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Rotation of a Spherical Top

  1. Dec 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid sphere of mass M and radius R rotates freely in space with an angular velocity ω about a fixed diameter. A particle of mass m, initially at one pole, moves with constant velocity v along a great circle of the sphere. Show that, when the particle has reached the other pole, the rotation of the sphere will have been retarded by an angle

    α=ωT(1-√[2M/(2M+5m)])

    2. Relevant equations


    3. The attempt at a solution
    Apologies for the ugly formatting, can someone please link me an article on how to type equations?
    For the problem, I can find the solution if I set the sphere's rotation about the z axis and I assume that angular momentum about the z axis is constant. My question lies with the assumption that angular momentum about the z axis is constant. Angular momentum is constant if there is no external torque. However, the particle's movement from the top pole to the bottom pole implies that there is a torque.

    What am I missing?
     
  2. jcsd
  3. Dec 13, 2014 #2

    Bystander

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    Angular momentum is the sum of what for this problem?
     
  4. Dec 13, 2014 #3
    Angular momentum of the sphere as it rotates about the z axis in the space frame and and a particle of mass m moves along the circumference of the sphere in the xz plane as this plane rotates in the body frame.

    ## L_z = I_z \omega_z ##
    ## L_{e2} = I_{e2} \omega_{e2} ##
     
    Last edited: Dec 13, 2014
  5. Dec 13, 2014 #4

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    Write the angular moments in terms of their masses and R.
     
  6. Dec 13, 2014 #5
    The moment of inertia about the e3 axis at t=0 is the moment of inertia of the sphere (since the particle lies on the e3 axis).

    ## I = \frac{2R^2M}{5} ##

    At time t the moment of inertia about the e3 axis is the moment of the sphere plus a contribution from the particle, a distance d from the z axis.

    ## I = \frac{2R^2M}{5} + d^2 ##
    It can be shown that the distance d is given by
    ## d = mR^2 sin^2 \theta ##
    ## \theta = \frac{vt}{R} ##

    Substituting into equation for I
    ## I = \frac{2R^2M}{5} + mR^2 sin^2 \theta ##


    The e3 axis should precess, should it not? I can't see how the e3 axis remains parallel to the z axis after t=0.
     
  7. Dec 13, 2014 #6

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    ... and θ (t)?
     
  8. Dec 13, 2014 #7
    By trigonometry,
    ## \theta (t) = \frac{vt}{R} ##

    I can solve the problem correctly if I assume that e3 does not precess, my question is why does this axis not precess while there is torque about e2 from the movement of the particle?
     
  9. Dec 13, 2014 #8

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    No torque. And, no, I'm not exactly happy with the appeal to the original problem statement myself.
     
  10. Dec 13, 2014 #9
    Constant linear velocity v, constant angular velocity
    ## \omega_{e2} = \dot{\theta}(t) ##
    ## \omega_{e2} = \frac{v}{R} ##

    In the body frame, the angular velocity introduces a centrifugal force which must be balanced by a centripetal force. The centrifugal force is given by
    ## F_{cf} = m ( \omega_{e2} \times r) \times \omega_{e2} ##
    ## F_{cf} = m R \omega_{e2}^2 \hat{r} ##

    The balancing centripetal force is in the negative ## \hat{r} ## direction
    ## F_{cp} = - m R \omega_{e2}^2 \hat{r} ##

    Taking torque to be
    ## \Gamma = r \times F ##
    ## \Gamma_{e2} = 0 ##
    since r and F both lie in the ## \hat{r} ## direction.

    Is this the reason torque is zero?
    The reason I thought torque was not zero is somewhere in my notes I have written
    ## \Gamma = r \times \omega ##
    Using this logic,
    ## \Gamma_{e2} = R \hat{r} \times \omega \hat{e_2} ##
    which is a nonzero quanitity.
     
  11. Dec 13, 2014 #10

    TSny

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    Hello, Ben.

    The problem states that "A solid sphere of mass M and radius R rotates freely in space with an angular velocity ω about a fixed diameter."

    It seems to me that this could be interpreted to mean that the sphere is mounted on a fixed axle like a globe but is otherwise free to rotate about this axle. If so, then there would be an external torque on the axle to keep it always aligned parallel to the z axis. As you noted, in this case you will get the answer stated in the problem.
     
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