Physics of a swing (pendulum?)

Factual Questions
1

OK, it’s been over 12 years since I’ve had anything resembling a physics class, so let’s try to keep this in layman’s terms so I can understand it.

Imagine a chain suspended taut between two trees. A swing hangs from the middle of the chain; the swing has only one rope.

How can I compute the force on the chain based on a person’s mass m? And while we’re at it, at what point in the swing’s cycle is the maximum force exerted on the chain? Does the force exerted on the chain increase when the swing goes higher?

2

Gee… now that school has started we’re starting to see all of these basic physics 101 questions. Just a coincidence?

3

Sitting on the swing: F = ma where m is the person’s mass and a is the acceleration due to gravity (watch your units!).

I can’t figure out how actually swinging (assuming a smooth swing without “cherry bumps”) would change that equation as the maximum downward force is always generated by gravity’s action on the person, increasing from zero at the peak of the swing to the total value at the bottom.

By the way, don’t make the chain too taught or you’ll lose the amount of force you can apply to it (well, given strong chains and normal people this isn’t a huge concern).

4

Yup.

Well, that’s what I was wondering. It seems like there should be more force exerted on the chain due to the swinging, but I don’t know for sure.

5

Cherry bumps?

6

If you make the simplifying assumption that the chain itself has no weight compared to the weight on the swing, then to find the force on the chain, you need to find the force on the swing’s rope first.

The force on the swing’s rope is from two things: 1) the weight of the swing, and 2) centrifugal force. If it’s swinging, its speed will be maximum at the bottom, so the centrifugal force will maximum, therefore the force on the rope will be maximum.

To convert the force on the rope to the force on the chain holding it, take one-half of the rope force (because it’s distributed on two chain segments) and divide that by the sine of the angle that the chain makes.

By the way, the force on the chain due to gravity won’t decrease to zero at the peak of the swing, unless you’re swinging all the way up to a horizontal position.

7

Now we’re getting somewhere. By rope force do you mean the force exerted on the rope due to the person’s weight (i.e. F=ma, as mentioned earlier)? And what angle are we referring to? The angle the chain makes with what?

8

By rope force, Curt is indeed talking about the force due to the person’s weight, but the total rope force would be the sum of the force due to gravity (F=mg, where g = acceleration due to gravity) and the force due to the swinging action (F=mv[sup]2[/sup]/L, where v = velocity at the bottom of the swing and L = length of rope, if my memory is correct). The angle the chain makes is the angle between the chain and a horizontal line. The more horizontal the chain, the smaller the sine of this angle, and the higher the tension in the chain.

9

Ok. After some calculations including the centrifugal force I find that the max force would be:

mg + [sup]pi[sup]2[/sup][/sup]/[sub]4[/sub]mg

Anyone say different?

10

Ok, ok, I’ll at least explain what I did so anyone can criticize as necessary.

equation for the period P of a simple pendulum = 2pi*sqrt(L/g)

Max velocity at 1/2 P which is at the bottom of the swing… ack! now that I’m typing it I think it shoild be at 1/4 P !! anyway, that doesn’t significantly change the answer.

We know that the centrifugal force is mv[sup]2[/sup]/L

So the total force is mg + mv[sup]2[/sup]/L

So what is v? v = gt
What is t? Why, 1/4 P (again, I used 1/2 P without thinking in the equation, so final answer will differ here).

So now we have
F[sub]T[/sub] = mg + ([sup]m[/sup]/[sub]L[/sub])v[sup]2[/sup] or equivalently
F[sub]T[/sub] = mg + ([sup]m[/sup]/[sub]L[/sub])(p/4)[sup]2[/sup]

which eventually works out through appropriate algebra to be the same as I gave above, but with pi[sup]2[/sup]/16 instead of 4.

So, roughly, the downward force on the swing is about 1.6mg. As for what it is at the chain where it is tied off, now that’s a different answer altogether (and depending on how taught the chain is, somewhat suprising!)

11

Where do you get v = gt? The centrifugal force should be independent of the other quantities in the problem. Although the period of a pendulum (for small displacements) is set by m, g, and L, the total arc the pedulum swings through (and thus the speed) can be a range of numbers.

12

Velocity = acceleration * time under constant acceleration.

The period of a pendulum is independent of the mass and is only set by the length of the arm and the height from which it starts. The speed goes through a range of numbers, of course, from 0 at the “start” to a maximum at the very bottom. The time it takes to do this is strictly a function of the pendulum length since all masses (disregarding wind resistence) fall at the same rate.

Since we’re only concerned with the velocity as it approaches and reaches the maximum, the velocity is a linear function of time, and time is a function of length and the (constant) acceleration due to gravity.

13

The centripetal force should be the resultant force of the weight of the person + the reaction force exerted on the swing by the person. To find the tension in the string, it shouldn’t be the summation of the 2 forces since the centripetal force and the weight of the person acts in different directions.

I think that the centripetal force shouldn’t play a role here. The tension on the rope depends on the force due to gravity. W=mg. And the resultant tension on the rope should differ at various angles because of the effective angle. For example, a swing inclined at 35deg shld have a tension given by mg(cos(35deg)).

14

The accelaration is not constant. The accleration is at its max at the top of the swing and is zero at the bottom. Look at it this way, the period is constant. The speed at the bottom must be slower for a smaller arc.

15

True in general, but the maximum centripetal acceleration occurs when the swing is at the bottom of the arc, in which case the forces are aligned. Since the maximum is likely the most interesting case, I’ve been assuming the simplification of calculating tension when the swing is vertical.

And also what DrMatrix said: constant period + varying arc length = varying velocity.

16

I’m not trying to analyze the entire period, however. Velocity is monotonic and increasing from t=0 to t= P/4. The rest of the period is useless for determining the maximum force.

17

I think you’re missing the point. Let’s suppose, just to pick a number, that we’re working with a pendulum that has a period of four seconds (and a length of roughly 13 feet). If I pull back the pendulum one degree and let go, it will swing through one degree in one second, with an average velocity of pi/180*L/1 = 2.72in/sec. The maximum velocity, of course, is greater (ummmm… by a factor of sqrt(2)? Doesn’t really matter). If I then pull back the pendulum five degrees and let go, it will swing through the five degrees in one second, with an average velocity of 13.61in/sec; the maximum velocity is greater by the same factor. So velocity, and thus acceleration, are dependent on how far back you pull the pendulum.

18

erislover, I think your problem is that you are using a constant value for acceleration (g), and from that deriving the force of gravity on the pendulum. The problem is the equation “F=ma” is a vector equation, and since the pendulum is constrained to move in an arc, the actual force on the pendulum in the direction it can accelerate depends on the sine of the angle from vertical. At the bottom of the pendulum, the acceleration due to gravity is zero, because the bob can only move side-to-side.

Does that help?

19

What would help is if someone answered the OP instead of me flailing around here.

I thought it was implied that we are discussing the maximum force this chain would feel from the swing. Thus, the “pull back” amount would be completely horizontal. Anything greater would result in a different effect and anything less won’t have the maximum velocity.

I understand force and velocity are vectors. In my mind, the direction of the swing is totally unimporant for one part of the equation: the mass is always going to be exherting a downward force of mg. The rope will feel cos(theta)*mg as the normal component when the swing and the direction of the force are not the same. However, at the bottom when the normal force is equal to the gravitational force, theta=90 and costheta = 1, so the downward force = mg.

What is important to the tension would then have to be any force felt by the rope/chain due to cetrifugal action. Some here have led me to believe this would be additive. I do not know; I am willing to do the calculations when everyone else is clearly not willing to, but I need to know what the hell to calculate when I have pretty much demonstrated in my first post that I didn’t know what to include.

So using the mv^2 / L i tried to come up with the extra term. Since we know the pendulum’s period is dependent only on the length of the arm in our scenario we should be able to figure out velocity from the period.

Using the trig and dealing strictly with vector math I see that the component of gravity which results in the velocity is F = mgsin(theta). If this is true then I was correct in my very first post which stated that we would only feel the weight of the person at most.

If anyone would care to draw themselves a diagram they would see that when one tries to include centrifugal force, or any other force they might conjure up, they would see that the only force one finds are angular components of the mass’s weight, and no matter what the velocity of the swing, no matter what angle it is at relative to anything, the resultant forces must equal mg, can never be more than mg because mg is the hypotenuse of the vector triangle.

PLEASE either correct me once by demonstrating the real answer, or stop leading me on goose chases and say "You were right in your first post erislover. I don’t care if I’m wrong, but for once I’d like to see it.

20

Perhaps you are making a different assumption about the problem than I am? I’m gathering this from the last sentence quoted here. Let me lay out what I’m assuming and then see where we differ:

  1. A (weightless) chain is connected to two trees.
  2. From the center of the chain a rope of length L hangs vertically.
  3. On the end of the rope is a weight of mass m.
  4. When everything is stationary, the weight causes the chain to pull into a V shape; each leg of the V is at an angle theta with the horizontal.
  5. In the stationary condition, the rope is vertical, and the tension in the rope is = mg.
  6. With that in mind, the tension in the chain can be calculated geometrically: F = (mg/2)/[sin(theta)] (the 2 from the fact that there are two “legs” of the V.

OK, is that reasonable? Now, let’s add in the forces due to the swinging of the weight:

  1. I assume that, when the weight swings, the chain stays stationary and the rope/weight rotates around the connection point between the rope and chain. (I realize this isn’t really quite true, and there are better assumptions, but just to keep things simple…) erislover, did you make a different assumption here?

  2. I further assume that the weight acts like a simple pendulum, with a period, T = 2pisqrt(L/g).

  3. Thus, the weight travels in a circular arc, with zero velocity at each end of the arc and maximum velocity in the middle, when the rope is vertical.

  4. The centrifugal force on the weight is =mv[sup]2[/sup]/L, where v is the linear velocity of the weight along the arc of its travel. This force is in the radial direction.

  5. Since the velocity is maximum when the rope is vertical, the centrifugal force is also. I’ll simplify by looking at the tension in the rope only at this position.

  6. The tension in the rope is simply the sum of the gravitional and centrifugal forces, F = mg + mv[sup]2[/sup]/L; the tension in the chain is 1/2 of this divided by sin (theta).

  7. OK, now, what is v? Consider: If the weight is not swinging, v=0. Further: If I pull back the weight one degree and let go, it will take T/4 seconds for the weight to reach the vertical position. If I pull back the weight five degrees and let go, it will still take T/4 seconds for the weight to reach the vertical position. Thus: If the weight travels five times as far in the same time, it must travel five times as fast.

  8. Thus, the v term depends on how far I pull the weight back.

  9. With this in mind, there’s really no simpler answer than the one I give above in step 12. The velocity term could be rewritten in terms of pullback distance, or angle, or some other measure, but it can’t be simplified to be solely in terms of m, g, and L.