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Rotational motion pendulum problem

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data

    peQlS.png

    2. Relevant equations
    ƩF = ma


    3. The attempt at a solution
    I know how to break this problem up into the x and y direction

    Y Direction:
    Ft cosθ - Fg = 0

    X Direction:
    Ft sinθ - Fg = mv2 / r

    The hint says to find the inward radial component. I'm guessing this means to find r, because I think that is the only thing missing for me to solve this. To find r I think I would use some combination of the length of the string and theta, but don't exactly know.

    Thanks for any help!
     
  2. jcsd
  3. Oct 21, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The "inward radial component of the string's tension" is just the horizontal component of the the tension force (i.e., what you called the X component of Ft).

    You've got -Fg acting in both the Y-direction and X-direction???

    You should be able to express r in terms of d and theta using trig on an appropriate right triangle.

    You'll have two unknowns, Ft and v. But you'll also have two equations. Try eliminating Ft and solving for v.
     
  4. Oct 21, 2012 #3
    When it was talking about the vertical and inward radial components, it was talking about finding Fty and Ftx, which is what you have already done. You know r because you know d and θ. Just solve the two equations for v, and you're all set.
     
  5. Oct 21, 2012 #4
    There is no -Fg in the x direction, silly me.
    r = Ft sinθ?

    Y Direction:
    Ft cosθ - Fg = 0
    Ft = Fg / cosθ ??

    X Direction:
    Ft sinθ = mv2 / r
    (Fg / cosθ)sinθ = mv2 / (Fg / cosθ)sinθ

    ???? lol im so confused
     
  6. Oct 21, 2012 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Get an expression for r in terms of d and theta, not in terms of Ft and theta. Consider the right triangle where the string is the hypotenuse and the radius r is the side opposite theta.
    yes
    For r, use the expression derived from the right triangle mentioned above.
     
  7. Oct 21, 2012 #6
    Ohhh. r = dsinθ. Whick makes....

    ((mg / cosθ)sinθdsinθ) / m = v2

    Then I would just take the square root of both sides.

    Thanks for all the help again guys!!!
     
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