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Central force

  • Thread starter raman911
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  • #1
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Answers and Replies

  • #2
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Can Any One Help Me?
 
  • #3
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I Am Wating From Last Hour
 
  • #4
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Plz Help Me
 
  • #5
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Plz Help Me
 
  • #6
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Can Any One Help Me?
 
  • #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)
 
  • #8
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what to do in d ?
 
  • #9
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Can Any One Help Me?
 
  • #10
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what to do in d ?
 
  • #11
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plz help hurry its due after 3 hours
 
  • #12
For D, draw a Free body diagram. Just draw the force vectors that act on a person. Please don't keep bumping this.
 
  • #13
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For D, draw a Free body diagram. Just draw the force vectors that act on a person. Please don't keep bumping this.
than u . can u help me in e. what formula i need to use in e.
 
  • #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]
 
  • #15
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what is w is that v^2
 
  • #16
[tex]\omega[/tex] is the angular velocity:
[tex]\omega = \frac{v^2}{r}[/tex]
 
  • #17
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[tex]\omega[/tex] is the angular velocity:
[tex]\omega = \frac{v^2}{r}[/tex]
ok

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

why u wrote one more r
 
  • #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]
 
  • #19
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plz hep
 

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