# Calculate angle of swing in a carousel

• whatdoido
In summary: In general, an iteration procedure xn+1=f(xn) converges in a region where |df/dx| <1. See picture.In the figure, the iteration converges to A but it does not converge to point C.
whatdoido
Hello, this problem has given me headaches for days

1. Homework Statement

The diameter of a carousel is 15 m and it rotates 8,0 rounds per minute. Calculate how large of an angle is formed with respect to the vertical plane by the swing that hangs by the ropes fixed to the edge of carousel. The length of ropes is 2,5 m.

So I mark:
$n=8,0 r/min=0,1333... r/s$
$r=(15/2) m=7,5 m$
$x=2,5 m$
Angle between the vertical plane at the edge of carousel and swing is $\theta$

## Homework Equations

$F=m\frac {v^2}{r}$
$F=mg$
$\omega=n2\pi$


## The Attempt at a Solution

[/B]
The angular velocity:

$\omega=n2\pi=0,1333... r/s*2\pi=0,8377... 1/s$

Then I thought that I need the distance of swing from the edge of carousel. I mark this distance as R

Thus, $R=x*sin\theta$
Then the total distance of swing from the middle is $r+R$, when tangential velocity $v=\omega(r+R)$

I figured that the angle should be solvable from vertical and horizontal forces which act on the swing:

$\left\{\begin{array}{cc} Fsin\theta=m\frac{v^2}{r+R}\\Fcos\theta=mg\end{array}\right.$

Those divided gives:

$\frac{sin\theta}{cos\theta}=\frac{v^2}{g(r+R)}=\frac{\omega^2(r+R)^2}{g(r+R)}$

$tan\theta=\frac{\omega^2r+\omega^2x*sin\theta}{g}$

$g*tan\theta-\omega^2x*sin\theta-\omega^2r=0$

and this is where I got stuck.. I tried solving this equation with substituting $tan\theta=\frac{sin\theta}{\pm\surd(1-sin\theta)}$ and $sin\theta=u$ which did not help. So maybe I got some idea wrong and my equation is wrong or I am just missing some trick by how to solve that equation. On the other hand I checked that does the correct angle (32) work on that equation and it gives about 0=0, so my equation should be correct (I think?).

whatdoido said:
$g*tan\theta-\omega^2x*sin\theta-\omega^2r=0$

and this is where I got stuck..

Yes, solving that exactly looks hard.

Here's one approach that will lead to a good approximation in just a few steps.

Rearrange the equation as ##tan \theta = \frac{\omega^2 r}{g} + \frac{\omega^2 x \sin \theta}{g}## .

Start with a guess for ##\theta##. You can even start with ##\theta = 0## if you want. Plug this value into the right hand side of the equation to calculate a value for ##\tan \theta## and hence a new value for ##\theta##. Now plug this new value into the right hand side to find a new value of ##\tan \theta## and hence ##\theta##. If you repeat this a few times you will see that you will begin to converge to the answer. It only takes about three cycles to get 3 significant figures for this particular equation.

TSny said:
Yes, solving that exactly looks hard.

Here's one approach that will lead to a good approximation in just a few steps.

Rearrange the equation as ##tan \theta = \frac{\omega^2 r}{g} + \frac{\omega^2 x \sin \theta}{g}## .

Start with a guess for ##\theta##. You can even start with ##\theta = 0## if you want. Plug this value into the right hand side of the equation to calculate a value for ##\tan \theta## and hence a new value for ##\theta##. Now plug this new value into the right hand side to find a new value of ##\tan \theta## and hence ##\theta##. If you repeat this a few times you will see that you will begin to converge to the answer. It only takes about three cycles to get 3 significant figures for this particular equation.

That method gives me the correct result, which is fantastic! But I don't know how I could have known to use it. Does this 'iteration' work because ##tan \theta## and ##sin \theta## are trigonometric functions (periodic)? Since not all functions have fixed points to which they converge

I find that this method often converges to the answer. I don't know a general way to tell if it will work before trying it.

TSny said:
I find that this method often converges to the answer. I don't know a general way to tell if it will work before trying it.
Okay, but thanks for helping me. I'm glad that your method worked and I got an answer to the problem. I was only thinking algebraic methods..

whatdoido said:
Okay, but thanks for helping me. I'm glad that your method worked and I got an answer to the problem. I was only thinking algebraic methods..
You didn't post what happened when you tried your u substitution. It produces a quartic, no? In principle, you can solve that algebraically.

TSny said:
I find that this method often converges to the answer. I don't know a general way to tell if it will work before trying it.

In general, an iteration procedure xn+1=f(xn) converges in a region where |df/dx| <1. See picture.
In the figure, the iteration converges to A but it does not converge to B.

whatdoido and TSny
haruspex said:
You didn't post what happened when you tried your u substitution. It produces a quartic, no? In principle, you can solve that algebraically.
Yeah and I took it as a warning sign that I must be doing something wrong. I have never solved a quartic function and nobody have taught me that in school. So surely I should not be meant to solve it like that, which is why I stopped trying to solve it. But there really seems to be no good trigonometric identity which I could use to make equation simpler

Well anyway, after solving it numerically, I got curious about solving that quartic function and found a quartic equation calculator.. so I can show the equation I got and the answers

$tan\theta=\frac{\omega^2r+\omega^2x*sin\theta}{g}$

$g~tan\theta=\omega^2r+\omega^2x*sin\theta$

$g\frac{sin\theta}{\pm\surd(1-sin^2\theta)}=\omega^2r+\omega^2x*sin\theta$

$g\frac{u}{\pm\surd(1-u^2)}=\omega^2r+\omega^2x~u$

$g^2\frac{u^2}{1-u^2}=(\omega^2x~u+\omega^2r)^2$

$g^2\frac{u^2}{1-u^2}=\omega^4x^2u^2+2\omega^4~x~r~u+\omega^4r^2$

.
.
.

$x^2u^4+2~x~r~u^3+(r^2-x^2+\frac{g^2}{\omega^4})u^2-2~x~r~u-r^2=0$

Then I used this site's http://www.1728.org/quartic.htm calculator which gave:

$x_1=0.5343342065691381$

$x_2=-3.05761046744197 +i* 5.553303328218349$

$x_3=-3.05761046744197 -i* 5.553303328218349$

$x_4=-0.41911327168519863$

$x_2$ and $x_3$ can be dismissed since answer should be a real number. $x_1$ is correct, but how to get rid of $x_4$? Because it is negative? But anyway.. I just used a calculator and certainly this method should not be preferred even though it is interesting.

You mean u when writing x1...x4, don't you? And these are values for sinθ. How would the swings stand if sinθ was negative? Outward or inward?

Yes I mean u

Okay yeah, those values are for $sin\theta$. So $\theta$ should be positive to make sense physically (because of centripetal force), for swings to be outward. Thus it leaves only $x_1$ ($u_1$)

whatdoido said:
Yes I mean u

Okay yeah, those values are for $sin\theta$. So $\theta$ should be positive to make sense physically (because of centripetal force), for swings to be outward. Thus it leaves only $x_1$ ($u_1$)
Yes, the angle should be outward, so positive, and it also can not be greater than pi/2, so the negative sign in front of your square root cancels.

ehild said:
Yes, the angle should be outward, so positive, and it also can not be greater than pi/2, so the negative sign in front of your square root cancels.
I see, so $\frac{sin\theta}{\pm\surd(1-sin^2\theta)}$ needs to be positive → $\frac{sin\theta}{\surd(1-sin^2\theta)}≥0$ because $tan\theta≥0$ and $\theta≤\frac{\pi}{2}$ by definition. So $0≤\theta≤\frac{\pi}{2}$

## 1. How do you calculate the angle of swing in a carousel?

To calculate the angle of swing in a carousel, you will need to know the radius of the carousel and the speed at which it is rotating. The formula for calculating the angle is: angle = arctan(speed^2 / radius * 9.8). This will give you the angle in degrees.

## 2. Why is it important to know the angle of swing in a carousel?

Knowing the angle of swing in a carousel is important for safety reasons. It helps determine how much force is being exerted on the riders and can help prevent accidents or injuries.

## 3. Can the angle of swing in a carousel be controlled?

Yes, the angle of swing in a carousel can be controlled by adjusting the speed at which it rotates. Slower speeds will result in a smaller angle of swing, while faster speeds will result in a larger angle of swing.

## 4. Does the weight of the riders affect the angle of swing in a carousel?

Yes, the weight of the riders will affect the angle of swing in a carousel. Heavier riders will cause the carousel to swing at a larger angle compared to lighter riders.

## 5. Are there any other factors that can affect the angle of swing in a carousel?

Other factors that can affect the angle of swing in a carousel include the design and construction of the carousel, wind speed and direction, and any external forces acting on the carousel. These factors should be taken into consideration when calculating the angle of swing.

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