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Uniform circular motion and tension of a string

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data
    a 0.60 kg sphere rotates around a vertical shaft supported by 2 strings, as shown. if the tension in upper string is 18N calculate.

    a) tension in lower string?
    b) rotation rate (in rev/min) of the system.


    2. Relevant equations



    3. The attempt at a solution

    there is image of the picture.

    i have part a.
    Fnvertical=18Ncos53 = 10.8N
    Fn2=mg-Fn1
    Fn2 = (0.60kg*9.8m/s^2)-10.8
    Fn2 = 4.92

    T2= 4.92/cos53 = 8.2N

    Now part b is where i am confused. I did this so far but then i get lost.
    Fn= 18Nsin53 + 8.2Nsin53
    Fn = 21N --> centripetal force

    Now what?? :S
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2013 #2
    Mass and centripetal force give you centripetal acceleration. Centripetal acceleration and radius give you angular speed.
     
  4. Nov 22, 2013 #3
    so you saying... since Fn= 21N = centripetal force, i then... r=6.371X10^6m
    Fr =maR
    Fr = m (v^2/r)
    21N = 0.60kg (v^2/6.371X10^8m)
    V^2 = 2.223X10^8 ---> take tje square root
    v = 1.49X10^4 m/s

    from this i plugg into

    v = 2(3.14)r / T

    get T.. and that is the answer?
     
  5. Nov 22, 2013 #4
    OOPS! sorry i got my questions mixed up.. i used the incorrect r... but is the concept behind what i said correct?
     
  6. Nov 22, 2013 #5
    in the first answer how you have considered Fnvertical=18Ncos53 = 10.8N ?
    isnt fn the vertical component of tension t1
     
  7. Nov 22, 2013 #6
    yes i have that for part a... if yu scroll up you will see it there in my original post...

    my trouble i am having is for part B
     
  8. Nov 22, 2013 #7
    i mean Isn't Fn the vertical component of tension t1
     
  9. Nov 22, 2013 #8
    yes... not sure im quite following you.
     
  10. Nov 22, 2013 #9
    You have taken the horizontal componet Fnvertical=18Ncos53 = 10.8N
    it should be sin instead of cos
     
  11. Nov 22, 2013 #10
    when u look at the image, theta =53 is close to the vertical component, so Fnvertical = a/h = cos 53
    to find the vertical component the angle is adjacent to the vertical side thats why it is cos.
     
  12. Nov 22, 2013 #11
    Have you figured out the answer?
     
  13. Nov 22, 2013 #12

    haruspex

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    The confusion may be because the diagram shows theta as angle to the vertical but there is also a '53' written as the angle to the horizontal.
    Anyway, it is not a good idea to find the numerical value of the angle. It introduces extra rounding errors. You only care about the trig functions of the angle, and you can get those directly from the triangle dimensions.
    It would help greatly if you would define your variables. it's a pain having to deduce what they mean from the equations, especially since the equations might be wrong. And which directions are you defining as +ve?
     
  14. Nov 22, 2013 #13
    centripetal force =mω2r

    m is the mass
    ω is the angular velocity
    r is the radius

    from this you can find ω
    from ω you can easily find rev/min
     
  15. Nov 22, 2013 #14
    w = 2pi(r)/T is that the formula you are talking about????
     
  16. Nov 22, 2013 #15
    then. T= 1/f ...
     
  17. Nov 23, 2013 #16

    haruspex

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    I wouldn't think so. What formula is that?
    What formulae do you know for centripetal acceleration? (If you don't know any, search the net.)
     
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