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anonymous
 one year ago
I need help with this circular motion problem
anonymous
 one year ago
I need help with this circular motion problem

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BAdhi
 one year ago
Best ResponseYou've already chosen the best response.0so what have you tried so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0idk, I've tried so many different methods that i'm just lost and I have no idea what i'm doing anymore.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know that Ac is toward the center

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433905966296:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here are the external forces acting on each seat: dw:1433906060256:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why is centripital force outward and not toward the center?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the magnitude of the centrifugal force is: \[m\frac{{{v^2}}}{R}\] where R is the radius of the rotating platform

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since I measure those forces with a reference system located outside of the rotating platform

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, so you need to find Fc, centripetal force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then you can find your V

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433906399481:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is simple, since, using trigonometry, we ca write: \[\Large m\frac{{{v^2}}}{R} = mg\tan \theta \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2after a simplification, we get: \[\Large v = \sqrt {gR\tan \theta } \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433906745625:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so It would look like this then?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, since the direction of the centrifugal force, not centripetal, is horizontal dw:1433906906053:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433906970477:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433907024481:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please, more precisely R is given by the subsequent equation: \[\Large R = \frac{d}{2} + l\sin \theta \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Since 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2I have applied the theorem of Pitagora to this triangle: dw:1433907439233:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah this is a centripetal force problem

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, 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
 one year ago
Best ResponseYou've already chosen the best response.2the centripetal force is the force applied by the chain on the seat

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got V = 3.6156 m/s

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for example, if I cut the chain, then the seat will go outside the rotating platform, along a radial direction

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.0From 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
 one year ago
Best ResponseYou've already chosen the best response.2you are right! @BAdhi
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