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Giant Swing, Uniform Circular Motion 
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#1
Feb1208, 10:55 AM

P: 609

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 cable 5.00 long, the upper end of the cable being fastened to the arm at a point 3.00 from the central shaft. A) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of with the vertical. B) Does the angle depend on the weight of the passenger for a given rate of revolution? 2. Relevant equations R = Lsin(\theta) v = [tex]\sqrt{gtan(\theta)R}[/tex] T = 2[tex]\Pi[/tex]R/v 3. The attempt at a solution tried using L = 3+5m*sin([tex]\theta[/tex])) to get 4m R then equals = 2m v then equals = [tex]\sqrt{9.8*2*tan\theta}[/tex] = 11.3m/s T then equals 2[tex]\Pi[/tex](2m)/11.3m/s = 1.1s wrong I tried a few other combinations where I used 3+(5sintheta) as the L and got T=4.4s which was wrong as well. I'm not sure what I'm doing wrong, I think I have the correct equations and I know I have the right angle and distances, I guess I'm just not sure how to derive the right length and and radius. I'm also thinking that for part B the weight will determine the angle that the seat swings, but I don't want to risk losing the only chance I have on that part of the problem (masteringphysics). any and all help is greatly appreciated. 


#2
Feb1208, 11:55 AM

P: 280

You might want to go back and look at this bit again:



#3
Feb1208, 11:56 AM

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It appears that you are confusing L with R. If L is the length of the cable, then what is R, which is the radius measured from the passenger to the central shaft? Also, if you draw a free body diagram and use Newton's laws, it will help to see whether or not the the mass comes into play. You shouldn't blindly be using a formula..



#4
Feb1208, 12:36 PM

P: 609

Giant Swing, Uniform Circular Motion
I was pretty sure it was the length of the cable alone (which would've been 5) I just wasn't sure where the distance of the angle of the seat from the vertical shaft (3m) played in. if it's only length of cable alone, then R = 5(sin30) so R = 2.5m, does that seem more correct? on the angle's dependance on the weight, I've drawn a free body diagram and it makes the sum of the forces be (sigma)Fx = marad + mgcos30 = 0 (sigma)Fy = T  mgsin30 = 0 so I would imagine the angle is dependant on the weight of the seat. 


#6
Feb1208, 03:51 PM

P: 609

thanks, I got the answer to the first part using that radius, I wasn't picturing in my head that the swing literally makes a circle when viewed overhead and the distance of 3m + (5sin30) was the radius of that circle.
now fro for the 2nd part, as I said I did a free body diagram and found the forces acting in each direction, but I'm actually still not certain, I believe the mass would cancel out from the 2 equations if I set them equal to each other, so the answer would be that it doesn't depend on the mass? 


#7
Feb1208, 04:04 PM

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#8
Feb1208, 04:57 PM

P: 609

so the angle is proportional to the mass? 


#9
Feb1208, 05:36 PM

P: 609

scracth that the angle is not proportionate to the mass, thanks for the help.



#10
Feb2808, 04:22 PM

P: 9

I have the same problem as well, and I don't quite understand how velocity was derived.



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