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Gyroscope Problem

  1. Nov 9, 2007 #1
    1. The problem statement, all variables and given/known data
    The rotor (flywheel) of a toy gyroscope has mass 0.130 kg. Its moment of inertia about its axis is 1.40 * 10^-4 kg·m^2. The mass of the frame is 0.0650 kg. The gyroscope is supported on a single pivot (Diagram: http://www.webassign.net/yf10/10-43.gif) with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.50 s.

    (a) Find the upward force exerted by the pivot.
    (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min.

    I know that I=(1/2)*M*R^2 so I found the radius of the rotor to be .0464 meters. I don't know where to go from there. Can anyone please help me with this?
     
  2. jcsd
  3. Nov 9, 2007 #2

    D H

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    What level of physics do you know? In particular, do you know what I mean by "Lagrangian"?
     
  4. Nov 9, 2007 #3
    "Lagrangian" no I don't know. I'm taking AP Physics C in high school right now. If you don't know what it is, it's basic physics with a bit of calculus.
     
  5. Nov 9, 2007 #4

    D H

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    Ok then.

    Since the center of mass of the gyroscope is not centered over the support, gravity will exert a torque on the gyroscope about the support. Can you determine what this torque is? What is the relevant equation relating a force and a torque?

    Next set of questions relate to angular momentum, if you know what that is. Can determine the angular momentum of the gyroscope?
     
  6. Nov 9, 2007 #5
    Well I know that torque = (lever arm) * (force)
    So I set up my equation (for part a) and have torque = .04 * (.065+.13) * 9.8
    Once I found my torque, though, I'm not sure how that helps me find the upward force exerted by the pivot.
     
  7. Nov 10, 2007 #6

    D H

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    Sorry, I thought that part was obvious. Gravity is pulling the gyroscope downwards and the pivot is pushing the top upwards. Since the gyroscope isn't moving at all vertically, it isn't accelerating vertically. What does this tell you about the upward force?

    The torque will tell you about the precession. Getting back to the torque and angular momentum, (1) what is the angular momentum, and (2) what is the relevant equation relating angular momentum and torque?
     
  8. Nov 10, 2007 #7
    I have torque=I*alpha and L=I*omega
    Therefore I can say that torque/alpha = L/omega
    Solving for omega, i get omega=(L*alpha)/torque
    and since L=r*p and p=m*v, I get L=r*m*v
    Also, I know that torque=r*F
    Plugging the L and torque into the omega equation, I get omega = r^2*m*v*a/(r*F) which simplifies into omega=r*m*v*a/F

    Now I'm confused as to which rotation this equation gets its variables from. The force I'm guessing is the upward force that the pivot exerts on the frame (I correctly found it to be 1.91 N). Omega is the angular velocity of the rotor around the frame. The radius is the given radius of orbit of the center of mass around the pivot. And i have no clue which velocity or acceleration to use. This whole paragraph is the part I am lost at.
     
  9. Nov 10, 2007 #8

    D H

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    I was hoping you might give me the rotational equivalent of Newton's second law,

    [tex]\vec \tau = \frac{d\vec L} {dt}[/tex]

    Notice that the torque is at right angles to the angular momentum vector. When the time derivative of some vector is normal to the vector, the vector's length doesn't change but its direction does. It rotates or precesses. The precession is related to the torque by

    [tex]\vec \tau = \vec \omega_p\times \vec L[/tex]

    You computed the torque, and you already know the precession rate. Since the vectors are at nice ninety degree angles, you can divide the torque by the precession rate to determine the angular momentum. From that point you should be able to compute the angular velocity.

    If you continue in physics you will cover this subject at least two more times. When you learn about vector operations you will find out how to deal with cases where the gyroscope is at some angle other than horizontal. When you learn about the Lagrangian or Hamiltonian, you will learn even more.
     
  10. Nov 11, 2007 #9
    Thanks a lot for the help, I really appreciate it. I correctly got (angular velocity of rotor around its axis) = (gravitational force of entire system)*(radius around pivot)/((given moment of interia)*(angular velocity around pivot)). My final answer turned out to be 2074.55 rev/min.
    Thank you again for all the help.
     
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