the can at the bottom of the slope

    2. Relevant equations

    Pe = MGH

    3. The attempt at a solution

    a, because the car has travelled 100m up the slope i calculated the height that the car was at. 100*1/10 = 10m

    50 grams = 0.050gk

    potential energy = mgh = 0.050*9.8*10 = 4.905Joules,

    answer to a = potential energy at the top of 4.905Joules, this energy is changed to kinetic energy at the bottom and is also 4.905Joules

    b, Kinetic energy gained ...E=1/2mv^2+1/2Jw^2

    = E=1/2mv^2+1/2mk^2*w^2... where m = mass and K= radius of gyration

    motion without slip w=v/r
    = E=1/2mv^2+1/2mk^2*(v^2/r^2)

    because the question states that the radius of rotaion of the can =r, I set these terms to 1

    therefore E=1/2mv^2+1/2m*v^2


  2. jcsd
  3. Sep 7, 2007 #2


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    Everything looks good except I think should be J = 1/2 mk^2... I think you made a mistake when substituting this into 1/2 Jw^2. that'll change you numbers afterwards.
  4. Sep 7, 2007 #3
    thanks for your reply, not quite sure what you mean though.
  5. Sep 7, 2007 #4


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    Can you describe how you got from the first line to the second...
  6. Sep 8, 2007 #5
    yeah, point taken, yet another silly mistake that I am prone to making
  7. Sep 8, 2007 #6
    having a look thrugh a text book it turns out that I am not sure what is correct.

    the book says E=1/2mv^2+1/2Jw^2 = E=1/2mv^2+1/2mk^2*w^2

    please comment
  8. Sep 8, 2007 #7


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    I apologize munkachunka. I was wrong. :redface: The textbook is right. The moment of inertia of a thin hollow cylinder is I = mr^2 (not I = 1/2 mr^2 as I thought).

    So everything in your initial post is correct. Sorry about that.
    Last edited: Sep 8, 2007
  9. Sep 8, 2007 #8
    no problem, thanks again for your help
  10. Sep 8, 2007 #9


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    you're welcome.
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