I'm so behind in physics, Circular Rotation help.


by GRice40
Tags: circular, physics, rotation
GRice40
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#1
Oct2-11, 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?
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issacnewton
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#2
Oct2-11, 10:51 PM
P: 607
is there any figure given with this.......i have trouble imagining this....
gneill
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#3
Oct2-11, 11:10 PM
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Quote Quote by GRice40 View Post
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?
It's always best to start with a diagram. Draw a cross section showing the vertical pole, a cross beam, and one cable (at 30 degrees from vertical) with a mass on the end. Identify the center of rotation of the mass and the radius of the circle it will follow.

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

GRice40
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#4
Oct2-11, 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?
GRice40
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#5
Oct2-11, 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 =/
issacnewton
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#6
Oct2-11, 11:48 PM
P: 607
without figure I can't help..... difficult to see the situation
gneill
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#7
Oct2-11, 11:51 PM
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Quote Quote by GRice40 View Post
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 =/
Good so far! Now, ask yourself why the cables are hanging at an angle of 30 to the vertical rather than flying out horizontally.
GRice40
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#8
Oct2-11, 11:52 PM
P: 20
Hope this works, we'll see!
Attached Thumbnails
phys.jpg  
GRice40
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#9
Oct2-11, 11:52 PM
P: 20
Hmm, gravity pulling them down would be my guess?
issacnewton
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#10
Oct3-11, 12:03 AM
P: 607
that helps....... set up free body diagram for the girl........there is tension in the wire
GRice40
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#11
Oct3-11, 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)
PeterO
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#12
Oct3-11, 12:10 AM
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Quote Quote by GRice40 View Post
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 =/
It would be useful to know the other formula for centripetal acceleration

Ac = V2/R or 4∏2R/T2

Where T is the period.
issacnewton
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#13
Oct3-11, 12:18 AM
P: 607
which component of tension is acting as a centripetal force ?
GRice40
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#14
Oct3-11, 12:21 AM
P: 20
The centripetal force should be Tsin(30), I think?
PeterO
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#15
Oct3-11, 12:28 AM
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Quote Quote by GRice40 View Post
The centripetal force should be Tsin(30), I think?
That's true. Did you read post #12
issacnewton
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#16
Oct3-11, 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
GRice40
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#17
Oct3-11, 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!
issacnewton
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#18
Oct3-11, 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 psycho-analyse 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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