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Rotational motion: Expression for Angular Velocity

  1. Nov 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A solid flywheel of radius R and mass M is mounted on a light shaft of radius r so that the axis of rotation is horizontal. A light, inelastic rope is wound around the drive shaft and is connected, via a light, frictionless pulley to a mass m, which is suspended a height h above the floor. You can assume the moment of inertia of the fly wheel is I=1/2MR^2.
    (a) If the mass is released from rest and allowed to fall to the floor, find an expression for the final angular velocity of the flywheel.

    2. Relevant equations

    So for mass m, I have T[itex]_{1}[/itex]-mg=-ma.
    Now I'm not sure if I should assume the mass of the pulley is negligible (the question indicates
    light, so I'm not sure).
    But assuming its massless, I have the torque τ=-Iσ=-T[itex]_{1}[/itex]r, where σ is the angular acceleration, and clockwise is negative.
    Then, since the tension in the rope isn't changing, we have constant σ. So, σ=ω/t. Then since I=1/2MR^2, we have ω = 2mr(g-a)t/MR[itex]^{2}[/itex]
    This seems way to cumbersome, any ideas on what I did wrong?
     
    Last edited: Nov 23, 2013
  2. jcsd
  3. Nov 23, 2013 #2

    haruspex

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    That won't do as an answer because you have t in there, which is unknown. You should instead have an h in the answer.
    Work will be conserved, no?
     
  4. Nov 23, 2013 #3
    Okay but the question says the rope is inelastic so energy wouldn't be conserved.
    But taking what you said i have ω = √(2τθ/I) = √(2σθ), using work done by torque and the conservation of rotational kinetic energy. Although this is probably not what you had in mind?
     
  5. Nov 23, 2013 #4

    Doc Al

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    That just means the rope doesn't stretch, not that energy isn't conserved. (Don't confuse with inelastic collision.)
     
  6. Nov 23, 2013 #5

    haruspex

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    That's only a problem when an inelastic string goes from slack to taut suddenly. If it's taut throughout, there's no 'collision'.
     
  7. Nov 23, 2013 #6
    Okay so I employed the defn of W for constant linear force (T-mg) over distance h.
    That is W=ΔK + ΔU, and then used v=Rω, and I=1/2mv[itex]^{2}[/itex] and the fact that T=Iσ/r, from the defn of torque to obtain ω=√(σh/r). Did I do this correctly?
     
  8. Nov 24, 2013 #7

    haruspex

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    You mean 1/2 mR2, right?
    I think there's a factor 2 missing in there. But anyway, you need to get ω as a function of M, m, R h and g. No references to T, t v, a or σ remaining. Using energy, you should not even need to consider a, T or σ.
    What is the KE as the mass hits the ground? What is the relationship between v and ω?
     
  9. Nov 24, 2013 #8
    okay so I understand if we consider only m, then mgh=1/2mR[itex]^{2}[/itex]ω[itex]^{2}[/itex]
    And I believe the flywheel has final rotational kinetic energy given by K = 1/2Iω[itex]^{2}[/itex], but I'm unsure how the energy of the flywheel relates to the energy of the mass hanging.
    If I don't consider the rotational motion of the flywheel and just focus on the mass hanging, than I obtain ω=√(2hg/R[itex]^{2}[/itex])
     
  10. Nov 24, 2013 #9

    haruspex

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    The total KE+PE does not change. What is it at the start, and what is it at the end?
     
  11. Nov 24, 2013 #10
    Then, you mean to say that the potential energy of the mass hanging gets converted into kinetic energy of rotation of the fly wheel and kinetic of mass before it hits the ground? (I assuming the fly wheel has no initial energy)
     
  12. Nov 24, 2013 #11

    haruspex

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    Exactly.
     
  13. Nov 24, 2013 #12
    Okay so just to clarify the intuition for why this is true is it that the mass hanging does work on the
    flywheel via the force of tension which causes the torque in the flywheel.
     
  14. Nov 24, 2013 #13

    haruspex

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    Yes.
     
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