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### Fun Force Question

09/04/2012 9:02 AM

What is the relationship between the amount of kinetic energy of a swinging pendulum at the bottom of its swing and that of a free falling object of the same mass.

To put it another way: If you dropped two heavy weights at the same time, only one of them was tied to a fixed point by a horizontal massless string, what would their conditions be at their mutual lowest point?

Do they have the same speed?

Do they have the same kinetic energy?

This is not actually a homework question. I was just now pondering building a hypothetical machine that would drop a weight, pendulum style, that would simulate different gun recoils. You know, for fun. And, I was thinking of the appropriate force equations to use and I couldn't remember ever have worked something like this in school (maybe I skipped that day). Anyway, I thought it would be fun so I'm sharing. I think this also might be a speed trap for trolls who don't read posts.

Thanks!!

-A-

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#1

### Re: Fun Force Question

09/04/2012 9:44 AM

Given that the item tying the mass to the point of rotation is massless, yes the two objects would have the same kinetic energy, hence the same speed. However, given the restraint of motion, the velocity vector of the rotating mass would be horizontal while that of the falling mass would be vertical, but still with the same speed.

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#2

### Re: Fun Force Question

09/04/2012 10:41 AM

One might consider the aerodynamics of the two objects, which appear from the sketch to be mallet-shaped, assuming they are falling in still air of equal physical properties:

• The pivoted object travels a path that is pi/2 longer than the free-falling one and turns in the air as it travels, exposing itself to more form drag and profile drag then the free-falling one as it does so. Therefore one might reasonably expect it to arrive at the stop point with lower kinetic energy than the one that is free-falling.
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#3

### Re: Fun Force Question

09/04/2012 11:00 AM

I think it is reasonable to assume the "massless" handle on the rotating hammer experiences less drag than the falling hammer.

If the rotating hammer travels tau/4 further, then is there a delay? Or, do they come to the same "bottom" at the same time, but the rotating hammer is moving faster?

I feel an experiment coming on . . .

-A-

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#4

### Re: Fun Force Question

09/04/2012 11:30 AM

Go for it.

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#6

### Re: Fun Force Question

09/04/2012 12:23 PM

Time is a tricky one - think (from discussions in a previous thread) you can ony solve for time using numerical integration.

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#5

### Re: Fun Force Question

09/04/2012 12:04 PM

Different but related: I had to cut a decent sized branch from an ash tree. It was very awkward and I had to do it from a ladder. I asked myself what could go wrong and I decided the risk was that the smaller, springy, branches could hit the ground and throw the butt end back at me and knock me off the ladder. Not likely, but what if it does happen? Since it would push me to my left, I decided that I should throw my chainsaw to my right, and stay with the ladder as it would give me more time and I would be slower when I touched down.

The branch did spring back and hit the ladder. I followed my routine, but when I rolled on landing, I rolled backward over a rock that fractured a vertebrae in my lower back, giving the hospital an excuse to put me through every test imaginable.

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#7

### Re: Fun Force Question

09/04/2012 12:36 PM

If you can wade through all of this, it might help.

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#8

### Re: Fun Force Question

09/04/2012 1:03 PM

I have followed the links provided and come up with the following.

The pendulum equation is: T[sec]=2pi()*sqrt(L[feet]/g[ft/sec2])

Assuming the pendulum length to be 3 feet:

T=2pi()*sqrt(3/32.2) ==> T=1.918[sec]

Taking half of that: T/2=.9589[sec] (For half the swing)

Using my equation of motion calculator in excel, I find that a dropped object requires 0.43[sec] to fall 3 feet.

Does this look right? The rotating hammer requires TWICE the amount of time to reach bottom?

-A-

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#9

### Re: Fun Force Question

09/04/2012 4:35 PM

Yes it does. They start with the same acceleration, the falling hammer maintains that acceleration but the swinging hammer has almost no vertical acceleration towards the end.

Here is an analogous case. An Earth clone, not rotating about the sun, released just as the Earth goes by. The clone would splash into the Sun well before the Earth had traveled for one quarter turn, three months. Not quite an analog because the Sun's pull would increase as the clone approached.

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#12

### Re: Fun Force Question

09/04/2012 5:51 PM

The pendulum equation (as you've expressed it) is an approximation for small angles. 90° is not a small angle.

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#17

### Re: Fun Force Question

09/05/2012 8:26 AM

According to a table on wikipedia (on a link I just deleted) the period error on a 90deg swing is slightly larger than .1%. Therefore, if this is correct, I need to add .05% to my final answer? I can live with that.

-A-

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#22

### Re: Fun Force Question

09/05/2012 9:09 AM
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#14

### Re: Fun Force Question

09/05/2012 12:05 AM

Without actually doing any math, yes, that sounds right, because the hinge is exerting an upward (variable) vertical force which slows the vertical motion.

The hinge is also exerting a variable horizontal force, which has no effect on the vertical motion.

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#10

### Re: Fun Force Question

09/04/2012 4:45 PM

I just load 'em up till the primer starts extruding back into the firing pin hole.

Bent the firing pin on my 686 w/8 3/8" barrel doing that, once.

I backed off, just a hair on the load, and called it good.

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#11

### Re: Fun Force Question

09/04/2012 5:14 PM

Oooops!

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#13

### Re: Fun Force Question

09/04/2012 6:13 PM

I hope that wasn't yours.

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#20

### Re: Fun Force Question

09/05/2012 9:06 AM

Nope, not mine.

I never packed the cases. I knew a fellow who loaded his .300 Win Mag cartridges like that... similar results. The value of the rifle or pistol drops dramatically with a compromised breech.

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#21

### Re: Fun Force Question

09/05/2012 9:09 AM

There are rifle powders that are intended to be compressed by seating the bullet.

Never saw the need for these myself.

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#23

### Re: Fun Force Question

09/05/2012 9:37 AM

Huh... Okay. Something I haven't seen. Been years since I reloaded cartridges.

Only powder I ever packed was black powder or Pyrodex. Shootin' those frontstuffers was quite a bit of fun, actually.

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#24

### Re: Fun Force Question

09/05/2012 9:44 AM

Been too long for me too. I think some of the larger stick powders were compressed.

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#15

### Re: Fun Force Question

09/05/2012 8:08 AM

When freefall position hits zero (asterisk), freefall velocity is same as maximum velocity of pendulum (at bottom of swing).

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#19

### Re: Fun Force Question

09/05/2012 8:59 AM

Was that using the "pendulum equation" (small angle approximation)?

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#29

### Re: Fun Force Question

09/07/2012 4:41 PM

Was that using the "pendulum equation" (small angle approximation)?

No, angular acceleration equals 9.8 m/s2 times sin(θ), where θ is the angle of the pendulum with respect to vertical. Small angle approximation assumes θ = sin(θ).

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#30

### Re: Fun Force Question

09/07/2012 5:44 PM

Quite. Thankyou.

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#16

### Re: Fun Force Question

09/05/2012 8:17 AM

use point mass make less confusing schematics

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#18

### Re: Fun Force Question

09/05/2012 8:36 AM

OK, so it appears what I need to do to build my machine is to:

1. Get the published recoil force generated by different firearms [in lbf]

2. Pick a mass and calculate the drop distance

3. Build a rig that varies the length of the pendulum according to the height of the victim subject's shoulder

4. Adjust the mass to match the desired force calculated from my drop calculation

6. Charge teen-agers for the chance to see how tough they are in front of their friends

Got it!!

-A-

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#25

### Re: Fun Force Question

09/05/2012 9:46 AM

I'm sure you've already thought way past this but the rotating hammer is the same principle as dropping except you have to use the rotational energy equations.

http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html

I don't have it with me right now but you should get an ME Reference Manual (the blue/yellow book used for taking the PE). I'm sure it will have that in there.

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#26

### Re: Fun Force Question

09/05/2012 10:12 AM

Maybe I'm lazy but, looks alot like mv at point of impact for a semetrical body.

Besides, I'm thinking now I'm going to transfer the energy through a sled, that I can vary the weight of, to simulate different rifles. This way I can use a real butstock and optic (with the hopes of getting the occational "scope kiss").

-A-

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#27

### Re: Fun Force Question

09/05/2012 10:39 AM

is supposed to look a lot like mv.

It's the rotationl version of energy lol. Without doing a calculation, though, I don't know how far off you'll be by just using translational.

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#28

### Re: Fun Force Question

09/05/2012 12:40 PM

Or you could use the principle of conservation of (linear) momentum: along the line of the bullet's path and the rifle's recoil, bullet mass x muzzle velocity = rifle mass x recoil velocity.

Linear to angular momentum and then back to linear, as in your proposed hinged hammer experiment, is interesting by itself, but it might not be the simplest way to get what you want.

An accelerometer attached to the rifle would give you the recoil force (F = ma).

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#31

### Re: Fun Force Question

09/11/2012 12:41 PM

The problem is quite interesting and since I was not home I come a bit late into the discussion I shall try however to bring my 2 pence results.

If no resistance is taken into consideration then since the mass moves in the potential field of gravity its velocity value will be the same and only its angle will be either vertical or horizontal.

The Two cases can be described by the following equations :

vertical fall z"= g- Fr where Fr/M = ξ*A*ρf*(z')^2/2/M with ξ= form coefficient;ρf= specific mass of the fluid in which the body moves;A= area perpendicular to the velocity vector; M= mass of the body = A*h*ρm with h= body thickness and ρm= specific mass of considered body.

Making the simplifications one obtains the differential equation: z"=g- P*(z')^2 where P= ξ*ρf/(2*h*ρm).

If the equation is integrated for several values of "P" (0:0.1/0.5/1/5/10) the results are:

The red curves show the velocity versus time for same fall height of 1m. The displacement z is represented by the green curves.

For the circular movement the equation is similar but not identical:

φ" = (g/R)*cos(φ)- P*R*(φ')^2 where φ= angle between body position on the arc and horizontal (its initial position!) R= arc radius MUST be equal to the fall linear height (1m) and P is the same parameter as above for the linear, vertical, fall.

Proceeding as for the linear case one obtains following results:

The integration was lead up to φ=Pi/2 (φ is in radians thus the value 1.57 on the limit) since the 2 fall heights has to be equal. The P-values have been considered same as for the 1st case and by comparison one sees that if P is small final velocities are equal but that evolution versus time and values are VERY much different if the P has important values i.e. when the fluid is thick. Interesting is that the mass thickness is the parameter which makes the difference.

The form modification in case of a circular movement is easy to explain by the fact that the body moves over a longer stroke in the high speed domain: velocity is function of fall height and an higher angles the velocity is already high but the travel is still long to do.

One could make the comment that if P has higher values the equation for the resistance is not any more valid. The answer will yes and no since the ξ values depend on Reynolds and it can be made usage of correction factors. from the other point of view P≈1/h which means that even if the fluid is not very "thick" for thin plates the values can be important.

A last remark is the time to reach the fall height which is a lot more important in the case of a circular movement.

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