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Help with circular motion

  1. Mar 11, 2005 #1
    A 100gm bead is free to slide along a 0.8m long piece of string ABC. The ends of the string are attached to a vertical pole at A and C which are 0.4m apart. When the pole is rotated about its axis, BC becomes horizontal and equal to 0.3m.
    (a) Find the tension in the string
    (b) Find the speed of the bead at B.

    I don't know where to start with this problem, could someone please help?

  2. jcsd
  3. Mar 19, 2005 #2

    Andrew Mason

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    The tension in the string provides two forces on the bead. The forces have the same magnitudes but different directions. The horizontal component of these forces provides the centripetal force and the difference in vertical components provides the normal force (mg)

    [tex]Tsin\theta + T = T + .3T/5 = 1.6T = mv^2/r[/tex]

    [tex]Tcos\theta = .4T/.5 = .8T = mg[/tex]

    So: [itex]2g = v^2/r[/itex] where r = .3 m
    and: [itex]T = 1.25mg[/itex]

  4. Apr 17, 2005 #3
    The answer is 66.67 ms^1
    contact me for mre info.

    a hint- T sin ά=mg

    T cosά+T=mv*v/r
  5. Apr 17, 2005 #4
    The answer is 66.67 ms^1

    T sin ά=mg

    T cosά+T=mv*v/r
    If not for a care less mistake this should be correct.
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