- #1
garygary
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When you have a spinning top undergoing forced precession (due to the torque created by its weight trying to tip it), you seem to get two precession rates. Why?
I get this result from using Euler's equations, specifically that the sum of moments tipping the top is (in the case of steady precession):
[tex]\sum[/tex] M0 = [tex]\psi[/tex]' sin([tex]\theta[/tex]) (I ([tex]\psi[/tex]' cos([tex]\theta[/tex]) + p) - I0 [tex]\psi[/tex]' cos([tex]\theta[/tex]))
Where:
[tex]\psi[/tex]' = angular velocity of precession
I = mass moment of inertia of top about spin axis
I0 = mass moment of inertia of top about transverse axis
p = angular velocity of top about its spin axis, its spin
[tex]\theta[/tex] = angle of precession
The only moment acting on the top is its weight, which can be written as:
mgl sin([tex]\theta[/tex])
Where:
m = mass of top
g = gravity
l = distance from tipping axis to centre of mass of the top
This gives me the quadratic equation:
0 = ([tex]\psi[/tex]')2 cos([tex]\theta[/tex]) (I - I0) + I p [tex]\psi[/tex]' - lmg
Which when solved for [tex]\psi[/tex]' gives two precession rates. Which one will the top actually precess at, and why? One precession rate is much smaller than the other. If instead of a top it is a cylindrical shape, with I0>I, both precession rates have the same sign. I'm having trouble working out exactly what this means in reality. So the top can precess stably at both of these precession rates? What makes it go at one rate rather than the other?
In most other examples I can find it is typically assumed that the precession rate is much less than the spin rate, so the squared term of the quadratic is ignored and you only get one answer. This answer is always closest to the larger precession rate that I find above.
This is my first post, so hopefully you can understand my technique and exactly what I'm asking, any help is greatly appreciated.
I get this result from using Euler's equations, specifically that the sum of moments tipping the top is (in the case of steady precession):
[tex]\sum[/tex] M0 = [tex]\psi[/tex]' sin([tex]\theta[/tex]) (I ([tex]\psi[/tex]' cos([tex]\theta[/tex]) + p) - I0 [tex]\psi[/tex]' cos([tex]\theta[/tex]))
Where:
[tex]\psi[/tex]' = angular velocity of precession
I = mass moment of inertia of top about spin axis
I0 = mass moment of inertia of top about transverse axis
p = angular velocity of top about its spin axis, its spin
[tex]\theta[/tex] = angle of precession
The only moment acting on the top is its weight, which can be written as:
mgl sin([tex]\theta[/tex])
Where:
m = mass of top
g = gravity
l = distance from tipping axis to centre of mass of the top
This gives me the quadratic equation:
0 = ([tex]\psi[/tex]')2 cos([tex]\theta[/tex]) (I - I0) + I p [tex]\psi[/tex]' - lmg
Which when solved for [tex]\psi[/tex]' gives two precession rates. Which one will the top actually precess at, and why? One precession rate is much smaller than the other. If instead of a top it is a cylindrical shape, with I0>I, both precession rates have the same sign. I'm having trouble working out exactly what this means in reality. So the top can precess stably at both of these precession rates? What makes it go at one rate rather than the other?
In most other examples I can find it is typically assumed that the precession rate is much less than the spin rate, so the squared term of the quadratic is ignored and you only get one answer. This answer is always closest to the larger precession rate that I find above.
This is my first post, so hopefully you can understand my technique and exactly what I'm asking, any help is greatly appreciated.