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Cylinder Rolling Down an Inclined Plane

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    A 3.0kg solid cylinder (radius=.15m, length=.7m) is released from rest at a top of a ramp and allowed to roll without slipping. The ramp is .9m high and 5m long. Find the rotational and translational kinetic energy.

    2. Relevant equations

    krot=1/2Iw^2
    Ktrans=1/2mv^2

    3. The attempt at a solution

    I already determined that the final kinetic energy is 26J

    v=rw
    v=(.15)w
    Ktrans=1/2(3)(v^2)
     
  2. jcsd
  3. Mar 30, 2012 #2
    i set up
    mgH=1/2mv^2+1/2Iw^2
    and got
    26=1/2(3)v^2+1/2(.1225)w^2
    26=.03375w^2+.06125W^2
    26=.095w^2
    w=16.54
    with this i got
    Krot=16.76
    but it is saying this is not correct
     
  4. Mar 30, 2012 #3
    Wrong moment of Inertia!!
     
  5. Mar 30, 2012 #4
    would it be 1/2 MR^2
     
  6. Mar 30, 2012 #5
    If i tell that, then i would solve your homework.
    Think about it, how the cylinder will roll down (about which axis)?
     
  7. Mar 30, 2012 #6
    i got the rotational one and plugged in the w and r into v=rw
    then squared v and multiplied by the mass
    and divided by two, but i didn't get the right answer
     
  8. Mar 30, 2012 #7

    ehild

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    Homework Helper
    Gold Member

    The moment of inertia of the cylinder with respect to the axis through its centre is 1/2 MR^2. It is not 0.1225. How did you get that value?

    (The in a plane motion of a rigid body is equivalent with a translation of its CM and rotation around the axis through the CM.)

    ehild
     
  9. Mar 30, 2012 #8
    And can you tell us what value of g you used?
    I don't get 26 neither by 10 nor 9.8.
     
  10. Mar 31, 2012 #9
    i used 9.8 and it has already told me that Kf=26 and Krot=8.8 are correct.
    using this, i used v=rw to get v=3.399 and plugged that into 1/2mv^2
    doing that i got 17 but it is saying this is incorrect.
     
  11. Mar 31, 2012 #10
    Seriously...why 17J is not correct answer? If total energy is 26.486J and rotational energy is 8.8J then the kinetic energy will be 26.486J-8.8J=17.687J~18J. I was always taught to use the value 9.81m/s^2 and not to use rounded values like 26J in calculations. :wink:
     
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