# Clashing approximations for this precession problem

• B
• etotheipi
In summary, the conversation discusses a problem involving a spinning disk connected to a light axle and string, and the determination of its precession speed. Two different approaches are considered, and the discrepancy between the two is found to be due to a mistake in the calculation of angular momentum. The correct solution is obtained by considering the torque and applying the Newton equation.
etotheipi
I have a bit of a strange puzzle I can't work out. Let's say, we have a thin cylindrical disk of mass ##m## and radius ##r## connected on one side to a light axle of length ##d## through its centre. The axle is itself connected to a light string of length ##l##, the other end of which is connected to a fixed point. The string is taut and makes an angle of ##\theta## to the downward vertical, whilst the axle is horizontal and points in the radial direction, like this:

If the disk spins about its own central axis at ##\boldsymbol{\omega} = \omega_S \boldsymbol{e}_r##, and the configuration precesses about the dotted axis at ##\boldsymbol{\Omega} = \Omega_z \boldsymbol{e}_z##, then under the usual gyroscopic approximations ##\omega_S \gg \Omega_z## and ##d\Omega_z/dt = d\omega_S/dt = 0##, we want to determine ##\Omega_z##.

One way to solve it might be to take a coordinate system with origin at the hinge at the very top left hinge, and consider the system to be the string + rod + disk, in which case the angular momentum of the configuration is$$\mathbf{L} = \frac{1}{2}mr^2 \omega_S \boldsymbol{e}_r + [m(l\sin{\theta} + d)^2\ + \frac{1}{4}mr^2] \Omega_z \boldsymbol{e}_z \, \, \overset{\text{gyro approx}}{\implies} \, \, \boldsymbol{\tau} = (l\sin{\theta} + d)mg \boldsymbol{e}_{\theta} = \frac{1}{2}mr^2 \omega_s \Omega_z \boldsymbol{e}_{\theta}$$where the equality on the right equation holds under the approximation that ##d\Omega_z / dt = 0##. Another way to solve it, however, would be to take a coordinate system with origin at the centre of the disk (translating with the disk, but not rotating with the disk - i.e. as viewed by this coordinate system, the disk is slowly rotating about its own vertical axis), and consider the system to be the rod + disk. In this analysis, then, the string tension ##T## is an external force. In this coordinate system the angular momentum of the system is$$\boldsymbol{L} = \frac{1}{2}mr^2 \omega_S \boldsymbol{e}_r + \frac{1}{4} mr^2 \Omega_z \boldsymbol{e}_z \quad \overset{\text{gyro approx}}{\implies} \quad \boldsymbol{\tau} = dT\cos{\theta} \boldsymbol{e}_{\theta} = dmg \boldsymbol{e}_{\theta} = \frac{1}{2}mr^2 \omega_s \Omega_z \boldsymbol{e}_{\theta}$$where we used that ##F_z = T\cos{\theta} - mg = 0##. This second approach implies that the precession speed has no dependence on the string length nor the angle.

I presumed that, if I didn't mess up anywhere (I mean, it's pretty likely that I did! ), the discrepancy has to do with the assumptions made. Can anyone see? Thanks!

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etotheipi said:
...the discrepancy ...
What is the discrepancy exactly? Your final terms look the same.

A.T. said:
What is the discrepancy exactly? Your final terms look the same.

The first approach gives$$\Omega_z = \frac{2g(l\sin{\theta} + d)}{r^2 \omega_s}$$whilst the second approach gives$$\Omega_z = \frac{2gd}{r^2 \omega_s}$$Apparently, the second answer is the correct one! I wonder if, the gyroscopic approximation only holds if ##\theta= 0##...

It's not clear to me, what the precise setup is. Is the angle ##\theta## fixed or not? Is the horzontally drawn rod fixed or can it rotate around the point it is fixed at the string? In any case the Lagrangian should uniquely give an equation of motion without ambiguities.

etotheipi
Oh yeah, I forgot about this problem, I figured out the resolution last night, and it was a pretty silly mistake

The issue was that the angular momentum I wrote down in the first equation (about the top left hinge) was incorrect, since one of the terms in the angular momentum should have been$$[-l\cos{\theta} \boldsymbol{e}_z + (l\sin{\theta} + d) \boldsymbol{e}_r] \times \Omega_z (l\sin{\theta} + d) \boldsymbol{e}_{\theta}$$instead of$$[(l\sin{\theta} + d) \boldsymbol{e}_r] \times \Omega_z (l\sin{\theta} + d) \boldsymbol{e}_{\theta}$$
vanhees71 said:
Is the horzontally drawn rod fixed or can it rotate around the point it is fixed at the string?
The rod is rotating around with the whole configuration

I might write up the solution later for future reference, but I'm a bit embarrassed not spotting the mistake earlier!

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vanhees71
Okay, so here's the tea. First of all I'll note that it's a lot easier to solve it by taking an origin at the centre of mass, or solving the variational problem with the Lagrangian.

We also note that the problem specifies ##\theta## is small, so the validity of the small angle approximation holds here. But anyway, about the top left hinge,$$\mathbf{L} = m[-l\cos{\theta} \boldsymbol{e}_z + (l\sin{\theta} + d) \boldsymbol{e}_r] \times \Omega_z (l\sin{\theta} + d) \boldsymbol{e}_{\theta} + \frac{1}{2} mr^2 \omega_s^2 \boldsymbol{e}_r + \frac{1}{4} mr^2 \Omega \boldsymbol{e}_z$$On differentiation this will then give$$\frac{d\mathbf{L}}{dt} = m(\Omega^2 l^2 \sin{\theta} \cos{\theta} + \Omega^2 l d \cos{\theta} + \frac{1}{2}mr^2 \omega_s \Omega) \boldsymbol{e}_{\theta}$$From the Newton equation ##\mathbf{F} = m\ddot{\mathbf{x}}_{\mathcal{CM}}## applied to the centre of mass we obtain, if ##T## is the tension,$$T\sin{\theta} = m(l\sin{\theta} + d)\Omega^2, \quad T\cos{\theta} = mg$$or in other words$$\tan{\theta} = \frac{1}{g} \left[ l\sin{\theta} + d \right] \Omega^2 \approx \sin{\theta} \implies mlg\sin{\theta} \approx ml^2 \Omega^2 \sin{\theta} + mld\Omega^2$$Clearly if we take ##\cos{\theta} \approx 1## then this is just what's inside the first two terms of ##d\mathbf{L}/dt##, or in other words$$d\mathbf{L}/dt = mlg\sin{\theta} \boldsymbol{e}_{\theta} + \frac{1}{2} mr^2 \omega_s \Omega \boldsymbol{e}_{\theta}$$The torque is evidently$$\boldsymbol{\tau} = (l\sin{\theta} + d)mg \boldsymbol{e}_{\theta}$$and on equality we obtain$$mgd = \frac{1}{2} mr^2 \omega_s \Omega$$which is, exactly what we obtained with the other approach.

Moral of the story is don't drop terms!

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vanhees71

## 1. What is the precession problem?

The precession problem refers to the observed phenomenon where the axis of rotation of a spinning object, such as a top or a gyroscope, appears to wobble or change direction over time.

## 2. What are clashing approximations?

Clashing approximations refer to the use of two or more different mathematical models or approximations to solve a problem, which may lead to conflicting or inconsistent results.

## 3. How does clashing approximations affect the solution to the precession problem?

Clashing approximations can lead to conflicting results when trying to solve the precession problem, as different models may produce different predictions for the direction and rate of precession.

## 4. What are some possible solutions to the precession problem?

One possible solution is to use more accurate and precise mathematical models to describe the motion of the spinning object. Another solution is to take into account external factors, such as friction and air resistance, which may affect the precession.

## 5. How can the precession problem be applied in real-world situations?

The precession problem has practical applications in fields such as physics, engineering, and astronomy. For example, understanding the precession of the Earth's axis is important in predicting changes in the Earth's climate and in navigation systems. It is also relevant in designing stable and accurate spinning objects, such as satellites and gyroscopes.

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