
#1
Oct211, 07:41 PM

P: 20

1. The problem statement, all variables and given/known data
The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a 5.00m long cable, the upper end of which is fastened to the arm at a point 3.00 m from the central shaft. Find the time of one rotation if the angle produced from the cable connecting to the arm is 30 degrees from vertical. 2. Relevant equations Circumference= distance = 2 (pi) R Acceleration = V^2/R F = m(V^2/R) 3. The attempt at a solution I honestly have no clue how to do this one. I keep working myself in circles, which doesn't seem to be getting anything accomplished. Any help on where I should start? 



#2
Oct211, 10:51 PM

P: 607

is there any figure given with this.......i have trouble imagining this....




#3
Oct211, 11:10 PM

Mentor
P: 11,428

After that you'll be drawing an FBD and working out the components of the accelerations acting. 



#4
Oct211, 11:39 PM

P: 20

I'm so behind in physics, Circular Rotation help.
There was a figure with the problem, but I can't save it to upload here.
I've been working through it, and it seems that if I could find the velocity, I'd be able to solve it. Any tips on how to go about finding velocity? 



#5
Oct211, 11:46 PM

P: 20

Here's what I have so far:
Radius: 3 + 5sin(30) = 5.5 m V = d/t = 34.56/t Circumference = 2(pi)(5.5)= 34.56 Acceleration = V^2/R = (34.56/t)^2/5.5 That's about as far as I can get without another variable that I know =/ 



#6
Oct211, 11:48 PM

P: 607

without figure I can't help..... difficult to see the situation




#7
Oct211, 11:51 PM

Mentor
P: 11,428





#8
Oct211, 11:52 PM

P: 20

Hope this works, we'll see!




#9
Oct211, 11:52 PM

P: 20

Hmm, gravity pulling them down would be my guess?




#10
Oct311, 12:03 AM

P: 607

that helps....... set up free body diagram for the girl........there is tension in the wire




#11
Oct311, 12:10 AM

P: 20

I assume that because she is staying in place in the y direction, that the y direction is in equilibrium.
So here's what I have: Tcos(30) = m * (9.8) (Another possibility that is coming to mind is: Force(out) + W = T) (Here's the other possibility running through my mind: Force(out) = Tsin(30) for the x direction) 



#12
Oct311, 12:10 AM

HW Helper
P: 2,316

A_{c} = V^{2}/R or 4∏^{2}R/T^{2} Where T is the period. 



#13
Oct311, 12:18 AM

P: 607

which component of tension is acting as a centripetal force ?




#14
Oct311, 12:21 AM

P: 20

The centripetal force should be Tsin(30), I think?




#15
Oct311, 12:28 AM

HW Helper
P: 2,316





#16
Oct311, 12:28 AM

P: 607

yes........ we also know that the centripetal force is
[tex]\frac{mv^2}{r}[/tex] and you already said [tex]T\cos(30)=mg[/tex] so can you manipulate these 



#17
Oct311, 12:34 AM

P: 20

Ok here's what I got:
F(out) = m*a1 = Tsin(30) W = M*a2 = Tcos(30) So if you divide the two you get: (m*a1/m*a2) = Tsin(30)/Tcos(30) which reduces to (a1/a2) = sin(30)/cos(30) or (a1/9.8) = .577 .577(9.8) = a1 = 5.66 m/s^2 So 5.66 = 4(pi)^2R/t^2, 5.66 = 217.13/t^2 t^2 = 217.13/5.66 t = sqrt(217.13/5.66) = 6.19 seconds That's what the answer is supposed to be. Thank you so much guys, I've been working on that problem for HOURS! 



#18
Oct311, 12:37 AM

P: 607

don't forget to put figures when they are provided......without the figures its difficult for people here to understand........always post the problem AS it appears in the text.....
don't psychoanalyse the problem and write what YOU think the problem statement is... you will likely get more help when you present the problem and your attempted work in organised manner..... good luck [tex]\smile[/tex] 


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