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Homework help part 2

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data
    According to the Guinness Book of world records. the highest rotary speed ever attained was 2010 m/s (4500 mph). the rotating rod was 15.3 cm (6 in.) long. assume that the speed quoted is that of the end of the rod.
    a) What is the centripetal acceleration of the end of the rod?
    b) if you were to attach a 1.0 g object to the end of the rod, what force would be needed to hold it on its own.


    2. Relevant equations
    ac=v^2/r
    fnet=m(ac)


    3. The attempt at a solution
    ac=2010m/s^2/7.56
    ac=528117 that just seems ridiculous for an answer.

    fnet=1.0g(528117)
    fnet=528117
     
  2. jcsd
  3. Oct 19, 2007 #2

    learningphysics

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    You didn't convert to m. Also, are you sure the rod is rotating about the center and not the end?

    I think you're supposed to use r = 0.153m
     
  4. Oct 19, 2007 #3
    ok so you think i should use r = 0.153m. So just wondering but what would be the difference between rotating about the center and not the end?
     
  5. Oct 19, 2007 #4
    ok so i got ac=2010^2/0.153
    ac=41312.970
     
  6. Oct 19, 2007 #5

    learningphysics

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    I'm actually not really sure if they want you to use r = 0.153m or r = 0.0765m. :redface:

    Just got the feeling they meant the distance from the pivot to the end of the rod is 0.153m... but I really don't know.
     
  7. Oct 19, 2007 #6

    learningphysics

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    how did you get that? I'm getting 2.64*10^7 m/s^2.
     
  8. Oct 19, 2007 #7
    Thats alright I can do them with both that way i cant go wrong thank you again. But do you mind showing me how you converted them.
     
  9. Oct 19, 2007 #8
    Im not sure how i got it. Now I get the 2.64*10^7 m/s^2. I must have just hit some button on accident on the calc.
     
  10. Oct 19, 2007 #9

    learningphysics

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    convert them? you mean the cm to m, g to kg?
     
  11. Oct 19, 2007 #10
    Yeah I should probably learn how to do it.
     
  12. Oct 19, 2007 #11

    learningphysics

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    no prob.

    [tex]1cm = 10^{-2} m [/tex]

    [tex]1g = 10^{-3} kg[/tex]

    I can get the inverse relations... multiply both sides of the first equation by 100. second by 1000.

    so:

    [tex]100cm = 1 m [/tex]

    [tex]1000g = 1 kg[/tex]

    so I can convert one way or the other...

    so [tex] 15.3cm = 15.3*(10^{-2}m) = 0.153m[/tex]

    what's really handy for converting units is the "factor label" method:

    http://en.wikipedia.org/wiki/Units_conversion_by_factor-label
     
  13. Oct 19, 2007 #12
    ok so the teacher said that we use r=0.153m so to find the centripetal acceleration. ac=v^2/r so to find velocity i do distance/time so would it be 0.153m/2010m/s. would that be how i find velocity. because i also have a circular motion equation for velocity which is v=delta r/ delta time
     
  14. Oct 19, 2007 #13

    learningphysics

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    I don't understand. We already got the answer. You said you got the same answer as me 2.64*10^7 in post#8.
     
  15. Oct 19, 2007 #14
    Yeah I remember now but yeah we used the whole diameter instead of the circumfrance.
     
  16. Oct 19, 2007 #15

    learningphysics

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    Hmmm... ac = v^2/r. Your teacher said we use r = 0.153 right?

    ac = 2010^2/0.153 = 2.64*10^7 m/s^2
     
  17. Oct 19, 2007 #16
    yep thats what i got thanks
     
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