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Angular Velocity/Motion Question

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

    A bead of mass m is free to slide along a frictionless ring of radius R=.5 meters that rotates about the vertical axis with angular velocity ω=6 rad/sec as shown below. What angle θ with the vertical will the bead make?

    The answer choices are:
    (A) 90
    (B) 61
    (C) 56
    (D) 34
    (E) 29

    This is from the Physics C Mechanics: Sample Exam III.

    2. Relevant equations

    I have no idea.

    3. The attempt at a solution

    I don't know where to start.

    Attached Files:

  2. jcsd
  3. Feb 21, 2012 #2
    Try to make Free body diagram , with all the forces
  4. Feb 21, 2012 #3
    Okay, I did. I have an Fc pointing in, and an mg pointing down, but I'm still not seeing it.
  5. Feb 21, 2012 #4
    see there will mg (weight )going downward and a normal reaction ,
    then you take components of Normal reaction and solve :D
  6. Feb 21, 2012 #5
    Here is the figure

    Attached Files:

  7. Feb 21, 2012 #6
    ohh. okay. so we can use wr=v, and when Fn is mg, that way when we set up, the masses cancel, and we can solve for v. Thanks!
  8. Feb 21, 2012 #7
    but remember in v=rw r is the radius of the motion of particle not the ring :O
  9. Feb 21, 2012 #8
    oh. that's the d in your diagram. So the two radii are different. So how do I get this d?
  10. Feb 21, 2012 #9
    simple trigo rsin(theta)= d
  11. Feb 21, 2012 #10
    but we don't have ø?
  12. Feb 21, 2012 #11
    I tried working it out multiple times, but no matter how I do it, with your d or without it, the fraction at the end never falls in the domain for the sin-1. Can you write out what you would do? Thanks for all your help so far, in my other posts too.
  13. Feb 21, 2012 #12
    Oh wait a minute, I think I got it. Perhaps d=rcosø? If so, then at the end, tan-1((w2r)/g)=ø=61, choice B.
  14. Feb 22, 2012 #13
    Yea , good going :)
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