The Swing Problem:How can you go completely around?

1. Jul 13, 2004

E_Man

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 . Im 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

2. Jul 13, 2004

Staff: Mentor

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. Jul 13, 2004

use simple circular motion physics, you dont 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 eachother, 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 eachother, 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. Jul 13, 2004

E_Man

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. Jul 14, 2004

eJavier

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

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

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

for E_Man:

6. Jul 14, 2004

Staff: Mentor

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:$F = mg = mv^2/r$, so the minimum KE at the top of the swing is $KE_{top} = mgR/2$. From conservation of energy, the KE at the bottom is $KE_{bot} = mgR/2 + 2mgR = 5/2 mgR$, so the minumum speed at the bottom must be: $v_{min} = \sqrt{5gR}$.

7. Jul 14, 2004

Galileo

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:
$$E = \frac{1}{2}mv_{bottom}^2$$

And at the top (at height 2R):

$$E = \frac{1}{2}mv_{top}^2+2mgR$$

Therefore (equating these and simplifying):
$$v_{bottom}^2=v_{top}^2+4gR$$

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

$$v_{bottom}^2 \geq gR+4gR=5gR$$
And so the minimum speed at the bottom must be $$v_{min}=\sqrt{5gR}$$

Last edited: Jul 14, 2004
8. Jul 14, 2004

rayjohn01

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. Jul 14, 2004

Staff: Mentor

Give me a break.

10. Jul 14, 2004

eJavier

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. Jul 14, 2004

BobG

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 One broken arm later, plus a physics lesson learned........

12. Jul 14, 2004

LURCH

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.

Last edited: Jul 15, 2004
13. Jul 14, 2004

Staff: Mentor

nit picking :-)

At the point where I apply Newton's 2nd law (the top of the motion) the tangential acceleration is zero.

14. Jul 14, 2004

E_Man

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. Jul 14, 2004

turin

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. Jul 14, 2004

Janus

Staff Emeritus
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.

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

$$v =\sqrt{2GM \left\{\frac{r}{2(R+2r)^{2}}-\frac{1}{R+2r}+\frac{1}{R}\right\}}$$

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.

17. Jul 15, 2004

E_Man

Ah ok could you tell me more about these differential equations? this is the problem I originally had. I was trying to aproach the problem by finding the functions for each variable. Darn.

18. Jul 15, 2004

rayjohn01

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

Last edited: Jul 15, 2004
19. Jul 15, 2004

rayjohn01

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. Jul 15, 2004

E_Man

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