The Swing Problem:How can you go completely around?

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In summary, the conversation discusses the minimum speed needed for a swing to make a full rotation. The main forces involved are gravity and centripetal acceleration, with the minimum speed at the bottom being equal to the square root of five times the acceleration due to gravity multiplied by the radius of the swing. The conversation also mentions the use of simple circular motion physics and energy conservation to solve the problem, and addresses potential concerns such as the varying rate of change of theta and the changing mass of the person on the swing.
  • #1
E_Man
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Every kid has asked the question on a swing, could I go all the way around? How fast would I have to go? I wanted to find the answer.I’ve taken the first year of calculus, and I have been having trouble with the math of this problem :confused: . I am not sure there is an algebraic solution to this problem. :surprise:

You are on a park swing with radius R. What must your minimum speed be at the bottom of your circular trajectory such that you perfectly traverse the circular path created by the rotating swing?

Note: There are several parts to this problem. First you must calculate the rate of change of theta. You must then integrate the deceleration of gravity over the swing’s circular path. At all times the upward vector of centrifugal force, (V^2)/R must be greater than the acceleration of gravity so that you stay on your circular path.

Thanks, Elias
 
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  • #2
The easy way is to examine what happens at the top of the motion, where the speed is lowest. Find the minimum speed needed at the top to maintain tension in the swing (and thus maintain the circular path). Then use energy conservation to find the corresponding speed at the bottom.
 
  • #3
use simple circular motion physics, you don't need any integration. The main fores actiong on you are gravity and centripital acceleration. At the top of the swing is the most miportant part, since the gravity and centripital acceleration are in opposition of each other, so if you want the kid to have a circular path and make it over the actual swing, you must set the two forces equal to each other, solve for v, and this will be your MINIMUM speed. Like so:

mg = mv^2/r (m's cancel out)
g = v^2/r
v = (sqrt(gr))
therefore the minimum speed of the swing must be: v = (sqrt(gr))
 
  • #4
v=(sqrt(gr)) is ONLY the minimum velocity at the apex of the trajectory. The original starting velocity is NOT v=(sqrt(gr)).

Mathematically, this is wrong because you neglect that V varies according to the gravitational vector you also neglect that V varies according to the varying rate of change of theta.

Experimentally we can also see this if we solve for a 10 foot swing. V=(sqrt(32X10))
That means at 12 MPH youve got enough speed to go all the way around. Not True. Try it yourself.
 
  • #5
Nenad said:
use simple circular motion physics, you don't need any integration. The main fores actiong on you are gravity and centripital acceleration. At the top of the swing is the most miportant part, since the gravity and centripital acceleration are in opposition of each other, so if you want the kid to have a circular path and make it over the actual swing, you must set the two forces equal to each other, solve for v, and this will be your MINIMUM speed. Like so:

mg = mv^2/r (m's cancel out)
g = v^2/r
v = (sqrt(gr))
therefore the minimum speed of the swing must be: v = (sqrt(gr))

The forces acting are the magnetic force and the string tension. Not the centr. acc.

Besides, you can't write
[tex]\sum \vec F = - m\frac{v^2}{R} \hat \rho[/tex]

Because there this is NOT a uniform circular motion since the speed ain't constant.

for E_Man:
Follow this
 
  • #6
Magnetic force?? And, true, the speed isn't constant, but so what?

This is an easy one. I thought I had explained how to do it in my earlier post. Here are the details:

At the top of the motion, we need the minimum speed so that the swing remains taut (assuming a flexible chain). From Newton's 2nd Law we get:[itex] F = mg = mv^2/r[/itex], so the minimum KE at the top of the swing is [itex]KE_{top} = mgR/2[/itex]. From conservation of energy, the KE at the bottom is [itex]KE_{bot} = mgR/2 + 2mgR = 5/2 mgR[/itex], so the minumum speed at the bottom must be: [itex]v_{min} = \sqrt{5gR}[/itex].
 
  • #7
Doc Al's right. I think the easiest way to do it is to find a relation between the speed at the bottom and the speed at the top by means of energy conservation.
Setting the gravitational potential energy zero at the bottom we find for the energy of the kid at the bottom:
[tex]E = \frac{1}{2}mv_{bottom}^2[/tex]

And at the top (at height 2R):

[tex]E = \frac{1}{2}mv_{top}^2+2mgR[/tex]

Therefore (equating these and simplifying):
[tex]v_{bottom}^2=v_{top}^2+4gR[/tex]

You want [tex]v_{top}^2/R \geq g[/tex], so you must have:

[tex]v_{bottom}^2 \geq gR+4gR=5gR[/tex]
And so the minimum speed at the bottom must be [tex]v_{min}=\sqrt{5gR}[/tex]
 
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  • #8
Like your answer Doc but you neglected the fact that g is not constant ,and his mass changes with speed -- if the guy neglected this he might have a bad accident. Ray.
 
  • #9
rayjohn01 said:
Like your answer Doc but you neglected the fact that g is not constant ,and his mass changes with speed -- if the guy neglected this he might have a bad accident. Ray.
Give me a break. :rolleyes:
 
  • #10
Doc Al said:
Magnetic force?? And, true, the speed isn't constant, but so what?

This is an easy one. I thought I had explained how to do it in my earlier post. Here are the details:

At the top of the motion, we need the minimum speed so that the swing remains taut (assuming a flexible chain). From Newton's 2nd Law we get:[itex] F = mg = mv^2/r[/itex], so the minimum KE at the top of the swing is [itex]KE_{top} = mgR/2[/itex]. From conservation of energy, the KE at the bottom is [itex]KE_{bot} = mgR/2 + 2mgR = 5/2 mgR[/itex], so the minumum speed at the bottom must be: [itex]v_{min} = \sqrt{5gR}[/itex].

Sorry, I meant gravitational force, I was thinking of something else.

And the fact that velocity ain't constant means that there is a tangentical acceleration as well, so you have to take that into account in Newton's 2nd Law.
 
  • #11
rayjohn01 said:
Like your answer Doc but you neglected the fact that g is not constant ,and his mass changes with speed -- if the guy neglected this he might have a bad accident. Ray.

I think this was the mistake my daughter made. She tied herself to the swing by the waist, which obviously made things safe enough to let go the chains.

Oooh :eek: One broken arm later, plus a physics lesson learned...
 
  • #12
Have you ever shortened the chain on a swing that was too low to the ground? You know method I'm talking about; you thrust the swing forward with enough velocity to go all the way around. The velocity at which you must make the swing travel when you push it forward (the velocity at the bottom of its ark, where you release it) is exactly the same whether the swing is empty or occupied.
 
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  • #13
nit picking :-)

eJavier said:
And the fact that velocity ain't constant means that there is a tangentical acceleration as well, so you have to take that into account in Newton's 2nd Law.
At the point where I apply Newton's 2nd law (the top of the motion) the tangential acceleration is zero. :smile:
 
  • #14
THANKS DOC AL AND EVERYONE ELSE! Thanks too fo the link given by eJavier!

I pose a new challenge:
Can anyone find an explicit function for the velocity as a function of time?
 
  • #15
E_Man said:
Can anyone find an explicit function for the velocity as a function of time?
I'm pretty sure that the diff. eq. is nonlinear. Diff. eqs. of this kind are not generally solvable analytically (which basically means it is not generally possible to find the explicit solution, even in principle). The solution is probably some obscure special function defined by the diff. eq. that you can find.
 
  • #16
rayjohn01 said:
Like your answer Doc but you neglected the fact that g is not constant ,and his mass changes with speed -- if the guy neglected this he might have a bad accident. Ray.

Well you can neglect the mass increase because even with a near inifinite swing length the velocity of the swing won't exceed escape velocity (11205 meters/sec) and at that velocity the mass would only increase by a factor of 1.0000000007. :wink:

As for the varience in "g", if you absolutely have to, you could use the formula:

[tex]v =\sqrt{2GM \left\{\frac{r}{2(R+2r)^{2}}-\frac{1}{R+2r}+\frac{1}{R}\right\}} [/tex]

Where r is the radius of the swing and R is the distance of the bottom of the swing from the center of the Earth.

Now for a swing with a raduis of 4 meters, we would get an answer of:

14.03029073 m/s or 31.56815415 mph, using Doc Al's equation.

Using the new equation, we get an answer of:

14.03028017 m/s or 31.56813039 mph.

Thus we see that Doc Al's equation, though slightly less accurate, fudges a bit to the high side and still gives an answer that would safely get you over the top of the bar. :smile:
 
  • #17
Ah ok could you tell me more about these differential equations? this is the problem I originally had. I was trying to approach the problem by finding the functions for each variable. Darn.
 
  • #18
janus

Thank you Janus , I wondered if Doc would pick up on the fact that the guy was still safe -- but he just rolled his eyes instead -- clearly not a physicist of the first water - whereas you have genius quality . Ray
Now about length contraction -------------------.
 
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  • #19
If I remember correctly large pendulum swings involve integral(1/sqr(cos())) functions at least for 'period' and belong to eliptic functions -- angels fear to tread --
Ray.
 
  • #20
Would the function for velocity along a pendulum even fit? I mean it seems like the bottom half of the circle would have one equation and the top half would have another, because of the fact that on the bottom hemisphere of motion the deceloration of gravity is on an inclined plane, and on the top half it is straight down
 
  • #21
If you require that you should "safely" go all the way around, then your "orbit" in phase space would be well outside a particular orbit that is known as the "separatrix" (and, incidently, the 1-D pendulum, which is basically this problem, has been subjected to a tiresome share of analysis in classical mechanics). In phase space, the orbit is quite simply a sinusoid (I'm pretty sure). You can read the velocity off of the location in phase space directly, as velocity (or, more appropriately, momentum) is essentially one of the coordinates of phase space. Based on this argument, I would wave my hands and say that the velocity is a well defined function in the sense that it obtains a particular single finite value at any given point in time, but the function of time does not correspond directly to the function of position as a sinusoid. As rayjohn has mentioned, I smell elliptic integrals (which are defined with trig functions).
 
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  • #22
I thought this was a high school level problem of a simple swing, I didn't know we would end up at elliptic integrals. :smile:
 
  • #23
Gza said:
I thought this was a high school level problem of a simple swing, I didn't know we would end up at elliptic integrals.
At the high school level, they usually make the simplifying approximation of small oscillations. That allows one to approximate the motion as simple harmonic motion. On a slightly deeper level, you can thing of the problem as a 1-D problem in configuration (the angular position of the pendulum bob) which leads to a 2-D phase space (the other D is for the momentum or velocity). Then, you can define a potential energy and expand it in a Taylor series:

V = a0 + a1x + a2x2 + a3x3 + ...

The gauge invariance allows you to set a0 = 0 without changing the physical analysis.

For small oscillations, the pendulum stays close to the straight down position. If you consider the expansion about that point, then the potential increases in both directions, and thus a1 = 0. This gives an expasnsion of the potential as:

V = a2x2 + a3x3 + ...

Since the displacement is small, the lowest order term in the expansion dominates, so you can approximate the potential as:

V ~ a2x2.

This is the potential of the harmonic oscillator, which has sinusoidal solutions. Notice that this expression for the potential depended on the small oscillation approximation. Going all the way around the swing is not a small displacement, so you must include all of the even order terms which are signigicant:

V = a2x2 + a4x4 + a6x6 + ...

It turns out that the potential itself is actually a sinusoid:

V = Acos(kx).

When you stick this in the diff. eq., it gets nasty. Also notice that this is somewhat related to the fact that there is no analytical way in general to find the roots of an infinite order polynomial.
 
  • #24
Could you explain more about why you neglected the odd powers of Velocity's Taylor series?

Second question: would it be worthwhile to try and solve this by using series? would i still have to solve using differential equations?
 
  • #25
Sorry had to amend -went too fast
V^2 = r.g.(1 + 2.(1 - cos(theta)))
where theta is angle 0 at top
r is radius
g gravity
and V is the peripheral speed ( not angular)

Barring errors this is simply derived from energy
BUTTTTT the period is given by the integral of 1/v which is really nasty
and that's where I was told by a mathematician one gets into eliptical
integrals . So the above does not involve complicated maths but the concepts of energy exchange ( kinetic to potential in gravity) may be unfamiliar to some. I'm not going post a proof unless specifically asked because I do not know how to post diagrams. Ray
The period can be obtained either by numerical integration or by setting up a simulation for those that enjoy programming
This problem can be extended to long period pendulums by using two minutely different masses at both ends of a beam. have a good day Ray.
 
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  • #26
E_Man said:
Could you explain more about why you neglected the odd powers of Velocity's Taylor series?
If you are referring to the V in my previous post, then it does not represent Velocity; it represents potential energy, V. I was able to drop the odd powers of the potential energy based on the symmetry of the pendulum about the hanging-straight-down position.




E_Man said:
Second question: would it be worthwhile to try and solve this by using series? would i still have to solve using differential equations?
What is it that you have in mind? I almost always run into a combination of both series and diff. eq., specifically, expanding the function into an infinite series of basis functions. I don't know of any way around the diff. eq.; it is a fundamental part of the physical description.
 
  • #27
Back to my original question, is there a mathematical(analytical) way of solving this? or not?
 
  • #28
E_Man said:
Back to my original question, is there a mathematical(analytical) way of solving this? or not?

Sure, you could do the work integral for gravity on the swing, which is essentially what the conservation of energy stuff is all about.
 
  • #29
ray john came up with "V^2 = r.g.(1 + 2.(1 - cos(theta)))" but the reason i don't thing this can be evaluated is because theta is equal to the (average velocity X time)/radius of swing

If anyone sees a way to solve this quasi algebraiclly A post would be greatly appreciated
 
  • #30
Eman

Yes your correct -- but I'm not saying you cannot solve algebraically ---
someone showed me an Elliptic integral type solution but the answer seemed as complex as the problem, so I solved numerically because although it looks a bit nasty it is actually a well behaved function ( like it does not shoot off to infinity ). It's absolutely vital to recognize that the average required for T
is NOT ave(v), but ave(1/v) which is totally different.
That is ave(1 / v ) <> 1 / ( ave(v) )
I had originally tried to solve a long period pendulum with nearly equal masses at either end of a beam for angles starting near vertical , since I was not sure of the result , I simulated this using Qbasic and could not get it to agree with the numeric integration until I finally saw the above inequality.
Ray.
in case of confusion
The equation I gave is the instantaeous peripheral speed ( not angular ) at any point in the swing at angle theta from the top assuming that the initial speed is just sufficient to keep the chains taught all the way round.
So the initial speed is given for theta=0 and max speed for theta = pi.
the derivation is straight forward -- it's the period T which is problematic.
However for a 6ft rope T ~ 1.74 secs. and the g forces are ~5-6 g
Ray
 
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1. What is "The Swing Problem"?

"The Swing Problem" is a physics problem that involves a swinging motion, where an object is attached to a string and is swinging back and forth. The goal is to determine how to make the object go completely around in a circular motion.

2. What are the key factors that affect "The Swing Problem"?

The key factors that affect "The Swing Problem" are the length of the string, the mass of the object, and the initial angle at which the object is released. These factors affect the centripetal force and the speed at which the object is moving.

3. How does centripetal force play a role in "The Swing Problem"?

Centripetal force is the force that pulls an object towards the center of a circular motion. In "The Swing Problem", centripetal force is responsible for keeping the object in a circular motion and preventing it from flying off in a straight line.

4. What is the formula for calculating the speed needed to go completely around in "The Swing Problem"?

The formula for calculating the speed needed to go completely around in "The Swing Problem" is v = √(g * L), where v is the speed, g is the acceleration due to gravity, and L is the length of the string.

5. How can you increase the chances of successfully completing "The Swing Problem"?

To increase the chances of successfully completing "The Swing Problem", you can increase the length of the string, decrease the mass of the object, and release the object at a smaller initial angle. This will decrease the centripetal force and increase the speed, making it easier for the object to go completely around. Additionally, making sure the string is taut and the object is released with a slight push can also help.

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