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Rotational Kinematics of a turntable

  1. Dec 10, 2013 #1
    Good afternoon! I've been mulling over this question for a bit and I can't seem to understand what it is asking. This is a question for an introductory calculus-based physics university course.

    1. The Problem:

    A uniform disk, such as a record turntable, turns 8.0 rev/s around a frictionless spindle. A non-rotating rod of the same mass as the disk is dropped onto the freely spinning disk so that both turn around the spindle. Determine the angular velocity of the combination in rev/s.

    2. Equations used:

    I interpreted this as a conservation of angular momentum problem where the radius remains constant:

    m r^2 ω = m(disc and rod) r ^2 ω(final)

    3. The solution:

    Since the radius remains constant and the mass doubles, both the mass and radius^2 can be removed from both sides leaving:

    ω(initial) = 2ω(final)

    and since the initial angular velocity was 16π Rad/s the final angular velocity would be 8π Rad/s.

    Am I in the ballpark here assuming that this question is concerning the conservation of angular momentum? I don't see any other way to incorporate mass other than using Newton's laws, but I'm not sure on that.
     
    Last edited: Dec 10, 2013
  2. jcsd
  3. Dec 10, 2013 #2
    Is the radius of the rod and the disc necessarily equal?
     
  4. Dec 10, 2013 #3
    The problem doesn't state it unfortunately.
     
  5. Dec 10, 2013 #4
    The first thing that comes to my mind is to try solving it with energy, since the moments of inertia of a rod and a disc are different. Have you covered rotational kinetic energy yet?
     
  6. Dec 10, 2013 #5

    Mentz114

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    Gold Member

    Conservation of angular momentum gives
    ##m_dr_d^2\omega_0= \omega_1(m_dr_d^2+m_r r_r^2)## so ## \frac{\omega_0}{\omega_1}=\frac{I_d+I_r}{I_d}##. The subscripts are 'r' for the rod and 'd' for the disc and ##I## is a moment of inertia. I'm assuming the rod and the disc have ##I=mr^2/2##.
     
  7. Dec 10, 2013 #6
    We have, but I'm not sure how to imply it in this case without any information concerning the rod other than that it has the same mass as the disc and it is now a part of the system.
     
  8. Dec 10, 2013 #7
    This is a very good analysis of it, and this is what I would break it down as. I simply solved it for final angular velocity or "omega 1"
     
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