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Giant Swing, Uniform Circular Motion

by clope023
Tags: circular, giant, motion, solved, swing, uniform
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clope023
#1
Feb12-08, 10:55 AM
P: 618
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.
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lavalamp
#2
Feb12-08, 11:55 AM
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P: 280
You might want to go back and look at this bit again:
Quote Quote by clope023 View Post
[b]L = 3+5m*sin([tex]\theta[/tex])) to get 4m

R then equals = 2m
sin(30) = 0.5, so 5*0.5 + 3 = 5.5m, and you don't need to divide it by 2, because that is the radius, not the diameter.
PhanthomJay
#3
Feb12-08, 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..

clope023
#4
Feb12-08, 12:36 PM
P: 618
Giant Swing, Uniform Circular Motion

Quote Quote by PhanthomJay View Post
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..
I wrote it before, R = Lsin(30)

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.
lavalamp
#5
Feb12-08, 03:16 PM
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P: 280
Quote Quote by clope023 View Post
if it's only length of cable alone, then R = 5(sin30) so R = 2.5m, does that seem more correct?
As I already said:
Quote Quote by lavalamp View Post
You might want to go back and look at this bit again:
Quote Quote by clope023 View Post
[b]L = 3+5m*sin([tex]\theta[/tex])) to get 4m

R then equals = 2m
sin(30) = 0.5, so 5*0.5 + 3 = 5.5m, and you don't need to divide it by 2, because that is the radius, not the diameter.
clope023
#6
Feb12-08, 03:51 PM
P: 618
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?
PhanthomJay
#7
Feb12-08, 04:04 PM
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Quote Quote by clope023 View Post
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?
Your FBD's are not correct. In the y direction, mg acts straight down, and Tcos30 acts straight up. In the x direction, only Tsin30 acts.
clope023
#8
Feb12-08, 04:57 PM
P: 618
Quote Quote by PhanthomJay View Post
Your FBD's are not correct. In the y direction, mg acts straight down, and Tcos30 acts straight up. In the x direction, only Tsin30 acts.
so then in that case Tcos30 = mg

so the angle is proportional to the mass?
clope023
#9
Feb12-08, 05:36 PM
P: 618
scracth that the angle is not proportionate to the mass, thanks for the help.
nightlaei71
#10
Feb28-08, 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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