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Drawing a free body diagram on the mass

  1. Nov 11, 2006 #1
    Hi, I need some help with a question involving forces. I will appreciate greatly if someone can please have a look at this. Thanks.

    Question: A 2.00 kg mass, which is attached to a string of radius [tex]r[/tex], travels in a vertical circle. When the string makes an angle of [tex]\theta=52^o[/tex] with the horizontal, the speed of the mass is [tex]2.31m/s[/tex] and the tension in the string is [tex]31.6N[/tex]. Calculate the radius of the string, [tex]r[/tex].

    Diagram
    Code (Text):

              Center
     ------------+--------------
                  \  52 degrees
                   \
                    \
                     \ string
                      \
                       \
                        \
                         O mass
     
    I tried by drawing a free body diagram on the mass, labelling two forces acting on it, the force of gravity and the tensile force.
    I'm really not sure on how to solve it, but I'm guessing the vertical component of the force of tension - the weight = the vertical component of the centripetal force. Am I close? If not, may someone please give me a hand. Thank You.
     
  2. jcsd
  3. Nov 12, 2006 #2
    I'm not sure if this is right, but I'll give it a shot.

    If you do assume that the vertical components of tensile and centripal forces are the same, then you can calculate the magnitude of the centripetal force, which ends up being the same as the tension. Then use F = mv^2/r and solve for r.

    I have no idea if this is right or not. I get an answer of about 0.34m.
     
    Last edited: Nov 12, 2006
  4. Nov 12, 2006 #3
    i did not calculate but it is just like this,

    Tsin(theta)=mv^2/r,

    then find r
     
  5. Nov 12, 2006 #4

    radou

    User Avatar
    Homework Helper

    Note that T - mg cos(90-52) = mv^2 / r.
     
  6. Nov 12, 2006 #5
    Wow thank you so much radou, your method worked!!

    So the free diagram should look something like this:
    Code (Text):

    Center
      +
       \ F_T|
        \   |
         \  |
          \ | F_Tsin52
           \|
            O
            |\
            | \ F_gcos(90-52)
      F_g   |  \
            |   \
            |  /
            |/

    The two components which supply the centripetal force are
    F_T - F_gcos(90-52) = F_c
     
     
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