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Kinetic Energy and Rolling motion

  1. Oct 22, 2007 #1
    A 500g, 8.0-cm-diameter can rolls across the floor at 1.0 m/s. What's the can's kinetic energy.

    I solved this using:

    Code (Text):
    K = [tex]\frac{1}{2}[/tex]*I*([tex]\frac{v}{r}[/tex])[tex]^{2}[/tex]
    and for some reason I get 1/3 the answer. So I multiplied it by 3 at the end. Is there something wrong with the formula I'm using?


    If i do the same thing with a similar problem I don't get the right answer. I think the difference lies in the 1.50 m/s.

    Similar problem:
    Code (Text):

    A 600 g, 6.50-cm-diameter solid cylinder rolls across the floor at 1.50 m/s.
     
  2. jcsd
  3. Oct 22, 2007 #2
    Can you show the formula u used for I(in terms of r) ?
     
  4. Oct 22, 2007 #3
    Oops sry,

    I = (.5*m*r^2)
     
  5. Oct 22, 2007 #4
    This is the value of I about COM, the cylinder is not rolling about the COM.
     
  6. Oct 22, 2007 #5

    rl.bhat

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    Since the can is hollow, you must know the inner and outer diameter of the can to find the MI.
     
  7. Oct 22, 2007 #6

    rl.bhat

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    If the walls of the can are very small, the moment of inertia = MR^2
     
  8. Oct 22, 2007 #7
    Hey karnataki, specify the axis (coz here I is 3/2MR^2)
     
  9. Oct 23, 2007 #8

    rl.bhat

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    Moment of inertia of a hollow cylinder about the axis parallel to the axis of cylinder is = M(R1^2 + R2^2)/2
    If the can is thin, in that case R1 is nearly equal to R2.
    and MI = MR^2. You have taken the axis tangential to the cylinder and parallel to the axis of the cylinder.
     
  10. Oct 23, 2007 #9
    Thats the axis along which pure rotation is taking place, I suppose.
     
  11. Oct 23, 2007 #10

    rl.bhat

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    K.E. of rolling can = 1/2*I(com)(v/r)^2 + 1/2*Mv(com)^2
    = 1/2*Mr^2(v/r)^2 + 1/2*Mv(com)^2
    = 1/2*Mv^2 +1/2*Mv^2
    = Mv^2
    In such problems radius of the object is not needed.
     
  12. Oct 23, 2007 #11
    Hey I(com) = 1/2Mr^2.
     
  13. Oct 23, 2007 #12
    Just mv^2 does not work.


    My buddy has a different values and he got 0.443 for the question:
    Code (Text):
    A 500g, 8.0-cm-diameter can rolls across the floor at 1.0 m/s. What's the can's kinetic energy.
     
  14. Oct 24, 2007 #13

    rl.bhat

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    I (com) = 1/2(MR^2) is true for solid cylinder.

    Total KE of rolling body = KE of translation + KE of rotaion
    If possible go through Halliday & Resnic Book of Physics
     
  15. Oct 24, 2007 #14
    Oh ya, I'm sorry for the confusion.
    So, K = Mv^2 = 0.5 * (1)^2 = 0.5 J

    and for the other ques. K = 0.6 * (1.5)^2 = 1.35 J
    Are your numerical values same, Nitrag?
     
  16. Oct 24, 2007 #15
    Yes, I got 1.35 J also but it is not correct.
     
  17. Oct 24, 2007 #16

    rl.bhat

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    For the other question K = 1/2*1/2*(MR^2)(V/R)^2 + 1/2*MV^2
    = 1/4MV^2 + 1/2MV^2 =3/4MV^2 = 1.0125 J
     
  18. Oct 25, 2007 #17
    ah, I see the connection now. thanks for the help!
     
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