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Lawsuit II

  1. Apr 11, 2009 #1
    As you approached the I-74 exit in your uncle's little red sports car convertible you happened to notice a cat crossing the road. Not that you have any particularlove of cats (especially this one whose snooty disposition seems to beckon for a game of chicken) you swerve out of the way spilling scalding hot coffee all over your new summer duds.

    Unknown to you at the time, the auto mechanic, having had an uneventful social life and being consequently somewhat incoherent, had inadvertently forgotten to fasten the lug nuts on one of your uncle's tires.

    Given your velocity on the exit ramp as 2.8 m/s, a tire mass of 10 kg and radius 30 cm;

    (a) What is the angular momentum of uncle's tire immediately upon its departure? (Assume the tire is a solid cylinder) in kg m2/s?


    (b) Next, the tire cruises up an embankment at 30° for 1 meter beforelaunching across Piranha Lake. What is the velocity of the tire at the topof the embankment in m/s?

    not sure where to start for this one
     
  2. jcsd
  3. Apr 11, 2009 #2
    i figured out that angular momentum is 4.2
     
  4. Apr 11, 2009 #3
    Inital KE of the tire is 1/2*M*V2 + 1/2*I*(V/R)2; both rotational and translational
    Final KE = Initial KE - M*g*h where h = 1m * sin(30°).

    i just plugged in the numbers and got 2.42...wrong.
     
  5. Apr 11, 2009 #4

    tiny-tim

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    Hi mattmannmf! :smile:
    Nooo … how did you get that? :confused:
     
  6. Apr 11, 2009 #5
    its rmvsin (30), i just plugged in the numbers and got 4.2...im sure its correct because it told me it was correct
     
  7. Apr 11, 2009 #6

    tiny-tim

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    No, the 30º is in part (b) …

    this is for part (a), where the road is level …

    you have the speed of the tyre, so what is the angular momentum?
     
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