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A Sphere Oscillating in the Bottom of a Cylinder

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    "A small ball with radius r and uniform density rolls without slipping near the bottom of a fixed cylinder of radius R. What is the frequency of small oscillations, assuming r<<R?"

    [​IMG]

    2. Relevant equations

    [tex]
    F=ma
    [/tex]

    [tex]
    \tau=I\alpha \\
    [/tex]

    [tex]
    I=\frac{2}{5}mr^2 \\
    [/tex]

    [tex]
    a = \ddot{\theta} R= \alpha r
    [/tex]

    3. The attempt at a solution

    The solution I can get, I just have a few questions about the details.

    First, writing out the forces on the CM of the ball: the force of friction points in the same direction as the velocity vector, due to the non-slipping constraint. The magnitude of the ball's velocity is decreasing the greater the angle between it and the bottom of the cylinder, and thus the angular frequency of its rotation must also decrease. Thus the friction force is opposite the rotation.

    The other relevant force is obviously gravity, scaled by a sine of theta:

    [tex]
    F_{\hat{\theta}}= F_f - mg sin\theta = ma
    [/tex]

    And there is a also a torque equation. Relative to the CM of the ball, we have a torque exerted by friction. Thus:

    [tex]
    -r F_f = \frac{2}{5}mr^2 \alpha
    [/tex]

    Combining these two equations, and assuming that, for small theta, [itex]\theta\approx sin\theta[/itex], we get the differential equation:

    [tex]
    \ddot{\theta}+\frac{5g}{7R}\theta =0
    [/tex]

    [tex]
    \omega = \sqrt{\frac{5g}{7R}}
    [/tex]

    What I can't understand...

    1. The Work Done By Friction
    The bottom of the ball is instantaneously at rest with respect to the bottom of the cylinder. The force of static friction, then, exerts a force over zero distance, and thus does no work. However, it also is responsible for the change in angular frequency. Without the friction, the ball would not rotate.

    The rotational kinetic energy of the ball is changing due to the friction force, but the friction force does no work. I don't get this.

    2. Non-inertial Frame?
    The CM of the ball is accelerating, yet the torque is still calculated using it as an origin. How is this legal?

    If it matters, this is problem 8.13 in Morin's Introduction to Classical Mechanics.

    Thanks for your help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 18, 2009 #2
    If you remember back to centripetal motion, recall that the centripetal force does no work as well. Since the force did not move the sphere, it just merely changes the direction of the sphere. Therefore it didn't do any "real" work.

    I guess it really goes back to how work is define. f (dot) x.

    I have a feeling that I am not making a lot of sense


    Edit:
    oops didn't see part 2.

    Well, frame of reference are relative. You can take any frame of reference to calculate physics (if you can, some frame of reference are a pain in the latus rectum). The earth is in motion (with respect to the sun) everyday, but we calculate most of physics without considering that motion.
     
  4. Jan 18, 2009 #3
    Yes, but a centripetal force does not change the magnitude of the velocity, thus there is no change is the total kinetic energy of the object it pulls on.

    The friction is transferring some of the translational kinetic energy of the ball to rotational. I have a hunch that the total kinetic energy of the ball does not change, though, so the friction does no work.

    But then again, in the angular direction, the friction force is in the f=ma equation.

    And I see what you mean about the non-inertial frame, but it just seems fishy to me. The size of the system of something on Earth is negligible compared to the Earth-Sun system, so that can be ignored. But taking the ball in the cylinder as a system, it seems that the effect of the non-inertial frame might be more noticeable.
     
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