1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Merry Go-Round Circular Motion

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.66 m/s and a centripetal acceleration a of magnitude 1.83 m/s2. Position vector r locates him relative to the rotation axis. a) what is the magnitude of r? What is the direction of r when a is directed b) due east and c) due south?


    2. Relevant equations
    a = v^2/r
    T = 2pi r/v



    3. The attempt at a solution
    a) no idea how to do this part.
    b) i thought since acceleration is east, which would be toward centre because circular motion, then wouldn't position be same direction??
    c) same as b
     
  2. jcsd
  3. Feb 20, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    You have a = v^2/r and T = 2pi r/v.
    Looking for r. Know v and a.
    Which of the formulas has r, v and a in it? Solve it for r, then put in the values for v and a.
     
  4. Feb 20, 2009 #3
    Hi I wish it was that easy. But they're looking for the position vector, not the radius, there should be a little arrow over the r that I don't know how to do. =)

    sorry for the misunderstanding.
     
  5. Feb 20, 2009 #4

    Delphi51

    User Avatar
    Homework Helper

    Part (a) asks for the magnitude of r .
    Okay, on part (b): What is the direction of r when a is directed b) due east
    sketch the thing and picture the man going around in circles with an acceleration vector on himself. Stop when his arrow goes east. What is the direction of the radius to/from the man at that point? Hmm, does r go toward the center or away from the center? Hope you know - I don't.
     
  6. Feb 20, 2009 #5
    Okay i understand b and c now.

    But a) is asking for the magnitude of the position vector? Not the radius because radius is just r, not r with an arrow above it. =S
     
  7. Feb 20, 2009 #6

    Delphi51

    User Avatar
    Homework Helper

    "Position vector relative to the axis" . . . sounds like radius to me!
    Ah, but it answers my question - you want the direction from the axis to the man in parts b and c.
     
  8. Feb 21, 2009 #7
    wow i'm an idiot.

    Thank you =)
     
  9. Feb 21, 2009 #8
    Another problem:

    A purse at radius 2.00 m and a wallet at radius 3.00 travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2)i + (4.00 m/s2)j, at that instant and in unit-vector notation, what is the acceleration of the wallet?

    equation:

    a = v^2/r

    attempt:

    i tried substituting v = square root (3a) into the equation a = v^2/r for the purse, because they both have same velocity? Uniform circular motion? Not working...
     
  10. Feb 21, 2009 #9

    Delphi51

    User Avatar
    Homework Helper

    I would use the other acceleration formula for this one, with v replaced by 2(pi)r/T.
    Knowing the magnitude of acceleration at r = 2, you can find the T. Then use the formula again to get the magnitude at 3. Or just look at the formula and ask yourself what increasing r by a factor of 1.5 does to acceleration.

    The direction of the acceleration will be the same for the wallet as it is for the purse.
     
  11. Feb 21, 2009 #10
    the answer should be (3m/s^2 i + 6m/s^2 j).. Now I did what you said but I'm not sure when calculating T how to keep everything in unit vector notation.. it all seems to be really weird.
     
  12. Feb 21, 2009 #11

    Delphi51

    User Avatar
    Homework Helper

    a = 4(pi)^2*m*r/T^2 so multiplying r by 1.5 makes acceleration 1.5 times bigger.
    The direction doesn't change, so both the x and y components must be multiplied by 1.5.
    So (2, 4) goes to (3, 6).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Merry Go-Round Circular Motion
  1. Merry go round torque (Replies: 3)

  2. Merry-go-round problem (Replies: 1)

  3. Merry Go Round Problem (Replies: 1)

Loading...