Precession of a Gyroscope

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Homework Statement


A top spins at 33 rev/s about an axis that makes an angle of 26° with the vertical. The mass of the top is 0.47 kg, its rotational inertia about its central axis is 5.0 x 10-4 kg·m2, and its center of mass is 3.1 cm from the pivot point. The spin is clockwise from an overhead view. (a) What is the precession rate? (b) What is the direction of the precession as viewed from overhead?

Homework Equations


Ω=Mgr/Iω

The Attempt at a Solution


20150713_225445_zpsgqmqlm6s.jpg


It looks like most of the variables were given in the problem statement, but im not sure which im getting wrong. i tried g=mgcosθ because the angled axis but that didnt work. Im not sure about my r value also.
 

Answers and Replies

  • #2
haruspex
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Ω=Mgr/Iω
A quoted equation doesn't mean much without the accompanying definition of the variables.
You may notice that there's no mention of an angle to the vertical here, so the question is, how is r defined?
The more basic equation is ##\dot L = \tau##, the torque. Certainly ##\tau = mgh \sin(\theta)##, where h is the distance from pivot to mass centre, but what about ##\dot L##? This is not simply ##I \omega \Omega##. That would be the case were it precessing in a horizontal plane, but here it is not. What do you think the equation for ##\dot L## would be?
 
  • #3
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A quoted equation doesn't mean much without the accompanying definition of the variables.
Ω=Mgr/Iω
where; Ω=precession rate, M=mass, g=9.8m/s/s, r=distance from com to pivot point, I=rotational inertia, ω=spin rate of top.

You may notice that there's no mention of an angle to the vertical here, so the question is, how is r defined?
I see a mentioned angle to the vertical in the problem statement as 26 degrees from vertical. So then r=0.031sin26.
##\dot L## would be the rotational inertia equation for a disk =½MR2 where R is the radius of the disk and M is mass.
 
  • #4
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What do you think the equation for ##\dot L## would be?
##\dot L## should be =Iω, because its a rigid body rotating about a fixed axis.
edit; or would it be equal to the torque because that is the only external force acting on the system?
20150714_132038_zpspt0fmvry.jpg
 
  • #5
Nathanael
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I see a mentioned angle to the vertical in the problem statement as 26 degrees from vertical.
Haruspex meant that equation did not mention any angle (not the problem statement).

##\dot L## should be =Iω, because its a rigid body rotating about a fixed axis.
edit; or would it be equal to the torque because that is the only external force acting on the system?
L is Iω, ##\dot L## (with a dot on top) represents the time derivative of it. Yes it would be ##\tau = \dot L## but the question is what is ##\dot L##? (Re-read post#2)


Imagine the torque vector in 3D. It is ##\vec R ×\vec {mg}## so it sticks out of your page. It represents the change in the vector L. How does this cause ##\vec L## to change?
Is there any part of the vector L which the torque doesn't change?
 
  • #6
haruspex
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Ω=Mgr/Iω
where; Ω=precession rate, M=mass, g=9.8m/s/s, r=distance from com to pivot point, I=rotational inertia, ω=spin rate of top.
Ok, so why did you interpret it as horizontal distance?
 

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