Physical pendulum with Pivot Point a little more

This is a subsequent post for this topic, and you should read this topic if you are interested in you and you are not used to any of the following: The same warning for mobile viewing is applied. All square roots are in both molecules and denominations, despite how they can appear on mobile devices.

Pentagon with mass middle And mass center (CM) inertial moment me_cm Pivot at one point d There is a period in cm tea given

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

With ~ G Acceleration due to gravity. The last post showed it tea There is the same value in one or two circles centered on the pendulum’s CM. One is a radius. d One with a radius d’, where d’ = L -D and

$$L=\frac{\mathrm{me}}{\mathrm{middle}d}$$

here I = I_cm+ MD². Cm center circle d and d’ It is composed of a back point. In other words, the period is pivotless regardless of the point of this circle.

This post is explored tea As a function d (or d’) And some attributes d.

how tea movement d Various

tea As a function d There is a shape as shown below the horizontal and vertical axis. d and teaEach.

This is a suggestion tea Approach to infinity d Approach 0 or infinite. This can be displayed by officially limiting, but it is not difficult to examine the expressions and to see it informally. tea.

like d Expressions approach 0 tea Similar or works with a certain amount of time of the squares. dThis is a jummy teaIt is reduced from the plot above and definitely goes indefinitely. like d Infiniti, approaching expression tea It acts like a certain time of square root dIt goes with infinity as shown in the graph.

This behavior tea Like ~ d Approaching 0 or infinite is meaningful. like d When you approach the zero, the pendulum’s pivot point approaches CM. CM does not have a restoration torque and the pendulum does not return from the displacement. Think about turning a uniform density disk fixed to the center (cm). If you turn it, it will not go back. It is an infinite time.

like d If you approach infinity, the situation is A Simple pendulumThe period is proportional to the square root of the lengthd), thus tea It also approaches infinity.

As you move away from this extreme tea Strictly monotonously decreases. This means that the minimum value must be located. d = d ‘. Otherwise, there will be one or two of them around CM. teaRadius d And one of the radius d’. This means that all pivot refers to the distance. R In cm dIt has a greater period of time, which contradicts the strict forged lines. tea 0 (or infinite) to the minimum value.

Or, we can take derivatives tea In relation d To find the minimum value, set it to 0 (omitted) tea Where is it

$$d^2 = \ frac {I _ {\ text {cm}}} {m}$$

In the previous post, this is also Dd ‘. It means the minimum value tea Where to be found d and d’ Matches. Value of this value d also Rotary radius. For justice, this is everyone middle If you live there, the moment of inertia is the same as an object (here) that is in a position other than the radius of the object.

Put this into the expression tea Minimum value given above tea (Or for convenience, tea²) am

$${\mathrm{tea}}_{\text{Minimal}}^2=\frac{8{\mathrm{blood}}^2}{G}\sqrt{\frac{{\mathrm{me}}_{\text{cm}}}{\mathrm{middle}}}$$

So when d = 0,,, D ‘= ∞ We have two (degeneration) equinoxes tea Infinite. like d Move outward in the radius of the cm d’ It moves inward and the circle they define always consists of isolation. If this happens tea Infiniti drops to a unique minimum value proportional to the quart roots. me_cm/middle. The minimum value is achieved d = d ‘In the radius of rotation, the two chambers become one (♥ cue violin ♥).

Where is the minimum value for pendulum?

where d Either tea Is it minimal for the boundaries of pendulum? There is never an external border of the pendulum. I will show you mathematically below, but it is intuitive. If all mass extends the simple pendulum at the end of the (assumed mass) support, the period increases. Therefore, there is a reason why the pivot point over the surface of the physical pendulum will have a greater period of time than the pivot point on the surface. Also, the definition of the rotary radius makes it clear that it is in the pendulum.

Anyway, let’s do math. The CM rotation inertia of the pendulum is depending on the definition.

$$I _ {\ text {cm}} = \ int R^2 \, dm$$

where R It is a vertical distance from CM to infinite mass elements. Dm. shift R r_MaxThe largest distance from CM to the boundary of pendulum, we get inequality.

$$I _ {\ text {cm}} = \ int r^2 \, dm \ leq r _ {\ text {max}}^2 int dm$$

Or briefly

$$I _ {\ text {cm}} \ leq r _ {\ text {max}}^2 \, m$$

Replace this with expression tea-solution d above, D ≤ R_Max. Equality is obtained only when the mass of all pendulum is when it is a certain radius of CM (ideal hoop of width). otherwise tea-solution d It is within the boundary of the pendulum. For all practical purposes (assuming something), D _Max. That is, not only tea-solution d Even on the surface, not outside the pendulum (except the ideal hoop). Strictly somewhere in the pendulum. It means both d and d’ It is within the boundary of the pendulum for some range.