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Work and a cylinder

  1. Nov 6, 2005 #1
    A homogeneous cylinder of radius 30cm and mass 40kg is rolling without slipping along a horizontal floor at 2.4m/s. How much work is needed to stop the cylinder?

    work = delta_KE in this case so oi figured .5 *40 * 2.4^2 = 115 J but this is not right, what should i do with the cylinder?
     
  2. jcsd
  3. Nov 6, 2005 #2
    i still can not figure this one out.
     
  4. Nov 6, 2005 #3

    lightgrav

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    Besides KE of the center-of-mass moving in translation (linear),
    you also have KE of the extended mass moving AROUND the c.o.m.

    KE_rotation = 1/2 I (omega)^2
     
  5. Nov 6, 2005 #4
    i did this and found omega to be 8 rad/s and I to be 1.8 then i found the KE to be 57.6 this is now right either.
     
  6. Nov 6, 2005 #5

    lightgrav

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    "Besides" the old linear KE there's now rotational KE, ALSO.
    The total KE is the SUM : KE_linear + KE_rotation
     
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