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Questions on Rotational Kinematics/Dynamics

  1. Jul 6, 2011 #1
    Some background information: I'm doing some reading up for PhO (so it's beyond what I'm supposed to learn), and so I'll post all my questions here (regarding both concepts and actual practice questions). Sorry if you feel there is a lack of effort on my part, but sometimes I'm really lost and don't know where to start.

    Questions:

    1. http://imgur.com/LEmOT

    Since it's a single length of rope, shouldn't tension be constant throughout the rope? That's my only question (at the moment, that is :D).
     
  2. jcsd
  3. Jul 6, 2011 #2

    Doc Al

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    No. That would be true if the rope were massless (which it presumably is) and the pulleys were massless (which they are not) and frictionless.
     
  4. Jul 6, 2011 #3
    Hi, then can you explain this: T2 * r - T1 * r = 0.5 * m * r^2 * (a/r)

    I'm guessing that it means clockwise moment - anticlockwise moment = something?

    But I'm not too sure.
     
  5. Jul 6, 2011 #4

    Doc Al

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    It's applying Newton's 2nd law for rotation: The net torque = I*alpha.
     
  6. Jul 7, 2011 #5
    Alright thanks a lot.

    Back to the question on the pulley. The pulley with a mass if fixed, so the mass shouldn't affect the tension? The movable pulley is massless.
     
  7. Jul 7, 2011 #6

    Doc Al

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    It has mass and thus rotational inertia, so the tension in the string must be different on each side.
    For that pulley the tension is the same on either side.
     
  8. Jul 7, 2011 #7
    Okay I get it, thanks! :D

    Next question (2): A father of mass [itex]m_f[/itex] and his daughter of mass [itex]m_d[/itex] sit on opposite ends of a seesaw at equal distances from the pivot at the centre. The seesaw is modeled as a rigid rod of mass [itex]M[/itex] and length [itex]d[/itex] and is pivoted without friction. At a given moment, the system rotates in a vertical plane with an angular velocity [itex]\omega[/itex]. Obtain an expression for

    a) the magnitude of the system's angular momentum

    According to the answer given,

    Modelling the father and daughter as particles, the moment of inertia of the system is
    [itex]I = \frac{1}{12}Md^2 + \frac{1}{4}m_fd^2 + \frac{1}{4}m_dd^2 = \frac{1}{12}d^2(M + 3m_f + 3m_d)[/itex]

    I get that a rod has a moment of inertia of (Md^2)/12, but how is the M.I. of the father and daughter derived? Manual derivation using calculus?
     
  9. Jul 9, 2011 #8
    Never mind I got it :D

    For anybody who needs the answer, it's because [itex]I = r^2 \times dm = (\frac{d}{2})^2 \times m_{f} = \frac{1}{4}m_fd^2[/itex],
     
  10. Jul 9, 2011 #9

    Doc Al

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    Oops... I didn't even see your second question. Sorry about that!
     
  11. Jul 9, 2011 #10
    It's okay, I feel great being able to solve a question completely out of my syllabus! :D

    Okay, but now I have another question. :X

    A uniform solid disk of radius R and mass M is set into rotation with an angular speed [itex]\omega_i[/itex] about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and immediately released as shown.

    What is the angular speed of the disk once pure rolling takes place?

    My solution:

    [itex]K_{rotational initial} = K_{rotational final} + K_{linear final}[/itex]
    [itex](\frac{1}{2})(\frac{1}{2}MR^2)(\omega_i)^2 = (\frac{1}{2})(\frac{1}{2}MR^2)(\omega_f)^2 + (\frac{1}{2})(M)(R\omega_f)^2[/itex]
    [itex]\frac{1}{2}\omega_i^2 = \frac{1}{2}\omega_f^2 + \omega_f^2[/itex]
    [itex]\omega_f = \sqrt{\frac{1}{3}\omega_i^2}[/itex]

    But the answer is [itex]\frac{1}{3}\omega_i[/itex]. I suspect a careless mistake somewhere, but I've been through my workings and I found no mistakes.
     
  12. Jul 9, 2011 #11

    Doc Al

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    Do not assume that kinetic energy is conserved. There is friction involved.
     
  13. Jul 9, 2011 #12
    But there is no mention of frictional forces in the question, so how do I go about compensating for energy losses due to friction?

    Can you give me a clue? :D
     
  14. Jul 9, 2011 #13

    Doc Al

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    There must be friction, otherwise it would just sit there rotating. Just assume that the coefficient of friction is μ and see what happens.
     
  15. Jul 9, 2011 #14
    So,

    energy lost to friction = [itex]\mu Mg \times \theta[/itex]?

    Since W = F x s?

    Sorry I'm wild guessing, I have no idea how to apply friction to rotational dynamics.
     
  16. Jul 9, 2011 #15

    Doc Al

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    Sure you do. It's a force, like any other. Apply Newton's 2nd law for translation and for rotation.

    Hint: The rotational speed is decreasing while the translational speed is increasing. Find the point where the condition for rolling without slipping is met.
     
  17. Jul 9, 2011 #16
    Okay... so...

    [itex]\tau = I\alpha[/itex]
    [itex]\mu MgR = \frac{1}{2}MR^2\alpha[/itex]
    [itex]\mu g = \frac{1}{2}R\alpha[/itex]
    [itex]R\alpha = 2\mu g[/itex]

    [itex]F = Ma[/itex]
    [itex]F = MR\alpha[/itex]
    [itex]F = M2\mu g[/itex]

    But since F is provided for by the friction,

    [itex]\mu Mg = M2\mu g[/itex]
    [itex]1 = 2[/itex]

    HUH?

    Condition for pure rolling is when [itex]\omega = v[/itex]. I could probably sub [itex]\alpha = \frac{d\omega}{dt}[/itex] and [itex]a = \frac{dv}{dt}[/itex], but why did I get 1 = 2? Where did I go wrong? :/
     
  18. Jul 9, 2011 #17

    Doc Al

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    Now write ω as a function of time.

    Then write v as a function of time.

    Solve for when the condition for rolling without slipping is met. (You can't assume it's met for all times; that was your error.)
     
  19. Jul 9, 2011 #18
    So my formula [itex]\tau = \mu MgR[/itex] is correct?

    [itex]\omega_f = \omega_i + \alpha t[/itex], [itex]v = at[/itex]

    When the disk is rolling without sleeping, i.e. when [itex]\omega_f = v[/itex],

    [itex]\omega_i + \alpha t = at[/itex]
    [itex]\omega_i + \frac{2\mu g}{R}t = \frac{\mu Mg}{M}t[/itex]
    [itex]\omega_i = \frac{\mu Mg}{M}t - \frac{2\mu g}{R}t = \mu gt (1 - \frac{2}{R})[/itex]

    So from here, do I find an expression for t, then apply to [itex]\omega_f = \omega_i + \alpha t[/itex]?
     
  20. Jul 9, 2011 #19

    Doc Al

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    Yes.

    Yes, but don't forget that alpha is negative, since the rotational speed is decreasing.

    Almost: [itex]R\omega_f = v[/itex]
     
  21. Jul 9, 2011 #20
    Hardly a eureka moment, but still :D

    Thanks a lot!
     
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