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Centripetal/Centrifugal Motion on a merry-go-round

  1. Oct 17, 2015 #1
    1. The problem statement, all variables and given/known data
    You are standing 2.7 m from the centre of a spinning merry-go-round holding one end of a string tied to a 120g mass. The merry-go-round has a period of 3.9 s.
    • Draw a system diagram of the situation.
    • Draw an FBD of the mass in Earth's frame of reference.
    • Draw an FBD of the mass in the merry-go-round's rotating frame of reference.
    • What angle does the string make with the vertical?
    • Determine the magnitude of the tension in the string.
    My issue arises with the final question only (which is at the end of this admittedly long post), however, I would appreciate it if you folks could look over my FBDs as well, the rest is only ancillary.


    2. Relevant equations
    • Fac = (mv^2)/r
    • V = d/t
    • circumference = pi * diameter
    • Fg = mg

    3. The attempt at a solution

    A.
    I drew a person on the outside of the merry-go-round, and the string he holds follows a path some amount below the horizontal where it connects to the 120g mass. Like this:
    0001GuyHoldingThang.jpg

    ----------------------------------------------------------------------------------------

    B.
    0001GuyHoldingThangFBD1.jpg

    ----------------------------------------------------------------------------------------

    C.
    0001GuyHoldingThangFBD2.jpg

    ----------------------------------------------------------------------------------------

    D: (My solution matches the textbook's)

    FNET_Y
    1. Fnet_Y = Fg - Ft sin x = 0
    2. Ft sin x = Fg
    3. Ft = mg / sin x
    FNET_X
    1. mv^2/r = Ft cos x
    2. mv^2/r = (mg/sin x) * cos x
    3. V^2 = g/sin x * cosx
    4. rearrange for x
    5. rg/v^2 = tan x
    SOLVE FOR VELOCITY
    1. v = d/t
    2. v = circumference / 3.9s
    3. v = 2 * pi * r / 3.9s
    4. v = ~4.34m/s
    SOLVE FOR THETA
    1. tan-1(rg/v^2)
    2. tanx-1[(2.7*9.8)/4.34m/s^2]
    3. x = 54 degrees
    4. since they're asking for the angle from the vertical, it would be the complementary to this one
    5. 90- ~54 = 36degrees
    ----------------------------------------------------------------------------------------

    E:

    1. Using one of my previous equations to solve for Ft
    2. Ft = mg/sinx
    3. Ft = 9.8*0.12/sin(54)
    4. Ft = 1.45N
    The bold is what I get wrong, the textbook's solution reads 0.96N, which would not even be equivalent to the Fg which (I presumed) the string counter-acted.

    Where did I go wrong in my solution?

    EDIT: The textbook's solution is wrong. I found the solution manual and they did the following operations
    1. cos x = Fg / Ft (x being the angle from the vertical)
    2. Ft = cosx * Fg
    They should have obviously had Fg/cosx as Ft.

    Ah well, good learning experience
     
    Last edited: Oct 17, 2015
  2. jcsd
  3. Oct 17, 2015 #2

    TSny

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    Homework Helper
    Gold Member

    Good work on getting the tension. Your free body diagram for (b) looks good, although I don't understand the "[toward center]" label. For the diagram in (c), why did you change the direction of the tension force. Also, in (c) there should be a third force in the diagram representing the centrifugal force.
     
  4. Oct 17, 2015 #3
    The towards center is actually incorrect, now that I look at it again. Thanks for catching my mistake! The acceleration towards the center is a component of the Ft, not Ft itself.

    My intent with the changing of direction was that from the frame of reference of the merry-go-round, the tension was pulling it outwards.

    But what it should be is the following, no?
    0001GuyHoldingThangFBD3.jpg
    My issue with this is that wouldn't that mean that Fc balances the x-component Ft?

    Which perhaps makes sense because the object is technically moving at a constant velocity tangential to the circle, correct?
     
  5. Oct 17, 2015 #4

    TSny

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    Homework Helper
    Gold Member

    The string should point in the same direction in both diagrams; that is, up and to the left if the central axis of the merry-go-round is to the left. Therefore, FT should point in the same direction in both diagrams. In the frame of the merry-go-round, the 120 g mass is at rest. So, the three forces FT, mg, and Fcentrifugal add to zero.
     
  6. Oct 17, 2015 #5
    Right! Makes sense, thanks again!
     
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