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

1. Oct 2, 2011

### GRice40

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. Oct 2, 2011

### issacnewton

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

3. Oct 2, 2011

### Staff: Mentor

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.

4. Oct 2, 2011

### GRice40

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. Oct 2, 2011

### GRice40

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. Oct 2, 2011

### issacnewton

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

7. Oct 2, 2011

### Staff: Mentor

Good so far! Now, ask yourself why the cables are hanging at an angle of 30° to the vertical rather than flying out horizontally.

8. Oct 2, 2011

### GRice40

Hope this works, we'll see!

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9. Oct 2, 2011

### GRice40

Hmm, gravity pulling them down would be my guess?

10. Oct 3, 2011

### issacnewton

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

11. Oct 3, 2011

### GRice40

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. Oct 3, 2011

### PeterO

It would be useful to know the other formula for centripetal acceleration

Ac = V2/R or 4∏2R/T2

Where T is the period.

13. Oct 3, 2011

### issacnewton

which component of tension is acting as a centripetal force ?

14. Oct 3, 2011

### GRice40

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

15. Oct 3, 2011

### PeterO

That's true. Did you read post #12

16. Oct 3, 2011

### issacnewton

yes........ we also know that the centripetal force is

$$\frac{mv^2}{r}$$

and you already said $$T\cos(30)=mg$$ so can you manipulate these

17. Oct 3, 2011

### GRice40

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. Oct 3, 2011

### issacnewton

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 $$\smile$$

19. Oct 3, 2011

### PeterO

You have the right answer, but I do not like the bit I highlited red. It looks like you think there is an outward force - the centrifugal force. There is no such force. The net force on the people on the device is INWARDS - the centripetal Force.

20. Oct 3, 2011

### vela

Staff Emeritus
From what you've written, it also looks like you don't understand how to apply ƩF=ma. What are a1 and a2 supposed to represent? The horizontal and vertical components of acceleration?

Back in post #11, you said correctly that the acceleration in the vertical direction was 0 because the girl wasn't moving vertically, so the equations should be of the form:
\begin{align*}
\sum F_x &= ma_x = m\left(\frac{v^2}{r}\right) \\
\sum F_y &= ma_y = m(0)
\end{align*}
where ax=v2/r is the centripetal acceleration and ay=0.