Angular speed ratio and gear teeth ratio

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Discussion Overview

The discussion revolves around the relationship between angular speed ratios and gear teeth ratios, specifically addressing the assumptions made about torque in gear systems and the application of Newton's third law in this context. Participants explore the derivation of the relationship between gear speeds and the implications of torque equality.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant attempts to derive the relationship n=N1/N2=omega2/omega1 but questions the assumption that T1=T2 is valid, suggesting it leads to confusion regarding Newton's third law.
  • Another participant clarifies that the energies of the gears are equal (E = T ⋅ ω), leading to the conclusion that Ta ⋅ ωa = Tb ⋅ ωb, and notes that Ta does not equal Tb due to the gear ratio.
  • A participant expresses an intuition that torque should be symmetrical like force, questioning why T1 does not equal T2 at the contact point of two gears.
  • Another participant responds by affirming that Newton's third law is valid for torque but highlights that comparing torques from different axes of rotation is inappropriate, suggesting that moment arms must be equal for a valid comparison.

Areas of Agreement / Disagreement

Participants show some agreement on the validity of Newton's third law for torque but disagree on the implications of torque equality in gear systems and the assumptions made regarding torque calculations.

Contextual Notes

Participants express uncertainty regarding the application of Newton's third law to torque and the conditions under which torque comparisons are valid. There are unresolved questions about the assumptions made in deriving the relationships between gear speeds and torques.

kidsasd987
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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.
 
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kidsasd987 said:
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.
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.
 
SteamKing said:
equal
SteamKing said:
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.
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:
kidsasd987 said:
Could you tell me why Newton's 3rd law is not valid for toruqe?
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.
 
jbriggs444 said:
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.

Thanks. I like your analogy of apples and oranges. I must go grab some.
 

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