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Homework Help: Circular motion

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A mass of 6.10 kg is suspended from a 1.51 m long string. It revolves in a horizontal circle as shown in the figure
    .http://capa2.cc.huji.ac.il/res/msu/physicslib/msuphysicslib/11_Force_Motion_Adv/graphics/prob03_pendulum.gif
    The tangential speed of the mass is 3.24 m/s. Calculate the angle between the string and the vertical.
    2. Relevant equations

    F=mv^2/R

    3. The attempt at a solution

    The F component at y is:
    Tcosa-mg=0
    Tcosa=mg
    T=mg/cosa
    The F component at x is:
    Tsina=ma
    Tsina=mv^2/R
    R=lsina
    mg*sina/cosa=mv^2/lsina
    lmgsin^2a=cosamv^2
    lgsin^2a=v^2cosa

    sin^2a=1-cos^2a
    after putting all the numbers i get a final equation which is
    10.798cos^2+10.4976cosa-14.798
    the solution of the equation was
    1-0.78
    2-1.753

    ofcourse i take the first one
    BUT!!!!its not correct!
    pleasew tell me where i was wrong
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Dec 10, 2009 #2

    kuruman

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    Your method is correct, but you made a mistake solving the quadratic. I recommend that you solve it symbolically and put in the numbers at the very end.
     
  4. Dec 10, 2009 #3
    I am doing my calculations over and over again but am getting the same answer !
    I cant see where i was wrong!
     
  5. Dec 10, 2009 #4

    kuruman

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    I should be able to help you if you show the equation (in symbols) before you put in the numbers and then what numbers you put in all laid out.
     
  6. Dec 10, 2009 #5
    the equation is:
    l*g(1-cos^2a)=v^2cosa
    l*g*cos^2a+v^2cosa-l*g=0
    that is befor i put in the numbers
    after:
    1.51*9.8*cos^2a+3.24^2cosa-1.51*9.8=0
    14.798cos^2a+10.4976cosa-14.798=0
    that is the final equation

    did i make any mistake here?
     
  7. Dec 10, 2009 #6

    kuruman

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    Not here, but you made a mistake in your first posting of the quadratic.


     
  8. Dec 10, 2009 #7
    I solved the equation with the correct one:
    I get two answers
    1-0.706
    2--1.415

    the first one is incorrect
    could it be the second one?can it be negative?
     
  9. Dec 10, 2009 #8

    kuruman

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    How do you know it is incorrect? What is the correct answer?

    [/quote]could it be the second one?can it be negative?[/QUOTE]

    Don't forget you are solving for the cosine of an angle which cannot be less that -1. Also, a negative cosine means that the angle is greater than 90o which is unphysical.
     
  10. Dec 10, 2009 #9
    I know its incorrect-thats complicated-
    but I dont know what is the correct answer

    Do you mean after i get my answers for the equation
    I have to do SHIF COS A ->the solution i get
    and than i get my angle?
     
  11. Dec 10, 2009 #10

    kuruman

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    I don't know about SHIF COS A, but the 0.706 that you found is the cosine of the angle. You need to find the angle itself from it and be sure you have the degrees/radians switches set properly.
     
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