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Central force

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]





    3. The attempt at a solution

    A)
    i think force of Tension give them necessary centripetal force directed toward the center of the cylinder.

    C)
    Centripetal Force

    Plz help me
     
  2. jcsd
  3. Oct 14, 2007 #2
    Can Any One Help Me?
     
  4. Oct 14, 2007 #3
    I Am Wating From Last Hour
     
  5. Oct 14, 2007 #4
    Plz Help Me
     
  6. Oct 14, 2007 #5
    Plz Help Me
     
  7. Oct 14, 2007 #6
    Can Any One Help Me?
     
  8. Oct 14, 2007 #7
    First, please, please, do not self bump like this.
    Second, for part A it is asking what physical object is exerting the force.
    Part B: If the riders vertical velocity is zero (they're not falling or rising) and that doesn't change, what must be true of their acceleration and thus the vertical force?
    Part C: Centripetal force is perpendicular to gravity here. Think what is happening between the person and the ride. (Read Part E for a hint)
     
  9. Oct 14, 2007 #8
    what to do in d ?
     
  10. Oct 14, 2007 #9
    Can Any One Help Me?
     
  11. Oct 14, 2007 #10
    what to do in d ?
     
  12. Oct 14, 2007 #11
    plz help hurry its due after 3 hours
     
  13. Oct 14, 2007 #12
    For D, draw a Free body diagram. Just draw the force vectors that act on a person. Please don't keep bumping this.
     
  14. Oct 14, 2007 #13
    than u . can u help me in e. what formula i need to use in e.
     
  15. Oct 14, 2007 #14
    Well, in this case we need a frictional force equal to the gravitational force (atleast). And a frictional force is dependent on the normal force, which in this case will be the centripetal force. So:
    [tex]F_f=F_g[/tex]

    [tex]\mu F_C=F_g[/tex]

    [tex]\mu m \omega r = mg[/tex]

    [tex]\mu \omega r = g[/tex]

    [tex]\omega = \frac{g}{\mu r}[/tex]

    Then you just need to convert the angular frequency to a frequency using
    [tex]\omega = 2 \pi f[/tex]
     
  16. Oct 14, 2007 #15
    what is w is that v^2
     
  17. Oct 14, 2007 #16
    [tex]\omega[/tex] is the angular velocity:
    [tex]\omega = \frac{v^2}{r}[/tex]
     
  18. Oct 14, 2007 #17
    ok

    [tex]\mu m \omega r = mg[/tex]

    why u wrote one more r
     
  19. Oct 14, 2007 #18
    Whoops, I typo'ed twice
    It's:
    [tex]\omega = \frac{v}{r}[/tex]
    and
    [tex]F_c=m\frac{v^2}{r}=m\omega^2 r[/tex]
    So
    [tex]\mu m\omega^2r=mg[/tex]
    [tex]\omega=\sqrt{\frac{g}{\mu r}}[/tex]
     
  20. Oct 22, 2007 #19
    plz hep
     
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