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Disk with catch

  1. Mar 22, 2015 #1
    1. The problem statement, all variables and given/known data

    A disk rotates with constant angular velocity ##ω##, as shown (attached image). Two masses, ##m_A## and ##m_B##, slide without friction in a groove passing through the center of the disk. They are connected by a light string of length ##l##, and are initially held in position by a catch, with mass ##m_A## at distance ##r_A## from the center. Neglect gravity. At ##t = 0## the catch is removed and the masses are free to slide.

    Find ##\ddot{r}_A## immediately after the catch is removed, in terms of ##m_A##, ##m_B##, ##l##, ##r_A##, and ##ω##.

    2. Relevant equations

    $$\sum_{}^{} \vec{F}_r = m(\ddot{r} - r \dot{\theta}^2) \hat{r}$$

    3. The attempt at a solution


    The particles are constrained to move according to ##r_A + r_B = l##. Differentiating twice with respect to time, we get ##\ddot{r}_A = -\ddot{r}_B##. Each particle experiences a force of magnitude ##T## where ##T## is the tension in the string. Writing down the equations of motion:
    $$-T = -m_A(\ddot{r}_A - r_A \omega^2)$$
    $$-T = -m_B(\ddot{r}_B - r_B \omega^2)$$
    Eliminating ##T## and substituting ##r_B = l - r_A## and ##\ddot{r}_B = -\ddot{r}_A## we get:
    $$\ddot{r}_A = \omega^2 \frac{m_A r_A + m_B r_A - m_B l}{m_A + m_B}$$
    Is this answer correct?
    Also, before ##t = 0##, were ##m_A## and ##m_B## acted upon by an additional force ##F## such that ##|F - T| = mr\omega^2##? Or was the string fixed?
    It seems to me that my physical intuition was wrong (again). I don't understand why the ##\ddot{r}## terms appear in the equations of motion. Why can't the masses remain at a fixed distance from the circle?
    My last question is really about the polar coordinate system and the constraint I used above, namely ##l = r_A + r_B##.
    If the disk was big enough, and the string was entirely on one side of the circle (the origin no longer lies on the string), then we would be forced to interpret the radial coordinate of the particle closer to the pole as negative, otherwise the sum of the radial coordinates would not equal the length of the string, right?
    Sometimes, I find the non-uniqueness of polar coordinates really frustrating
     

    Attached Files:

  2. jcsd
  3. Mar 22, 2015 #2

    mfb

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    Looks correct.

    We don't know how the system was hold in place before t=0, I don't see how this would matter.

    They can, if their forces are in equilibrium?

    That, or change the equation for l (and for the force balances) to keep two positive radial distances.
     
  4. Mar 24, 2015 #3
    Why do the particles move around in a spiral, rather than a circle?
     
  5. Mar 24, 2015 #4

    haruspex

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    Assuming they move in a circle, you can consider the forces on one and deduce the tension in the string. Likewise the other. If the two tensions are not the same then your assumption must be false.
     
  6. Mar 25, 2015 #5
    The two tensions are the same, since the string is massless. What I'm asking is: why are there ##\ddot{r}## terms in the equations of motion in the first place?
     
  7. Mar 25, 2015 #6

    haruspex

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    You seem to be saying you want an explanation for the ##\ddot r## term in ##|\Sigma F_r| = m(\ddot r - r {\dot \theta}^2)##. Is that right? You should be able to find a derivation online.
     
  8. Mar 26, 2015 #7
    I'm quite familiar with the derivation of acceleration in polar coordinates.
    I'm just wondering why it appears in some cases, but not others.
     
  9. Mar 26, 2015 #8

    haruspex

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    Well, it will be zero for circular motion of course. Can you list some other case?
     
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