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

1. anonymous

so what have you tried so far?

3. anonymous

idk, I've tried so many different methods that i'm just lost and I have no idea what i'm doing anymore.

4. anonymous

I know that Ac is toward the center

5. anonymous

|dw:1433905966296:dw|

6. Michele_Laino

here are the external forces acting on each seat: |dw:1433906060256:dw|

7. anonymous

why is centripital force outward and not toward the center?

8. Michele_Laino

the magnitude of the centrifugal force is: $m\frac{{{v^2}}}{R}$ where R is the radius of the rotating platform

9. Michele_Laino

since I measure those forces with a reference system located outside of the rotating platform

10. anonymous

ah ok, got it

11. anonymous

oh ok, so you need to find Fc, centripetal force

12. anonymous

Then you can find your V

13. Michele_Laino

|dw:1433906399481:dw|

14. Michele_Laino

it is simple, since, using trigonometry, we ca write: $\Large m\frac{{{v^2}}}{R} = mg\tan \theta$

15. Michele_Laino

after a simplification, we get: $\Large v = \sqrt {gR\tan \theta }$

16. anonymous

ah ok i see

17. anonymous

|dw:1433906745625:dw|

18. anonymous

so It would look like this then?

19. Michele_Laino

no, since the direction of the centrifugal force, not centripetal, is horizontal |dw:1433906906053:dw|

20. anonymous

Oh, ok I see

21. Michele_Laino

|dw:1433906970477:dw|

22. Michele_Laino

|dw:1433907024481:dw|

23. Michele_Laino

please, more precisely R is given by the subsequent equation: $\Large R = \frac{d}{2} + l\sin \theta$

24. 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

@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

26. Michele_Laino

I have applied the theorem of Pitagora to this triangle: |dw:1433907439233:dw|

27. anonymous

yeah this is a centripetal force problem

28. 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

29. Michele_Laino

the centripetal force is the force applied by the chain on the seat

30. anonymous

I got V = 3.6156 m/s

31. Michele_Laino

for example, if I cut the chain, then the seat will go outside the rotating platform, along a radial direction

32. anonymous

oh ok

33. Michele_Laino

the requested value of V, is: $\Large \begin{gathered} v = \sqrt {g\left( {\frac{d}{2} + l\sin \theta } \right)\tan \theta } = \hfill \\ \hfill \\ = \sqrt {9.81\left( {4.13 + 3.9\sin \left( {17.9} \right)} \right)\tan \left( {17.9} \right)} \hfill \\ \end{gathered}$

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..

35. Michele_Laino