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Angular speed ratio and gear teeth ratio

  1. Feb 22, 2016 #1
    http://www.tech.plym.ac.uk/sme/mech226/gearsys/gearaccel.htm

    Hello, I tried to derive n=N1/N2=omega2/omega1

    with using T1=T2 but it is a wrong assumption.
    I thought they should balance to satisfy newton's 3rd law, but it is not.

    Could you explain me why.
     
  2. jcsd
  3. Feb 22, 2016 #2

    SteamKing

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    In the diagrams, it's not the torques which are equal but the energy of each gear,

    E = T ⋅ ω,

    such that Ta ⋅ ωa = Tb ⋅ ωb

    Ta ≠ Tb because of the ratio between gears a and b.
     
  4. Feb 22, 2016 #3

    Hello, yes. It makes sense that work of each gear must be equal.

    But I thought T1=T2 because torque is somewhat symmetrical to force. It changes moment of inertia, and I thought at the contact point of two gears, T1 and T2 must be the same just because of intuition that force and torque (angular momentum and momentum) are symmetrical.

    If I push a box with a force F, by distance D.
    Ive done FD amount of work to the box and vice versa, the box has done -FD amount of work to me.
    Therefore, Torque*theata is symmetrical to Force*Distance.

    However, unlike Force, T1=/=T2
    Could you tell me why newton's 3rd law is not valid for toruqe?
     
    Last edited: Feb 22, 2016
  5. Feb 22, 2016 #4

    jbriggs444

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    Newton's third law is valid for torque. But when you compute torque T1 based on one axis of rotation and torque T2 based on a different axis of rotation, that's comparing apples and oranges. If you want the torques from two third-law partner forces to be equal, the moment arms should be equal as well.
     
  6. Feb 22, 2016 #5
    Thanks. I like your analogy of apples and oranges. I must go grab some.
     
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