A Bit Extra on Bodily Pendulum Isochronal Pivot Factors

This can be a follow-up put up to this one, which it is best to (re)learn if this subject is of curiosity to you and any of the next is unfamiliar. The identical warning about cellular viewing applies: all sq. roots are over each numerator and denominator, regardless of the way it could seem on a cellular system.

A bodily pendulum with mass m and heart of mass (CM) second of inertia I_cm pivoting at a degree d from the CM has interval T given by

$$T = 2pi sqrt{frac{I_{textual content{cm}} + md^2}{mgd}}$$

with g being the acceleration as a consequence of gravity. The final put up confirmed that T has the identical worth on two circles centered on the pendulum’s CM, one with radius d and one with radius d’, the place d’ = L – d and

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

Right here I = I_cm+ md². The CM-centered circles of radius d and d’ are composed of isochronal factors. That’s, the interval is identical regardless of which level on these circles the pendulum is pivoted about.

This put up explores T as a perform of d (or d’) and a few properties of d.

How T Behaves as d Varies

T as a perform of d has a form as proven within the determine beneath, the place the horizontal and vertical axes signify d and T, respectively.

This implies T approaches infinity as d approaches zero or infinity. This may be proven formally by taking limits, however it’s not arduous to see informally by analyzing the expression for T.

As d approaches zero, the expression for T resembles (or behaves like) a relentless instances the inverse of the sq. root of d, which is asymptotic to the T-axis within the above plot and positively goes to infinity. As d approaches infinity, the expression for T behaves like a relentless instances the sq. root of d, which additionally goes to infinity, because the graph reveals.

This habits of T as d approaches zero or infinity is smart. As d approaches zero, the pendulum’s pivot level approaches its CM. There isn’t any restoring torque on the CM, and the pendulum wouldn’t return from a displacement. Consider turning a disk of uniform density pinned on the heart (its CM). If you happen to give it a flip, it gained’t flip again. That’s an infinite interval.

As d approaches infinity, the state of affairs resembles a easy pendulum, for which the interval is proportional to the sq. root of its size (d), therefore T additionally approaches infinity.

Shifting away from these extremes, T strictly monotonically decreases to a minimal. This itself implies that the minimal have to be the place d=d’. If this weren’t the case, then there could be two isochronal circles, centered on the CM, of minimal T, one in every of radius d and one in every of radius d’. This might indicate that every one pivot factors a distance r from the CM such that d<r<d’ have bigger durations, which contradicts the strict monotonicity of T from zero (or infinity) to the minimal.

Or, we might take the by-product of T with respect to d and set it to zero (steps omitted) to seek out that the minimal T is the place

$$d^2 = frac{I_{textual content{cm}}}{m}$$

From the earlier put up, that is additionally the expression for dd’. That implies that the minimal T is discovered the place d and d’ coincide. This worth of d can also be the radius of gyration. By definition, that is the radius such that if all of m resided there, the second of inertia could be the identical as the article’s (right here the pendulum) with distributed mass.

Plugging this into the expression for T given above, the minimal T (or, for comfort, T²) is

$$T_{\text{min}}^2=\frac{8{\pi }^2}{g}\sqrt{\frac{I_{\text{cm}}}{m}}$$

So, when d=0, d’=∞ we now have two (degenerate) isochronal circles for which T is infinite. As d strikes radially outward from the CM, d’ strikes inward, and the circles they outline are all the time composed of isochronal factors. As this occurs, T drops from infinity towards a novel minimal proportional to the quartic root of I_cm/m. The minimal is achieved when d=d’, on the radius of gyration, and the 2 isochronal circles turn into one (♥ cue violins ♥).

The place is the Minimal with Respect to the Pendulum?

The place is the d for which T is minimal with respect to the boundary of the pendulum? It’s by no means outdoors the boundary of the pendulum. I’ll present this mathematically, beneath, however it’s intuitive. Lengthening a easy pendulum, for which all of the mass is on the finish of the (assumed massless) help, will increase its interval. So, it stands to motive {that a} pivot level past the floor of a bodily pendulum would have a bigger interval than a pivot level on its floor. Additionally, the definition of radius of gyration makes it fairly clear that it’s throughout the pendulum.

Let’s do it with math anyway. The pendulum’s CM rotational inertia is, by definition

$$I_{textual content{cm}} = int r^2 , dm$$

the place r is the perpendicular distance from the CM to a infinitesimal mass aspect dm. Changing r with r_max, the biggest distance from the CM to the boundary of the pendulum, we get the inequality

$$I_{textual content{cm}} = int r^2 , dm leq r_{textual content{max}}^2 int dm$$

Or, merely,

$$I_{textual content{cm}} leq r_{textual content{max}}^2 , m$$

Substituting this into the expression for the T-minimizing d above, d ≤ r_max. Equality is obtained solely when all of the pendulum’s mass is at a relentless radius from the CM (an idealized hoop of zero width). In any other case, the T-minimizing d is strictly throughout the pendulum’s boundary. For all sensible functions (assuming there are any), d < r_max. That’s, not solely is the T-minimizing d not outdoors the pendulum, it’s not even on the floor (aside from the idealized hoop). It’s someplace strictly contained in the pendulum. That implies that each d and d’ are throughout the boundary of the pendulum for some vary.

If the physique of the pendulum consists of the CM (true for convex shapes like a disk or triangle), then all potential values of T, from its most of infinity to its minimal on the d outlined above are discovered at pivot factors inside it. For all such values of T, there are arcs* of 1 or two isochronal circles throughout the physique of the pendulum — one when d’ is outdoors the pendulum’s boundary and two when it’s inside, which, as famous above, it will likely be for some vary. Actually, that vary is when d’ is between the T-minimizing d (the radius of gyration) and r_max. 

If the physique of the pendulum doesn’t embrace the CM (as could be the case for some concave shapes like a uniform density banana or a washer), the utmost T for which a pivot level is contained in the physique of the pendulum is given by

$$T = 2pi sqrt{frac{I_{textual content{cm}} + m r_{textual content{min}}^2}{m g r_{textual content{min}}}}$$

the place r_min is the space from the CM to the purpose throughout the pendulum closest to it.​

Abstract

• The interval of a bodily pendulum, T, varies with pivot level from infinity (at pivot factors CM and infinity) to a minimal on the radius of gyration.
• ​A minimal reaching pivot level is all the time throughout the physique of the pendulum.
• If the physique of the pendulum consists of the CM, then for all potential values of T, from its minimal to infinity, there are arcs of 1 or two isochronal circles throughout the physique of the pendulum.​
• If the physique of the pendulum doesn’t embrace the CM, there are solely isochronal circle arcs for T much less then a finite most worth, an expression for which we discovered.

* I’m speaking arcs of circles and never essentially full circles right here as a result of there could also be some radial distances from the CM for which the physique of the pendulum exists for just some angles. Consider a triangular-shaped pendulum versus a disk-shape one. For the previous, there are some CM-centered circles of radius lower than the extent of the triangle that aren’t solely throughout the triangle. For the latter, this isn’t true.