How does warming the hoop affect the frequency of an escapement in a clock?

[bolzano95]

Homework Statement

I have an escapement in a clock. It given shape is hoop with radius 5mm, mass 0.01g and has a frequency 5Hz. For how much does the frequency change if we warm the hoop for 20K?

[tex]α= 1.2 ⋅10^{-5} K^{-1}[/tex]

Homework Equations

[tex]
l=2πr\\Δl=lαΔT\\w=\sqrt{\frac{2D}{mr^2}}⇒D\\Δ\nu=\nu_1- \nu= \frac{(\sqrt{\frac{2D}{mR^2}}-\sqrt{\frac{2D}{mr^2}})}{2π}=-1.2⋅10^{-3} s^{-1}\\\frac{Δ\nu}{\nu}=-2.4⋅10^{-4} s^{-1}
[/tex]

The Attempt at a Solution

Well, official result is
[tex]

\frac{Δ\nu}{\nu}=-1.2⋅10^{-4} s^{-1}
[/tex]

but I got a solution
[tex]

\frac{Δ\nu}{\nu}=-2.4⋅10^{-4} s^{-1}
[/tex]

I am not sure if official solution is correct, because I couldn't find an error in my result. I would like to ask you to check this problem over. Let me know if I missed something!

 

[BvU]

I couldn't find an error in my result

Neither can we. You left out your detailed working.

And your description fits a balance wheel in a watch, better than an escapement

 

[bolzano95]

I had problem with Latex processing. I added my solving process.

 

[BvU]

Don't see anything wrong with the calculation. Is there something else that depends on temperature ? D, for example ?

 

[bolzano95]

I suppose only hoop depends on temperature. Thanks for looking this over!

 

[Dr.D]

Is this hoop subject to a gravitational restoring force or is the hoop spring driven? It seems to me to be impossible to address this question properly without a figure.

Also, the notation seems foreign. What are w, v, and D? I might guess at their meanings, but I might be wrong too. I used to work in the watch industry, and I would like to address this problem for the OP, but I can't do that without a clearer definition of the system and the symbols used.

 

[kuruman]

I think it is more correct to use the 2D expansion equation. It's like the hole in a plate that expands upon heating. The circumference of the hole gets larger according to the 2D expression. The expansion of the area of the ring will be given by ##\Delta A = 2\alpha A_0 \Delta T##. This means that the new area is ##A=A_0+\Delta A = A_0(1+2\alpha \Delta T)##.

 

[BvU]

My point is that the restoring torque ##D## could hide an ##r## dependency ... even if the spring (assuming it is a spring) itself doesn't change temperature (as it more or less explicitly says in the problem statement)
[edit] doesn't fly. from the pictures in the link it's obvious the spring is attached to thee hub of the balance wheel; surely the exercise composer doesn't mean to refer to that hub expanding as well ? or does he/she ?I also wonder about the 2 in front of the D: where does that come from ? (not that it matters for the outcome, both a ring and a disc have ##I\propto r^2##)

 

[kuruman]

My point is that the restoring torque D could hide an ##r## dependency ...

I don't think ##D## is a torque. It's most likely the torsional spring constant, as in the SHO equation for torsional motion $$I\frac{d^2\theta}{dt^2}=-D \theta.$$ For a ring the moment of inertia is ##I=mr^2## in which case the frequency is$$\omega =\sqrt{\frac{D}{I}}=\sqrt{\frac{D}{mr^2}}.$$
As for the origin of the factor of 2 in ##2D##, I have no clue unless the ring is treated as a solid disk for some reason.

 

[Dr.D]

I don't think DD is a torque.

This is why I've said previously, we are wasting our time with no diagram, no variable definitions, etc. The OP owes us at least that little bit, and any further discussion is simply nonsense since we don't really know what the problem is.

 

[haruspex]

I think it is more correct to use the 2D expansion equation. It's like the hole in a plate that expands upon heating. The circumference of the hole gets larger according to the 2D expression. The expansion of the area of the ring will be given by ##\Delta A = 2\alpha A_0 \Delta T##. This means that the new area is ##A=A_0+\Delta A = A_0(1+2\alpha \Delta T)##.

Why is the area interesting? The mass is constant.
The D in the relevant equation has suitable dimension for torque or torque per unit angle, and that is divided by what looks like the moment of inertia of a uniform disk (mr²/2). That makes sense for a spring balance escapement. Consequently the frequency should be inversely proportional to r.

 

[BvU]

I don't think ##D## is a torque. It's most likely the torsional spring constant

Of course, sorry for my sloppiness.
Usually the spring is attached to the hub of the balancing wheel. If the wheel expands with temperature, so does the hub, I suppose.

I'm very interested in what happens to that ##D## if the radius of attachment point increases. Would it remain the same ? I think it would not.

(admittedly far-fetched -- but the problem statement leaves plenty room for speculation sloppy exercise composer too ?)

 

[BvU]

That makes sense for a spring balance escapement

Don't see that from the pictures. @bolzano95 ?

Hard to increase the temp of one part in an escapement/balancing wheel combo without affecting the other ! @bolzano95 ?

 

[haruspex]

Don't see that from the pictures.

The only mismatch I see is that the balance wheel would not be a uniform disc, so the 2 in the given equation is wrong. But for the purposes of the question that is immaterial.