Unbalanced wheel acting as a pendulum

[dmcmd]

TL;DR Summary

How do you calculate the period of a wheel that is asymmetric?

Hi all,

This is a question that has been bothering me and applies to kinetic sculptures.

Let's say you have a wheel. The wheel is situated in the vertical plane (like a clock hanging on the wall). For the sake of easy math, let's say the wheel has a radius of 1m, and has the entirety of it's 1kg mass evenly distributed about its circumference. You place a 1g weight at the11:00 position on the wheel. How long does it take the weight to cause the wheel to rotate such that the weight goes to the 1:00 position and then back to the 11:00 position (like a pendulum). Assume frictionless rotation.

Thank you for your responses!
Darren Chapman

 

[BvU]

Hello @dmcmd ,

$\qquad$ !
​

Large ampltude pendula are mathematically tough customers. Your exact setup was analyzed by Salvador Gil; Andrés E. Legarreta; Daniel E. Di Gregorio here (Reference 12 in the wikipedia lemma).
(i used researchgate to get a sneaky peek -- not sure if PF allows this... ):

Working out the period is rather a daunting task !

$\$

 

[Lnewqban]

Welcome!

Why does it bother you, regarding kinetic sculptures?

 

[kuruman]

For the sake of easy math, let's say the wheel has a radius of 1m, and has the entirety of it's 1kg mass evenly distributed about its circumference.

Believe it or not, the math would be easier to handle if you called the radius of the wheel $L$, its mass $M$, the additional mass $m$ and the angle where it's added relative to the 12 o' clock position $\theta.$

 

[dmcmd]

Hello @dmcmd ,

$\qquad$ !
​

Large ampltude pendula are mathematically tough customers. Your exact setup was analyzed by Salvador Gil; Andrés E. Legarreta; Daniel E. Di Gregorio here (Reference 12 in the wikipedia lemma).
(i used researchgate to get a sneaky peek -- not sure if PF allows this... ):

View attachment 339113​

Working out the period is rather a daunting task !

$\$

Thank you for pointing me in this direction

 

[dmcmd]

Welcome!

Why does it bother you, regarding kinetic sculptures?

I've tried to build them myself from scratch, and I'd like to understand their motion better rather than just guess at construction. While looking at kinetic sculptures, they appear, deep down, just to be clocks, with pendulums and escapements, with most of the pendulums rotating as I described above.

 

[dmcmd]

Believe it or not, the math would be easier to handle if you called the radius of the wheel $L$, its mass $M$, the additional mass $m$ and the angle where it's added relative to the 12 o' clock position $\theta.$

you're right. That would make more sense.

 

[BvU]

I've tried to build them myself from scratch, and I'd like to understand their motion better rather than just guess at construction. While looking at kinetic sculptures, they appear, deep down, just to be clocks, with pendulums and escapements, with most of the pendulums rotating as I described above.

I would expect a lot of balance springs mechanisms -- which can be treated as harmonic oscillators, much simpler !

$\$

 

[Lnewqban]

I've tried to build them myself from scratch, and I'd like to understand their motion better rather than just guess at construction. While looking at kinetic sculptures, they appear, deep down, just to be clocks, with pendulums and escapements, with most of the pendulums rotating as I described above.

I would consider those mechanisms to be closer to a rollercoaster than to a pendulum.
By doing so, I would use conservation of energy as a tool to do any calculation or estimate.
For your example, the 1 gram mass would increase its rotational velocity first at the expense of its potential energy, but not as in free-fall, because it would be fighting against the rotational inertia of the balanced part of the disc.
It would then reverse the process in its way up, reaching the same height (almost, depending on friction and air drag).

 

[BvU]

I would consider those mechanisms to be closer to a rollercoaster than to a pendulum.
By doing so, I would use conservation of energy as a tool to do any calculation or estimate.
For your example, the 1 gram mass would increase its rotational velocity first at the expense of its potential energy, but not as in free-fall, because it would be fighting against the rotational inertia of the balanced part of the disc.
It would then reverse the process in its way up, reaching the same height (almost, depending on friction and air drag).

True, true, true. But the OP specifically mentioned

TL;DR Summary: How do you calculate the period of a wheel that is asymmetric?

... How long does it take ...

 

[Lnewqban]

Point taken.

 

[kuruman]

It seems to me that the rotating clock-face would have the period of a physical pendulum which is given by $$T_0=2\pi\sqrt{\frac{I_P}{mgI_{cm}}}$$ where $I_P$ is the moment of inertia about the point of support P and $I_{cm}$ is the moment of inertia about the center of mass both of which can be easily calculated.

This reference conveniently summarizes and evaluates several large angle approximations to the period.
https://www.scielo.org.mx/pdf/rmfe/v54n1/v54n1a10.pdf
The authors conclude that the expression $$T=T_0\left[\frac{\sin\left(\sqrt{{3}\frac{\theta}{2}}\right)}
{\left(\sqrt{3\frac{\theta}{2}}\right)}\right]^{-0.5}$$is the most accurate. Here, $\theta$ is the starting amplitude.

 

[berkeman]

Large ampltude pendula are mathematically tough customers.

@dmcmd -- Are you familiar with how to model motion in Excel? That might be an option for you. You could set up the equations of circular motion for such a pendulum, and then let the simulation run to give you the motion data from the starting angle to the point where the weight is at the bottom. You could use cells for the weight of the wheel and the weight of the extra mass and for the radius of the wheel, in order to see how those factors affected the period of the assembly.

 

[Baluncore]

The simple small-angle pendulum derivation, cancels the bob mass term, m, so need only consider pendulum length, the radius to the mass of the bob.

Conceptually, a balanced ring of mass, M, with the same radius, increases the (kinetic) inertia, but maintains the same restoring (potential) force, due to pendulum bob mass, m.

I would therefore expect a term of the form (M+m)/m in the period of a pendulum when loaded with a balanced ring of mass M.

But should that term be on the inside or the outside of the square root?