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Rotational Question

  1. Apr 1, 2010 #1
    1. The problem statement, all variables and given/known data
    An astronaut on the surface of the Moon fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume that the free-fall acceleration on the Moon is one-sixth that on the Earth. Radius of moon = 1.74 * 10^6 m
    a) What must be the muzzle speed of the probe so that it travels completely around the Moon and returns to its original location? Found out to be 1.68km/s
    b) How long does this trip around the Moon take?


    2. Relevant equations
    [tex]a=\omega^2r[/tex]

    [tex]v=\omega r[/tex]

    3. The attempt at a solution
    I tired to solve with constant acceleration formulas.
    a=1.62m/s^2
    vi=0
    vf=1.68km/s => 1678.928m/s

    vf=vi+at
    16878.928=0+1.62t
    t=10419.09136s

    2)A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 35.0m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

    I have tried to do the following:

    i know the static friction: 0.600

    i need to know [tex]\omega[/tex]

    which is [tex]\sqrt{9.8}{35}[/tex], [tex]\omega=0.529[/tex]

    find velocity

    [tex]v=r\omega[/tex]
    v=0.529 * 35
    v=18.52m/s

    i know this is incorrect, someone tell me how i can approach this.
     
    Last edited: Apr 1, 2010
  2. jcsd
  3. Apr 1, 2010 #2

    Doc Al

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    This is a circular motion/orbit problem. Use Newton's 2nd law and the formula for centripetal acceleration.
     
  4. Apr 1, 2010 #3
    how can i find the time with the centripetal acceleration formula?
     
  5. Apr 1, 2010 #4

    Doc Al

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    OK. Looks like you solved for the speed (but didn't show your work). The speed is constant. What distance does it travel in going around once?
     
  6. Apr 1, 2010 #5
    do you mean [tex]\omega = 9.649 * 10^{4} rad/s[/tex]?
     
  7. Apr 1, 2010 #6

    Doc Al

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    You could certainly use that if you like. But that should be 10-4!

    I was thinking of distance = speed * time. You have the speed. And you should be able to calculate the distance, then solve for the time.
     
  8. Apr 1, 2010 #7
    How can i find the distance if i only have velocity, acceleration?
     
  9. Apr 1, 2010 #8

    Doc Al

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    What path does it take? Hint: Make use of the moon's radius.
     
  10. Apr 1, 2010 #9
    would you find the diameter of the moon, which is 3480000?
     
  11. Apr 1, 2010 #10

    Doc Al

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    Answer this: What is the shape of the package's orbit around the moon?
     
  12. Apr 1, 2010 #11
    well the shape of a car....
     
  13. Apr 1, 2010 #12

    Doc Al

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    :bugeye:

    It goes all the way around the moon in a big circle!
     
  14. Apr 1, 2010 #13
    i misunderstood your question. -_-
     
  15. Apr 1, 2010 #14

    Doc Al

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    Do you understand what I mean now? How can you find the length of that path? (Remember that this is a question about circular motion.)

    (You could also solve for the time using ω, which you've already calculated.)
     
  16. Apr 1, 2010 #15
    yes, i understand it took me awhile -_-.

    So from the equation: [tex]\omega = \frac{\theta}{t}[/tex]

    derived from the linear [tex]v=\frac{x}{t}[/tex]

    where:

    [tex]\theta = 2\pi[/tex] because it go around once.

    [tex]\omega = 9.649 * 10^{-4}[/tex]


    [tex]t=\frac{2\pi}{9.649*10^{-4}}[/tex]

    t=6511.74765s

    [tex]t=\frac{6511.74765}{3600}[/tex] turn into hours

    t=1.81hrs however book's answer has a slight difference in 1.80hrs. Edit** i know why because of my rounding off.
     
  17. Apr 1, 2010 #16
    would someone answer my second question?
     
    Last edited: Apr 1, 2010
  18. Apr 2, 2010 #17

    Doc Al

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    You used the first equation, but you could also have used the second one directly. When it goes around once, the package traces the circumference of a circle. So [itex]x = 2\pi r[/itex].
     
  19. Apr 2, 2010 #18

    Doc Al

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    How did you solve for ω?

    Again, this is a circular motion problem that can be solved using Newton's 2nd law and what you know about centripetal acceleration.

    What force acts on the eggs? What's the expression for its acceleration?

    FYI: ac = ω²r = v²/r
     
  20. Apr 2, 2010 #19
    the mass and the centripetal force.

    Don't i need to use the static friction?
     
  21. Apr 3, 2010 #20

    Doc Al

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    Yes. Static friction is the only force available to provide the centripetal acceleration.
     
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