anonymous
  • anonymous
I need help with this circular motion problem
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
BAdhi
  • BAdhi
so what have you tried so far?
anonymous
  • anonymous
idk, I've tried so many different methods that i'm just lost and I have no idea what i'm doing anymore.

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anonymous
  • anonymous
I know that Ac is toward the center
anonymous
  • anonymous
|dw:1433905966296:dw|
Michele_Laino
  • Michele_Laino
here are the external forces acting on each seat: |dw:1433906060256:dw|
anonymous
  • anonymous
why is centripital force outward and not toward the center?
Michele_Laino
  • Michele_Laino
the magnitude of the centrifugal force is: \[m\frac{{{v^2}}}{R}\] where R is the radius of the rotating platform
Michele_Laino
  • Michele_Laino
since I measure those forces with a reference system located outside of the rotating platform
anonymous
  • anonymous
ah ok, got it
anonymous
  • anonymous
oh ok, so you need to find Fc, centripetal force
anonymous
  • anonymous
Then you can find your V
Michele_Laino
  • Michele_Laino
|dw:1433906399481:dw|
Michele_Laino
  • Michele_Laino
it is simple, since, using trigonometry, we ca write: \[\Large m\frac{{{v^2}}}{R} = mg\tan \theta \]
Michele_Laino
  • Michele_Laino
after a simplification, we get: \[\Large v = \sqrt {gR\tan \theta } \]
anonymous
  • anonymous
ah ok i see
anonymous
  • anonymous
|dw:1433906745625:dw|
anonymous
  • anonymous
so It would look like this then?
Michele_Laino
  • Michele_Laino
no, since the direction of the centrifugal force, not centripetal, is horizontal |dw:1433906906053:dw|
anonymous
  • anonymous
Oh, ok I see
Michele_Laino
  • Michele_Laino
|dw:1433906970477:dw|
Michele_Laino
  • Michele_Laino
|dw:1433907024481:dw|
Michele_Laino
  • Michele_Laino
please, more precisely R is given by the subsequent equation: \[\Large R = \frac{d}{2} + l\sin \theta \]
Michele_Laino
  • Michele_Laino
Since we have no motion along the direction of a single chain, then we can apply the first principle of Newton, and we can write: \[\Large \tau = \sqrt {{{\left( {m\frac{{{v^2}}}{R}} \right)}^2} + {{\left( {mg\tan \theta } \right)}^2}} \] where \tau is the requested tension
BAdhi
  • BAdhi
@Michele_Laino. You are solving this problem with reference to a person who is on the the seat itself. Not on an inertial frame of reference. Am i right? Otherwise youll have to use centripetal force instead of centrifugal force
Michele_Laino
  • Michele_Laino
I have applied the theorem of Pitagora to this triangle: |dw:1433907439233:dw|
anonymous
  • anonymous
yeah this is a centripetal force problem
Michele_Laino
  • Michele_Laino
no, if I solve that problem using the seat as reference frame, then as you well said my new reference system is not inertial, so, I can consider it again inertial provided that I introduce a fictious force , well that force is the centrifugal force of the rotating platform, at the position of the seat
Michele_Laino
  • Michele_Laino
the centripetal force is the force applied by the chain on the seat
anonymous
  • anonymous
I got V = 3.6156 m/s
Michele_Laino
  • Michele_Laino
for example, if I cut the chain, then the seat will go outside the rotating platform, along a radial direction
anonymous
  • anonymous
oh ok
Michele_Laino
  • Michele_Laino
the requested value of V, is: \[\Large \begin{gathered} v = \sqrt {g\left( {\frac{d}{2} + l\sin \theta } \right)\tan \theta } = \hfill \\ \hfill \\ = \sqrt {9.81\left( {4.13 + 3.9\sin \left( {17.9} \right)} \right)\tan \left( {17.9} \right)} \hfill \\ \end{gathered} \]
BAdhi
  • BAdhi
From what i understand what we do is making the frame of the guy seated on that seat an inertial frame of reference by introducing a centrifugal force. and thats why i said you are looking at the problem wrt to the seated person. am I wrong? Also only the person seated will see himself moving along the radial axis when the chain is broken while an observer outside will see him moving tangently.. please correct me if im wrong..
Michele_Laino
  • Michele_Laino
you are right! @BAdhi

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