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Angular momentum question

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A small object with mass 3.20 kg moves counterclockwise with constant speed 6.30 m/s in a circle of radius 4.70 m centered at the origin. It starts at the point with position vector (4.70 i+ 0j ) m. Then it undergoes an angular displacement of 8.00 rad.
    What it its position vector?
    In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
    What is its velocity?
    What is its acceleration?
    What total force is exerted on the object?


    2. Relevant equations
    i believe its the kinematics equations just replaced with angular speed and acceleration variables



    3. The attempt at a solution
    honestly..i think i should know how to do this but i cant seem to figure it out
    any lil start would help..thanks
     
  2. jcsd
  3. Feb 21, 2009 #2

    Doc Al

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    Staff: Mentor

    How many radians in a circle?
     
  4. Feb 21, 2009 #3
    umm isnt like 2pi radians
     
  5. Feb 21, 2009 #4

    Doc Al

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    Staff: Mentor

    Yes. Use that fact to figure out where the object ends up on the circle after it moves 8.00 radians.
     
  6. Feb 22, 2009 #5
    ok so it goes around 4 times i guess?
    i still dont understand
     
  7. Feb 22, 2009 #6
    8 much less than 4*2Pi.
     
  8. Feb 22, 2009 #7
    ok so i did 8rad/2pi = 1.27
    what do i do with this information?
     
  9. Feb 22, 2009 #8
    So the object has undergone 1 full revolution and .27 of a full revolution, which means that the rotation of the object from its original location is .27*2Pi radians. Can you see why and how to get the resulting vector? It is very important to draw a picture since this is the first time you are doing this.
     
  10. Feb 22, 2009 #9
    hmm ok so im pretty sure this means the particle is in the second quadrant right?
    so .27*2Pi = 1.70...im not quite sure how to get the resulting vector tho...is it just 4.7*1.70?
     
  11. Feb 22, 2009 #10
    4.7 is the length of the vector along the x-axis, while 1.7 is the angle the vector we are looking for makes with the x-axis. Why are you multiplying a length with an angle?
    Draw a picture. We want to rotate the vector 4.7i + 0j, which luckily lies along the x-axis, 1.7 radians. The length of the vector is 4.7, so by the definition of sine and cosine, this gives us 4.7*sin(1.7) for the y-component and 4.7*cos(1.7) for the x-component. Remember that the 1.7 is in radians, not degrees. You will have to translate that if you are using a calculator that only works in degrees.
    If you have drawn a diagram, you might also notice that this is the same as 4.7*sin(Pi - 1.7) and -4.7*cos(Pi - 1.7).
     
  12. Feb 22, 2009 #11
    wow okay that makes so much more sense
    so for x component = -.61
    y component = 4.67

    how do i do part b
    i think part c for xcomponent its -6.30 m/s but i dont know the y component
     
  13. Feb 22, 2009 #12
    You already know the angle it makes from part a. From the signs of the components, or from your knowledge of angles, you should know the quadrant.
    Draw a diagram. The tangential velocity vector starts at the tip of the position vector for the particle, is perpendicular to the position vector, and you have its length of 6.3. Getting the components is geometry/trigonometry similar to what we did in part a.
     
  14. Feb 22, 2009 #13
    ok cool i think i got those parts anyways..but thanks for explaining them

    how about e and f..any hints
     
  15. Feb 22, 2009 #14
    You need to use the definition of centripetal force.
     
  16. Feb 22, 2009 #15
    o ok..well that means that a = v^2/r
    so 6.3^2/4.7 = 8.44m/s^2 right..how do i put that in x and y components

    and for f its just mv^2/r right?
     
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