Rotational Circular Motion with fixed ends and a spinning button in the middle

[rushy]

Homework Statement

Equations deriving the relationship between hanging mass and angular velocity

Relevant Equations

Rotational Motion equations with circular motion

This button has fixed ends and the string is twirled and on the fixed ends there is hanging masses. I have found out that if string twirls are constant, the hanging mass is directly proportional to angular velocity squared. But I want to understand how that is derived. Could anyone please help?

Thanks

 

[haruspex]

This button has fixed ends

The string has fixed ends?

I have found out that

Do you mean you worked that out for yourself or someone told you?

the hanging mass is directly proportional to angular velocity squared

The mass is? Do you mean the height of each mass, or the speed of the masses, or the height they have fallen by, or...?

 

[rushy]

The hanging mass can be inferred as a pulley. As the weight increases, it causes angular velocity to increase thus directly proportional.

This relationship I found from the internet by reading some articles but I want to understand how to derive it

Yes, the string is fixed.

 

[haruspex]

As the weight increases, it causes angular velocity to increase

I think you may be misreading the equation you found. Please quote it exactly or post a link.

 

[rushy]

I am not sure if the website is reliable but this is the relationship:

The hanging mass is directly proportional to angular velocity squared.

If that is not the case, could you suggest what the relationship could be with it derived from the equations:

 

[haruspex]

I am not sure if the website is reliable but this is the relationship:

The hanging mass is directly proportional to angular velocity squared.

If that is not the case, could you suggest what the relationship could be with it derived from the equations:

I still think you are not providing the whole story. Please post the link.

 

[rushy]

I can't seem to find the link anymore. I will try sending a setup of the experiment of what I mean

 

[kuruman]

I think OP's problem could be a variant of this with hanging masses, possibly over pulleys, providing the tension. There is no resonant driving force, but in the absence of dissipative forces mechanical energy is constant and distributed among kinetic energy of the button and masses and gravitational plus torsional potential energy.

 

[haruspex]

I think OP's problem could be a variant of this with hanging masses, possibly over pulleys, providing the tension. There is no resonant driving force, but in the absence of dissipative forces mechanical energy is constant and distributed among kinetic energy of the button and masses and gravitational plus torsional potential energy.

Yes, I understood that much. My problem is what the the quoted relationship refers to.

If we are considering the masses in a single descent, ω² will be proportional to the product of the combined mass and the distance descended.

If we are considering it as SHM, with the masses rising and falling and the spin reversing, then there needs to be something about maximum angular velocity and maximum descent.

Either way, it doesn't make a lot of sense to say it is proportional to mass without specifying "for a given descent".

 

[kuruman]

Yes, I understood that much. My problem is what the the quoted relationship refers to.

If we are considering the masses in a single descent, ω² will be proportional to the product of the combined mass and the distance descended.

If we are considering it as SHM, with the masses rising and falling and the spin reversing, then there needs to be something about maximum angular velocity and maximum descent.

Either way, it doesn't make a lot of sense to say it is proportional to mass without specifying "for a given descent".

Yes, the problem statement requires clarification from OP. The title doesn't help much either, "Rotational circular motion with fixed ends"?

 

[rushy]

here is the setup. the button is in the middle and the string is twirled and released to let it spin. The ends are fixed and are connected with hanging masses.

Since we know the relationship is that mass is directly proportional to angular velocity squared for a given descent, I want to understand how that proportionality is derived.

Thanks for the help

 

[haruspex]

for a given descent

Ok, that was the missing piece.
Just use energy conservation.

 

[rushy]

What do you mean by energy conservation?

Provide me with the relevant formulas that can conclude that proportionality.

 

[haruspex]

What do you mean by energy conservation?

https://en.m.wikipedia.org/wiki/Conservation_of_energyBut for the purposes of this question, the sum of the mechanical energies (kinetic energy, KE, and gravitational potential energy, GPE) is constant.
Do you know the equations for rotational KE and GPE?

 

[rushy]

I think these are the equations:
KE = 0.5 * moment of inertia * angular velocity squared
GPE = linear KE (0.5 * mass * velocity squared) + rotational KE (as above)

If I want to find the proportionality statement between the falling masses and the angular velocity of the button. Would I have to isolate the mass (does that mass equal the hanging masses or the whole system). I can also swap the velocities (v) to angular velocity by dividing it by the radius of the button.

Let me know if my approach is right?

 

[haruspex]

KE = 0.5 * moment of inertia * angular velocity squared

For rotational KE, yes.

GPE = linear KE (0.5 * mass * velocity squared) + rotational KE (as above)

No, change in gravitational potential energy equals height change x mass x gravitational acceleration.

 

[kuruman]

But for the purposes of this question, the sum of the mechanical energies (kinetic energy, KE, and gravitational potential energy, GPE) is constant.

Shouldn't there be torsional potential energy (TPE) also? If you compare two such systems, I think that for the same change in height $\Delta h$ from minimum to maximum GPE of the hanging masses, a button that has the higher "twist change number" $\Delta N$ will acquire a higher maximum rotational energy.

 

[jbriggs444]

Shouldn't there be torsional potential energy (TPE) also? If you compare two such systems, I think that for the same change in height $\Delta h$ from minimum to maximum GPE of the hanging masses, a button that has the higher "twist change number" $\Delta N$ will acquire a higher maximum rotational energy.

I am not clear on where you would expect such "torsional potential energy" to be stored.

The picture in my mind's eye is of two ideal inextensible, massless cords that individually have negligible resistance to twisting but which have non-negligible thickness so that longitudinal tension gains a torsional component when the cords spiral around one another. There is no place for energy to be stored in such cords.

 

[kuruman]

I am not clear on where you would expect such "torsional potential energy" to be stored.

The picture in my mind's eye is of two ideal inextensible, massless cords that individually have negligible resistance to twisting but which have non-negligible thickness so that longitudinal tension gains a torsional component when the cords spiral around one another. There is no place for energy to be stored in such cords.

If that's the model, yes I will agree with you, but what about friction? It is fair to imagine that as the strands of the double helix separate they don't rub against each other so no kinetic friction. However, I wonder if it is fair to ignore static friction. The strands of the double helix are pressed tightly against each other as they are wound around and the tension should increase as one moves away from the ends, sort of like a capstan. The tension at the button will be greater than the tension at the ends. Of course if this is the case, the string will have to be extensible and store energy. Would this gadget work as well as shown in the video if the contact between twisted strands were completely frictionless and didn't have the benefit of tension and torque augmentation at the button? I am not sure, one way or the other.

 

[haruspex]

The strands of the double helix are pressed tightly against each other as they are wound around and the tension should increase as one moves away from the ends, sort of like a capstan.

In a capstan, static friction acts in the same direction as the tension, so adds to it. In the twisted strands, I see no reason there should be any friction.

 

[CWatters]

I would start by ignoring things like friction or energy stored in the string. Try assuming that all the energy lost by the falling mass ends up in the spinning button. See what relationship that produces.