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Circular Motion- Vertical

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A 1.62 kg object is swung in a vertical circular
    path on a string 1.8 m long.
    The acceleration of gravity is 9.8 m/s^2.
    If the speed at the top of the circle is 5 m/s,
    what is the tension in the string when the
    object is at the top of the circle?
    Answer in units of N


    2. Relevant equations
    ac=mv^2/r
    Fnet=ma

    ac= Centripetal acceleration
    3. The attempt at a solution


    Ac=mv^2/2
    ac=5^2/1.8
    ac=13.89

    (1.63kg=.00162g)
    Fnet=ma
    Ft+Fg=ma
    Ft+(9.8*.00162)=13.89(.00162)
    Ft=.006624

    But that force of tension does not work :(
     
    Last edited: Dec 15, 2011
  2. jcsd
  3. Dec 15, 2011 #2
    You have the right idea but you have mixed up your units!!!
    The mass is 1.62kg.... no need to change it to anything else !!
     
  4. Dec 15, 2011 #3
    you should keep the mass as 1.63kg
    btw 1.63kg is 1630g
     
  5. Dec 15, 2011 #4
    Keep everything in kg, m and s

    Do you know the answer?
     
  6. Dec 15, 2011 #5
    Wow. I feel so stupid. That is like one of the basic parts of physics. I got 6.66488889 which was right. Thanks for your help :)
     
  7. Dec 15, 2011 #6
    well done.... I got 6.3 but I used g = 10 to make life easy !!!!
    A little tip..... dont give so many figures in an answer... round it off to 3 figures/
     
  8. Dec 15, 2011 #7
    haha! If only it was that easy, we are assigned problems online through UTexas, and so the answer has to be exact to get the question right.
     
  9. Dec 15, 2011 #8
    Is this the physics you're taking in college?
     
  10. Dec 15, 2011 #9
    that does not matter.... the important thing is that you KNOW how to do it:smile:
     
  11. Dec 15, 2011 #10
    Nope, normal high school physics.
     
  12. Dec 15, 2011 #11
    Yea that's what I thought as well because its really simple.
     
  13. Dec 15, 2011 #12
    Can someone help me with part 3 of the problem?
    For the acceleration I got 10.97670251m/s^2
    and for the Net force I got 603.7186381 N

    I did
    fn-fg=fnet
    and got
    fn=1142.718638 and it was wrong
     

    Attached Files:

  14. Dec 15, 2011 #13
    I got the acceleration to be 10.8.... which is close enough to your answer.... I do not have a calculator..... I am using a slide rule:grumpy::mad::wink:
    Do you know what a slide rule is !!!!!!
     
  15. Dec 15, 2011 #14
    Is it a circular slide rule or a normal one?
     
  16. Dec 15, 2011 #15
    the acceleration and netforce is correct! and what is the slide rule?
     
  17. Dec 15, 2011 #16
    Uhm.. it's a mechanical analog calculator used for multiplication, division, and also logarithms and roots.
     
  18. Dec 15, 2011 #17
    Pluv
    What a great question.... what is a slide rule.... I do have a calculator but the battery has gone flat.... useless.
    A slide rule is a great device for doing multiplications and divisions..... be grateful that you will probably never need to resort to one.
     
  19. Dec 15, 2011 #18
    Interesting! I am still at a loss as to how to find the force exerted by the pilot :(
     
  20. Dec 15, 2011 #19
    At the bottom of the loop the pilot is travelling at 210km/hr = 58.5m/s
    Therefore the acceleration at the bottom of the loop = v^2/r = 11m/s^2
    From this I got the force on the pilot to be 615N
     
  21. Dec 15, 2011 #20

    PeterO

    User Avatar
    Homework Helper

    Assuming your acceleration figure of 11 is correct [I have no reason to doubt, but I have not calculated it myself] then the force you have calculated is the net force on the pilot [or close to it].

    That force is achieved as the sum of two actual forces - gravity and the force from the seat.

    Gravity pulls down with a force approaching 550N [if g=10 it is exactly 550 N]

    The seat has to push strongly enough that the net force is your 615N up.

    Draw some vectors to see what you get [if necessary]

    EDIT: I just calculated it without rounding off on the way through and got 1143.3 N
     
    Last edited: Dec 15, 2011
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