The shape of the pivot point for physical pendulum

If you read this on a mobile device, some equations are not visible properly. In particular, one thing I noticed was that the square roots over the fountain seemed to be on the molecule. All such square roots are also in the denominator. In terms of devices that are not mobile (for example, my laptop), the equation looks right (e.g. my laptop).

This post is about the subject of physics, and I don’t think I’m interested in many regular tie readers. But you can be interested in irregular readers.

Why physics? I was my undergraduate major, and now I am helping my daughter through AP physics. I am going down the rabbit hole, exploring what I did not learn in college. Some are not in my fingertips or online textbooks (as far as I know). This post is one of those things. I used AI to check math and logic. Please let me know if there is a problem.

What this post showed
Our starting point is A’s well -known attribute. Physical (Also called compound pendulum), the discovery Christian Huygens: You can get the same period with the pivot at two different points (they are Back Crohn). One Pivot Point, Huygens, and other default physics texts that deal with this topic show how to find “other isocron pivotpoints”.

What this post shows is that in fact, there is an infinite number of alphabetical points in fact, so in the previous sentence, quotes around the “the“ the ”). Moreover, they are all placed in two circles about mass center (CM).

First tie
You can find derivatives for the period of physical pendulum. Online Or in many textbooks. period tea About Pivot Point blood Given

$$t = 2 \ pi \ sqrt {\ frac {I} {mgd}}$$

where me It is a moment of inertia for Pivot Point. middle The mass of pendulum is G This is acceleration due to gravity d It is a distance between blood And cm. at Parallel

$$I = I _ {\ text {cm}} + md^2$$

The first term is the moment of inertia for the center of mass. therefore,

$$t = 2 \ pi \ sqrt {\ frac {I _ {\ text {cm}} + md^2} {mgd}}$$

It is constant for all distances d In cm In any direction. Therefore, there is an infinite flooding pivot point placed in the radius. d Centered on CM. This is the first one. It is not a cron sphere. It is important to keep the pivot axis parallel because the inertia moment of mass is different for other (non -parallel) axis. In other ways, The moment of inertia is Tenser.​​

Second equivalent circle
As you have discussed in many places but not in a simple way (I saw it), there are pivot points. blood It is not a circle described above. It is called Vibration And we will point it out cue. At the end of this post is simple evidence that there is a point. cue It is a distance L From ~ blood Through cm where

$$l = \ frac {I} {md}$$

Define d’ = L -D Distance cue And cm. There is a moment of inertia in physical pendulum. me’ About points cue. We can stop d’ and me’ To the equation provided in the above section (the first round prototype). The conclusion is based on the same claim as the period is constant. tea At some point distance d’ From cm. We know that it is tea (The period as in the previous section) blood and cue This is a time of pivot point. Therefore, there is an infinite pivot point in the radius. d’ Centered on CM. This is the second round prototype.

Or summarize our eyes on the fact that Huygens and many other people have two lamps. blood and cue. There are also two corps systems for CM. blooddifferent cue.

This is consistent with the clear fact that the vibration period of the uniform density rods figures at one end is the same as the pivot of the other end (the ducts pass through both ends). Or the vibration cycle is the same for any point of the circular disk at a uniform density equal distance from the center (the light circle is concentrated with CM). This obvious fact is clear for us in symmetry.

What is given above shows that the circular symmetry also exists for asymmetrical (no uneven density) physical pendulum. This is not intuitive (not for me anyway). However, the only thing that is not fixed at all points related to the pendulum given to the expression tea It is a distance of pivot point at CM. As a result, the original form is inevitable.

conclusion
Pay attention d It can be a positive distance, causing a positive value of the amount during the period. tea. Thus, with the above debate, all the pivot points are located in the concentric pairs centered on CM. There is a small technology of (OK, infinite number). dSome homogeneous points may not be in the body of the pendulum. This does not mean that it is not lived at the same time as the other point of the relevant circle. d (or d’). It only means that pendulum actually pivotes that point is actually difficult. From that point on, we will receive humble and humble support from that point.

Evidence that the center of vibration is omnipotence (or Pivot cue There is the same period as Pivot blood)
From above, I promised this evidence. To trust Huygens and many other physics texts, skip.

To get the expression using all the terms defined in the post, start with a preliminary object. d’ From a perspective d It will be useful later. Remember it

$$L = \ frac {I} {md},$$

As mentioned in parallelism and the posts, as mentioned in the post

$$\mathrm{me}={\mathrm{me}}_{\text{cm}}+\mathrm{middle}{d}^2$$

therefore,

$$l = \ frac {I _ {\ text {cm}} + md^2} {md}.$$

And simplicity,

$$l = \ frac {I _ {\ text {cm}}} {md} + d.$$

Using this and definition d’ (= L -D) We have in the post

$$D ‘= \ left (\ frac {I _ \ text {cm}}}} {md} + d \ right) -d.$$

Make

$$D ‘= \ frac {I _ {\ text {cm}}} {md}.$$

Pendulum on the point cue Given

$$T ‘= 2 \ pi \ sqrt {\ frac {i’} {mg d ‘}},$$

Work is to show T ‘= t. We will connect and do that me’ and d’ And we are doing algebra. Here we go. The only thing we can do me’ Apply parallel axes.

$$I ‘= I _ {\ text {cm}} + m d’^2.$$

Connect the expression d’ From above. (This is why we did a preliminary work.)

$$I ‘= I _ {\ text {cm}} + m \ left (\ frac {I _ {\ text {cm}}} {md} right)^2.$$

cancellation middle Connect this to the molecule tea’ And use the expression d’ Even in the denominator.

$$T ‘= 2 \ pi \ sqrt {\ frac {I _ {\ text {cm}} + \ frac {I _ {\ text}^2} {mg \ cdot \ frac {I _ _}}}}}}}}}}}}$$

There are many things to simply be canceled. Continue to do so in paper pieces or heads. You can get

$$T ‘= 2 \ pi \ sqrt {\ frac {1 + \ frac {I _ {\ text {cm}}} {md^2} {\ frac {g} {d}}}.$$

Multiply the molecule and denominator to the square root MD² To get

$$t = 2 \ pi \ sqrt {\ frac {i} {mgd}} = 2 \ pi \ sqrt {\ frac {I _ {\ text {cm}} + mgd}}}}}}}.$$

It is recognized that the molecule is the complex in parallelism. me And this is an expression tea. so, T ‘= t And we are over.