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Circular Motion Problems

  1. Oct 8, 2005 #1
    I think I did the problem(s) right but my answers doesn't match up with the correct answer.

    1) Suppose a space shuttle is in orbit 400km from the earths surface, and circles the earth once every 90 minutes. Find the centripital acceleration of the space shuttle's orbit. Express your answer in terms of g, the gravitational acc. or the earth's surface.

    Okay, so my approach:
    First change 400km into m = 40,000m & 90min = 5400s

    So since a=v^2/r, I will find velocity first. So I used Vel=2(pie, 3.14)(r)/t
    (2 x 3.14 x 40,000m)/5400s = 465.4 m/s

    Then I took a=v^2/r. So ... a=465.4^2/40,000m and got .541 m/s^2

    Then convert into g's, .541/9.8 = .06 g
    But the correct answer is 0.9 g's What am I doing wrong!? :(


    2) What is the max speed w/ which a 1050-kg car can round a turn with a radius of 77m on a flat road if the coefficent of static friction b/t tires and road is 0.80?

    What I tried to do was find my acceleration so I could get my velocity but I have no idea how to do that? Since the velocity is not given... Can anyone help me? Correct answer: 25 m/s
     
  2. jcsd
  3. Oct 8, 2005 #2
    The shuttle is 400km from the surface of the earth. You need to find the distance from the center of the earth.

    Alex
     
  4. Oct 8, 2005 #3

    mezarashi

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    You forgot to add in the Earth's radius of about 6500km !!! That makes a big difference. Try it again.

    First of all, find the maximum frictional force that can act on the 1050kg car given the coefficient of 0.8. Now with that find the corresponding acceleration possible. From there you know the acceleration is radial, so you can apply it to circular motion :)
     
  5. Oct 8, 2005 #4
    That worked... but how was I suppose to know the radius of the earth?
     
  6. Oct 8, 2005 #5

    mezarashi

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    Haha, I'd guess it would more or less become common knowledge sooner or later. Physics major? Look at the back of your textbook's cover, such figures are usually available or else, try googling: Earth's diameter

    In anycase, it should have been clear that that information was missing.
     
  7. Oct 8, 2005 #6
    Umm... Sorry, this is my first semester of Physics. I am just trying to get through.
     
  8. Oct 8, 2005 #7
    Yep I got an a=0.95 g's. Now I'll try the other problem.
     
    Last edited: Oct 8, 2005
  9. Oct 8, 2005 #8

    mezarashi

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    Check your math: 6900km = 6,900,000m (just for the record, the better estimate for the Earth's radius is 6378km for an equatorial orbit)

    The answer of 0.9g is very reasonable. This means that moving 400km off the surface of the Earth, the gravitational force will feel will be 0.9 times what we feel here on the surface.
     
  10. Oct 8, 2005 #9
    Okay, now for the second problem. I have finished the first part but am still stuck... Okay...

    I get the max tension/force by first getting the Normal force like so...

    (1050kg) x (9.8 m/s^2) = 10290 N

    Then I take my static coeffiecent of .80 and times it by 10290N
    (.80) x (10290N) = 8232 N as my max frictional force

    But how can I get the velocity from this? The equations don't seem to work...
     
    Last edited: Oct 8, 2005
  11. Oct 8, 2005 #10
    Now how do I get the velocity? :S
     
  12. Oct 8, 2005 #11

    mezarashi

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    The equations don't work? Think carefully. What is this frictional force working against. Or rather, what does this frictional force represent? Why does it exist? We're talking about circular motion and centripetal acceleration here remember.
     
  13. Oct 9, 2005 #12
    So, I cannot use v= square root of (F/m x r) ?
     
  14. Oct 9, 2005 #13

    lightgrav

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    You might need to use parentheses in your calculator ... .

    by the way, the first problem can also be answered using
    gravity field strength g proportional to 1/r^2 :
    a/"g_surface" = R^2 / (R+h)^2
     
  15. Oct 9, 2005 #14
    So the equation will work for problem 2?
     
  16. Oct 9, 2005 #15

    lightgrav

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    Look, the only horizontal Force, the only one with a center-pointing component, is the friction Force. The sum of the central Force components provides m with a_central , which is v^2/r . Just like ANY OTHER known component of acceleration is caused by that Force component Sum.

    The diagrams are ALWAYS the key to Physics problems ...
    Your diagram should've shown the satellite with a circular path that had a radius of R+h. That's how you know to look up R inside the book cover, and suggests that you could solve the problem by looking at gravitational Force.

    This Free-Body Diagram is simple enough that you should have a lot of confidence in your result ... a = (.8)(9.8 m/s^2) = v^2/r
     
  17. Oct 9, 2005 #16
    I still do not understand, but thanks for trying.
     
  18. Oct 9, 2005 #17

    lightgrav

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    If you isolate "v" in the equation: a = v^2/r ,
    yes, you get v^2 = ar , so ... v = sqrt(ar).
    F/m ... which you had long back ... IS "a" .
     
  19. Oct 9, 2005 #18
    So it would be sqrt (F/m) x r ) (8232 N/1050kg) x 77 = 24.5 = 25.0 ! It worked!
     
    Last edited: Oct 9, 2005
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