Giant Swing, Uniform Circular Motion

In summary: Can someone please explain the velocity equation?[SOLVED] Giant Swing, Uniform Circular MotionIn summary, the conversation discusses the problem of finding the time of one revolution for a "Giant Swing" ride at a county fair. The conversation includes attempts at solving the problem using equations for uniform circular motion and a discussion on the dependence of the angle on the weight of the passenger. The correct solution for the first part is found by using the radius of the circle created by the swing and the second part concludes that the angle does not depend on the weight of the passenger.
  • #1
clope023
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[SOLVED] Giant Swing, Uniform Circular Motion

Homework Statement



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.

YF-05-57.jpg


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?


Homework Equations



R = Lsin(\theta)

v = [tex]\sqrt{gtan(\theta)R}[/tex]

T = 2[tex]\Pi[/tex]R/v


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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  • #2
You might want to go back and look at this bit again:
clope023 said:
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.
 
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  • #3
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
PhanthomJay said:
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 dependence 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.
 
  • #5
clope023 said:
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:
lavalamp said:
You might want to go back and look at this bit again:
clope023 said:
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.
 
  • #6
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
clope023 said:
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.
 
  • #8
PhanthomJay said:
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?
 
  • #9
scracth that the angle is not proportionate to the mass, thanks for the help.
 
  • #10
I have the same problem as well, and I don't quite understand how velocity was derived.
 

What is a Giant Swing?

A Giant Swing is an amusement ride that involves a person being strapped into a harness and swung back and forth on a large pendulum-like structure.

How does the Giant Swing work?

The Giant Swing uses the principles of uniform circular motion to create the swinging motion. The rider's weight and the force of gravity act as the centripetal force, keeping the rider in circular motion.

What is uniform circular motion?

Uniform circular motion is when an object moves in a circular path at a constant speed. This means that the object's velocity is always changing, but its speed remains the same.

What forces are involved in uniform circular motion?

In uniform circular motion, there are two main forces at play: the centripetal force, which keeps the object moving in a circular path, and the centrifugal force, which is the outward force acting on the object due to its inertia.

What factors affect the speed and frequency of a Giant Swing?

The speed and frequency of a Giant Swing are affected by the length of the pendulum, the mass of the rider, and the initial angle of release. The longer the pendulum, the slower the swing. A heavier rider will also result in a slower swing. The initial angle of release determines the height of the swing and can affect the frequency of the swings.

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