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Archived Centripetal Acceleration Question

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A 6.0kg object is attached to two 5.0m-long strings and swung around in a circle at 12 m/s. Determine the tension in the two strings, and explain why the tensions are not the same.

    Given:
    m=6.0kg
    v=12m/s
    length of string= 5.0m

    2. Relevant equations

    ƩF=m.a
    ac=v^2/R

    3. The attempt at a solution

    There was a diagram that goes with the question but i couldnt simply draw it. Basically its a 8.0m pole with two 5.0 m strings attached to a 6.0kg ball. the strings and the pole form a isosceles triangle. I used the 3,4,5 triangle rule to find the radius which is 3, then ac=v^2/R to find the acceleration which came out to be 48m/s^2. After, used sin law to find angle between ball and radius. When i wrote my ƩF statement one came out to be ƩFy=Tasin37°-Tbsin37°-Fg other statement is ƩFx=Tacos37°+Tbcos37°. Is it possible to substitute these two equations eventhough they're acting in different directions?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2016 #2

    QuantumQuest

    User Avatar
    Gold Member

    Diagram.gif

    The radius found in the OP is right (3 m). The centripetal acceleration is also correct (48 m / s2). The OP uses symbols like Ta, Tb etc. and takes sin law, without giving a diagram, so I'll not comment on that part. I really do not understand, what he/she means by

    "Is it possible to substitute these two equations eventhough they're acting in different directions?"


    Anyway, I'll use a Cartesian coordinate system rather than the sin law used in the OP.
    The diagram is shown above.

    We take the horizontal and vertical components of the two tensions T1 and T2 (see Fig. 2).
    We first calculate sin(θ) and cos(θ) [Fig. 1]:
    sin(θ) = 3 / 5 and cos(θ) = 4 / 5.

    For the vertical components, because there is no motion on the vertical direction, we'll have:
    T₁cos(θ) = T₂cos(θ) + mg, so: T1 - T2 = mg / cos(θ) = 6.0 x 9.8 / 0.8 ## kg \frac{m}{s^{2}}## = 73.5 N.

    For the horizontal components of tension, we see that their sum is the centripetal force for the circular motion of the sphere.
    So, T1 sin(θ) + T2 sin(θ) = m v2 / R , so, T1 + T2 = m v2 / R sin(θ) = 6.0 x 122 / 3 x 0.6 kgr (m/s)2 / m = 480 N.

    Now, having T1 - T2 and T1 + T2 we solve the simple system of equations:
    T1 - T2 = 73.5 N
    T1 + T2 = 480 N

    We get T1 = 276.75 N and T2 = 203.25 N.

    We see that the tensions are not equal. This is explained, if we take into account the equilibrium on vertical axis (i.e. the weight of the sphere).
     
    Last edited: Feb 7, 2016
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