Draft:Ball on a String, the Ultimate Summary

• Comment: Wikipedia is not a research paper site. AngusW🐶🐶F (bark • sniff) 23:22, 25 January 2024 (UTC)

• Comment: Wikipedia cannot be used as a source, which leaves just one source.
  Is this draft original research? Qcne (talk) 13:23, 25 January 2024 (UTC)

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1) Introduction

Ball on a String classic exercise in physics manuals is intended to show an example of Second Euler Law, often referred to as the "Law of Conservation of Angular Momentum":

(1) ${dL \over dt}=T$

Where $L$ is the Angular Momentum, $t$ time and $T$ the total external torque acting on the spinning (rigid) object.

In the Ball on a String classic exercise a ball of mass $m$ is attached to a string and rotates on a horizontal frictionless plane $\alpha$, balancing ball weight, at an angular initial speed $\omega _{0}$ around a vertical axis ${\vec {z}}$ while an arbitrary force $F$ is pulling the string downward causing the ball to close its distance from rotation axis and shortening same string from its $r_{0}$ initial lenght to some $r$ such as $r<r_{0}$ in a given time interval $t-t_{0}$:

Figure 1: Standard Ball on a String Pictorial

Along the ball the suppport plane further simplifying assumptions are made (along others):

1. Infinitely rigid, massless, unflexible string.
2. No air drag.
3. No friction along string axis.
4. Pulling force $F$ parallel (coincident) to rotation axis.

2) Conservation of Angular Momentum (CoAM), Perpetual Motion Assumption (PMA)

Only under the assumptions above thus being zero the total external torque $T$ exerted on the ball ($T=0$), (1) can be rewritten as:

(2) ${dL \over dt}=0$which defines CoAM as $dL=0$ and thus $L(r_{0})=L(r)$, stating $L$ equivalence between two distinct distances $r_{0}$ and $r$.

Being $J_{0}=mr_{0}^{2}$ the ball Moment of Inertia (MoI) with respect to rotation axis at a distance $r_{0}$ and $J=mr^{2}$ MoI after string retraction, thus angular momentum $L=J\omega$, (2) becomes:

(3) $dL=d(mr^{2}\omega )=0$, then $L$ not changing in the $t_{0}-t$ ($r_{0}\to r$) time interval and thus $mr^{2}\omega =mr_{0}^{2}\omega _{0}$ which is, skipping steps:

(4) $\omega ={\frac {r_{0}^{2}}{r^{2}}}\omega _{0}$ the final angular speed of the ball after string retraction.

We could have come to the same conclusion also by expressing (2) as:

(5) ${dL \over dt}={d(mr^{2}\omega ) \over dt}=2mr{\dot {r}}\omega +mr^{2}{\dot {\omega }}=2{\dot {r}}\omega +r{\dot {\omega }}=0$ (balance between Coriolis force torque with respect to the rotation axis and rotational inertia) which also is:

(6) ${d\omega \over \omega }=-2{dr \over r}$ that, after integration of both members between $(\omega _{0},\omega )$ and $(r_{0},r)$ is exactly (4).

Equations (2) $\to$ (6) expresses CoAM again in the ideal only case where any disturbance comes to alter ball dynamics thus any assumption like the support plane and (1.,2.,3.,4.) leading external torque to be $T=0$ are equivalent to a perpetual motion assumption. CoAM is instead and generally a tendency, when $T\to 0$, ${dL \over dt}\to 0$, $dL\to 0$, $L\to constant$: like in outer space.

3) Conservation of Rotational Energy (CoRE) within PMA

To better analyse the Ball on a String system still in the PMA case we shall account for both ball angular speed $\omega$ about the rotation axis z and its linear velocity $v$ along the string while the latter is retracted. Total kinetic energy $K$ being:

(7) $K={\frac {1}{2}}J\omega ^{2}+{\frac {1}{2}}mv^{2}$(which could be also written directly adding the kinetic energy from both ball linear speed components, $v_{r}=\omega r$ and $v$: $K={\frac {1}{2}}m\omega ^{2}r^{2}+{\frac {1}{2}}mv^{2}$ which is (7) ).

At the instant $t_{0}$ string is steady, thus $v=v_{0}=0$, the ball carries only rotational energy:

(8) $K_{0}={\frac {1}{2}}J_{0}\omega _{0}^{2}$

Instead at the generic instant $t$ the string is being retracted thus $v\neq 0$ and the Ball on a String system kinetic energy is still expressed by (7).

Energy conservation principle (CoE) can be thus written such as (7)=(8), $K=K_{0}$:

(9) ${\frac {1}{2}}J\omega ^{2}+{\frac {1}{2}}mv^{2}={\frac {1}{2}}J_{0}\omega _{0}^{2}$ meaning:

(10) $\omega ={\sqrt {(}}{\frac {r_{0}^{2}\omega _{0}^{2}-v^{2}}{r^{2}}})$ generic angular speed change during string retraction.

Equation (10) is different from classic (4), being the angular speed proportional not to the square of retraction ratio ${r_{0}^{2}} \over {r^{2}}$ but to its square root.

If $\omega _{K}$ is the final angular speed obtained through (10) CoE model and $\omega _{L}$ the one obtained by means of (4) CoAM model, it will always be:

(11) $\omega _{K}<\omega _{L}$

And it has to be further noted that:

(12) $\lim _{v\to 0}\omega _{K}={\frac {r_{0}}{r}}\omega _{0}$ << ${\frac {r_{0}^{2}}{r^{2}}}\omega _{0}=\omega _{L}$

and:

(13) $\lim _{v\to 0}({\frac {1}{2}}mv^{2}+{\frac {1}{2}}J\omega ^{2})={\frac {1}{2}}J\omega ^{2}$

So, for slow retracting speeds $v\simeq 0$ equation (9) becomes:

(14) ${\frac {1}{2}}J\omega ^{2}={\frac {1}{2}}J_{0}\omega _{0}^{2}$ expressing conservation of rotational energy (CoRE), otherwise not conserved for not negligible retracting speeds where (9) CoE applies. Equation (14) means a ball angular speed while string is retracted as:

(15) $\omega ={\frac {r_{0}}{r}}\omega _{0}$ that may be accepted instead of (4) under the proper $v\simeq 0$ assumption.

4) Differences between PMA CoAM / CoRE / CoE

Below plot shows a comparison between predicted angular speed from CoRE and CoAM model (equation (4) and (15)) when string is retracted:

Figure 2: Predicted Final Angular Veolicty from CoRE and CoAM Models after String Retraction

Figure 3: CoAM to CoRE Model Final Angular Velocity Prediction after String Retraction

Surely CoRE (15) equation shows closer to realistic predictions than CoAM cause the first accounts for the entire Ball on a String system kinetics (rotational and linear) while the latter only rotational, the same applies for the not negligible $v$ case, where we use (10) CoE instead of CoRE.

5) Realistic Ball on a String, Removing PMA, SAM Failed CoAM

Removing the ball support plane but keeping (1.,2.,3.,4.) so letting same ball to move also in the vertical (string bending angle $\phi$) plane resembles a more realistic Ball on a String system that can be modeled by a simple Spinning Atwood Machine (SAM):

Figure 4: Spinning Atwood Machine (SAM): negligible weight pulley P (to reduce string friction) spins around ${\vec {z}}$ axis within an inside lubricated and infinitely rigid (with respect to the string) pipe of appropriate shape to avoid windings: the pipe is hosting a negligible weight and always straight steel string of length $l$ much greater than P radius selected to withstand centrifugal forces: a counterweight mass $M$ constrained to prevent its rotations and connected to the string by a ball joint allowing the same string to spin while being pulled from $M$ forcing the ball to close its distance from rotation axis and shortening the string. An electric motor could be set on top of the assembly to support it.

Which can be assessed by a standard Lagrangeapproach to write SAM Equation of Motion (EoM):

(16)${\begin{cases}(M+m){\ddot {r}}-mr{\dot {\theta }}^{2}sin^{2}\phi -mr{\dot {\phi }}^{2}=mgcos\phi -Mg\\mr^{2}{\ddot {\theta }}sin^{2}\phi +2m{\dot {r}}{\dot {\theta }}rsin^{2}\phi +mr^{2}{\dot {\theta }}{\dot {\phi }}sin2\phi =0\\mr^{2}{\ddot {\phi }}+2mr{\dot {r}}{\dot {\phi }}-mr^{2}{\dot {\theta }}^{2}sin\phi cos\phi =-mgrsin\phi \end{cases}}$

Where ${\dot {\theta }}=\omega$ is same as above the angular speed of the ball around the z axis while ${\dot {\phi }}$ being the ball angular speed in the vertical plane and ${\dot {r}}=v$ the same speed of the ball along the string during retraction.

First equation shows ball dynamics along the string axis, second the ball rotational equilibrium about rotation axis z and the third same rotational equilibrium in the vertical plane. It may be noticed that if $\phi ={\frac {\pi }{2}}$ the second equation can be rewritten as $mr^{2}{\ddot {\theta }}+2m{\dot {r}}{\dot {\theta }}r=0$ then $r{\dot {\omega }}+2{\dot {r}}\omega =0$ exactly (6) and thus CoAM (4). But the latter CoAM shows only about the z axis, there's not linear momentum conservation (along string axis) nor CoAM in the vertical plane where ball weight is not equilibrated after removing the support plane assumption, so $T\neq 0$ in (1).

Which means that SAM model of the Ball on a String does not preserve angular momentum (it's only a tendency, as above) as it can be shown by numeric (16) EoM numeric integration plot:

Figure 5: SAM Missing CoAM, mass $m$ normalized, $L$ by numerical EoM integration.

And when introducing air drag (only on the z axis spinning motion), string friction and stiffness, staying the latter unflexible and massless, also z axis CoAM (second (16) equation) vanishes:

(17) ${\begin{cases}(M+m){\ddot {r}}-mr{\dot {\theta }}^{2}sin^{2}\phi -mr{\dot {\phi }}^{2}+c{\dot {r}}+K_{E}r=mgcos\phi -Mg\\mr^{2}{\ddot {\theta }}sin^{2}\phi +2m{\dot {r}}{\dot {\theta }}rsin^{2}\phi +mr^{2}{\dot {\theta }}{\dot {\phi }}sin2\phi +{\frac {3}{2}}K{\dot {\theta }}^{2}r^{3}=0\\mr^{2}{\ddot {\phi }}+2mr{\dot {r}}{\dot {\phi }}-mr^{2}{\dot {\theta }}^{2}sin\phi cos\phi =-mgrsin\phi \end{cases}}$

Where $K_{E}={\frac {EA}{l}}$ ($E$ string Young Modulus, $A$ string cross section, $l$ string lenght) and $K={\frac {1}{2}}\rho _{a}C_{d}A$ ($\rho _{a}$ air density, $C_{d}$ drag coefficient for a streamlined ball to reduce turbulences).

5.1) The Electric Motor

It has to be noted that equations (16) and (17) are written without considering the torque $T_{E}$ of the electric motor driving SAM: EoM describes the condition in which $T_{E}=0$ (motor off) and the counterweight $M$ is released, when the distance of the ball from the rotation axis could decrease (string to "shorten") or increase (string to "lenghten") following the balance between centrifugal force and same counterweight.

5.2) SAM EoM Details

First equation of system (16) is about forces equilibrium along the string: inertial force of two masses $(M+m){\ddot {r}}$, then $mr{\dot {\phi }}^{2}$ centrifugal acceleration in the vertical plane and $mr{\dot {\theta }}^{2}sin^{2}\phi$ = $mr{\dot {\theta }}^{2}sin\phi cos({\frac {\pi }{2}}-\phi )$ component of centrifugal force due to ball rotation around z axis projected on the string direction.

Second equation shows torques equilibrium around the z axis, where:

• $mr{\ddot {\theta }}sin\phi *rsin\phi$ is the inertial torque due to angular acceleration ${\ddot {\theta }}$.
• $C(\theta )=2m{\dot {\theta }}{\dot {r}}sin\phi *rsin\phi$ the torque of Coriolis force (velocity component ${\dot {r}}sin\phi$).
• $mr^{2}{\dot {\theta }}{\dot {\phi }}sin2\phi =2m{\dot {\theta }}r{\dot {\phi }}cos\phi (=C(\theta \phi ))*rsin\phi$ the torque of Coriolis force (velocity component $r{\dot {\phi }}cos\phi$).

Third equation represents torque equilibrium on the vertical plane:

• $mr^{2}{\ddot {\phi }}$ is the inertial torque.
• $C(\phi )=2m{\dot {r}}{\dot {\phi }}*r$ torque of Coriolis force.
• $mr{\dot {\theta }}^{2}sin\phi *rcos\phi$ torque of the centrifugal force.
• $mgrsin\phi$ weight of the ball.

Figure 6: Force Diagram -1: $\omega _{\theta }={\dot {\theta }}$ and $\omega _{\phi }={\dot {\phi }}$.

Figure 7: Force Diagram -2

Figure 8: Ball angular velocity $\omega$ around z axis evolution by EoM numerical integration.

Figure 9: Ball elevation angle $\phi$ evolution toward ${\pi } \over {2}$ while ball angular velocity $\omega$ increases, by EoM numerical integration.

5.3) Simplified SAM CoAM (PMA is Back).

Assuming back the ball's spinning on the support plane so that $\phi ={\frac {\pi }{2}}$, its weight is balanced, no air drag, and a new friction coefficient $c$ (16)-(17) EoM become:

(18)${\begin{cases}(M+m){\ddot {r}}-mr{\dot {\theta }}^{2}+c{\dot {r}}=-Mg\\mr{\ddot {\theta }}+2m{\dot {r}}{\dot {\theta }}=0\end{cases}}$

And below the plots of predicted dynamics of the simplfied SAM and CoAM condition met:

Figure 10: Evolution of the ball distance from rotation axis after counterweight $M$ is released (EoM numerical integration).

Figure 11: Evolution of the ball angular velocity after counterweight $M$ is released (EoM numerical integration).

Figure 12: Simplified SAM meet CoAM condition.

Figure 13: Counterweight $M$ and centrifugal force balance ($v=0$) vs ball distance from rotation axis ($\omega _{\theta }=\omega$).

5.3.1) Out of Curiosity: The Unreachable Simplified SAM (USS)

Considering the centrifugal force component (green in the force diagarm below) $m\omega ^{2}rsin\phi cos\phi$ balancing ball weigth $mgsin\phi$ on the tangent to the trajectory we get the equilibrium condition:

(19) $\omega ={\sqrt {\frac {g}{rcos\phi }}}$.

Figure 14: Centrifugal force for $\omega$ spinning ball balancing ball weight on the trajectory tangent.

For the elevation angle $\phi$ of the SAM ball to be exactly ${\frac {\pi }{2}}$ it must be $\omega \to \infty$ ( $\phi \to {\frac {\pi }{2}}$) in (19), thus same SAM ball can't be spinning on a horizontal plane circular trajectory and its string being orthogonal to ${\vec {z}}$ rotation axis regardless the magnitude of $\omega$.

It is also graphically evident that when $\phi \to {\frac {\pi }{2}}$ weight component $mgsin\phi \to 1$ while centrifugal force component $m\omega ^{2}rsin\phi cos\phi \to 0$ being progressively impossible for them to balance.

6) Keeled Ball on a String (KBS), Helped CoAM

As seen, the classic Ball on a String exercise is proposed being the string pulled by a vertical $F$ while the ball is spinning on the horizontal plane.

What if on the contrary we assume an horizontal $F$ and a vertical spinning plane?

Figure 20: Keeled Ball on a String

Below a force diagram of KBS:

Figure 21: KBS force diagram.

EoM of KBS are, always under (1.,2.,3.) assumptions:

(20) ${\begin{cases}{\ddot {r}}-r{\dot {\theta }}^{2}=-gsin\theta \\r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}=-gcos\theta \end{cases}}$

Which cleary resemble EoM (18) with the difference of the weight component $mgrcos\theta$ preventing CoAM: external torques on the ${\vec {x}}$ rotation axis are indeed different than zero ($T\neq 0$) thus $L$ won't preserve as a consequence of (1). It will preserve only when an electric motor is set along ${\vec {x}}$ to continuously balance with its torque $T_{E}$ the ball weight torque (thus it can't be switched off).

7) About Ball on a String Example

Simpler, more intuitive, mechanical systems needing less assumptions to prove CoAM are (have always been) available (online too). Both conservation of linear and angular momentum are just tendencies from the general principle stating that each object set in motion tends to preserve its state of motion, but obviously and due to the conservation (transformation) of energy principle, no state of motion can last forever (ad infinitum).